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QXs3r63kN8yLmLW88R12Y | maths | height-and-distance | height-and-distance | A tower T<sub>1</sub> of height 60 m is located exactly opposite to a tower T<sub>2</sub> of height 80 m on a straight road. Fromthe top of T<sub>1</sub>, if the angle of depression of the foot of T<sub>2</sub> is twice the angle of elevation of the top of T<sub>2</sub>, then the width (in m) of the road between the feetof the towers T<sub>1</sub> and T<sub>2</sub> is : | [{"identifier": "A", "content": "$$10\\sqrt 2 $$ "}, {"identifier": "B", "content": "$$10\\sqrt 3 $$"}, {"identifier": "C", "content": "$$20\\sqrt 3 $$"}, {"identifier": "D", "content": "$$20\\sqrt 2 $$"}] | ["C"] | null | Let the distance between T<sub>1</sub> and T<sub>2</sub> be x
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264796/exam_images/a99rugdbqpo21gvuvpmm.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Evening Slot Mathematics - Height and Distance Question 32 English Explanation">
<br><br>From the figure
<br><br>EA = 60 m (T<sub>1</sub>) and DB = 80 m (T<sub>2</sub>)
<br><br>$$\angle DEC = \theta $$ and $$\angle BEC = 2\theta $$
<br><br>Now in $$\angle DEC$$,
<br><br>$$\tan \theta = {{DC} \over {AB}} = {{20} \over x}$$
<br><br>and in $$\Delta BEC$$,
<br><br>$$\tan 2\theta = {{BC} \over {CE}} = {{60} \over x}$$
<br><br>We know that
<br><br>$$\tan 2\theta = {{2\tan \theta } \over {1 - {{\left( {\tan \theta } \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ $${{60} \over x} = {{2\left( {{{20} \over x}} \right)} \over {1 - {{\left( {{{20} \over x}} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ $${x^2} = 1200$$ $$ \Rightarrow $$ $$x = 20\sqrt 3 $$ | mcq | jee-main-2018-online-15th-april-evening-slot |
szA1nx6Gyh7SRMBqUuAbX | maths | height-and-distance | height-and-distance | A man on the top of a vertical tower observes a car moving at a uniform speed towards the tower on a horizontal road. If it takes 18 min. for the angle of depression of the car to change from 30<sup>o</sup> to 45<sup>o</sup> ; then after this, the time taken (in min.) by the car to reach the foot of the tower, is : | [{"identifier": "A", "content": "$$9\\left( {1 + \\sqrt 3 } \\right)$$ "}, {"identifier": "B", "content": "$$18\\left( {1 + \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$18\\left( {\\sqrt 3 - 1} \\right)$$"}, {"identifier": "D", "content": "$${9 \\over 2}\\left( {\\sqrt 3 - 1} \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267455/exam_images/gwuen82slkerzet1ztko.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 16th April Morning Slot Mathematics - Height and Distance Question 31 English Explanation">
<br><br>Assume height of tower = AB = h
<br><br>From $$\Delta $$ABD,
<br><br>tan45<sup>o</sup> = $${{AB} \over {AD}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{AB} \over {AD}} = 1$$ [ as tan45<sup>o</sup> = 1]
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ AB = AD
<br><br>$$\therefore\,\,\,$$ AD = h
<br><br>From $$\Delta $$BAC,
<br><br>tan30<sup>o</sup> = $${{AB} \over {AC}}$$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ $${h \over {AC}} = {1 \over {\sqrt 3 }}$$
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ AC = h $$\sqrt 3 $$
<br><br>As, AC = AD + DC
<br><br>$$ \Rightarrow $$$$\,\,\,$$ DC = AC $$-$$ AD = $$\sqrt 3 h$$ $$-$$ h
<br><br>Given that, time taken to reach from point C to D = 18 min.
<br><br>$$\therefore\,\,\,$$ Car speed = $${{distance} \over {time}}$$ = $${{CD} \over {18}}$$ = $${{(\sqrt 3 - 1)h} \over {18}}$$
<br><br>$$\therefore\,\,\,$$ Time taken to move from D to A
<br><br>= $${{Distan ce\,\,of\,\,DA} \over {speed}}$$
<br><br>= $${h \over {{{\left( {\sqrt 3 - 1} \right)h} \over {18}}}}$$
<br><br>= $${{18} \over {\left( {\sqrt 3 - 1} \right)}}$$
<br><br>= $${{18\left( {\sqrt 3 + 1} \right)} \over {3 - 1}}$$
<br><br>= 9 $$\left( {\sqrt 3 + 1} \right)$$ min.
| mcq | jee-main-2018-online-16th-april-morning-slot |
tkQWnPZ6nK5w9g4gfgFcZ | maths | height-and-distance | height-and-distance | Consider a triangular plot ABC with sides AB = 7m, BC = 5m and CA = 6m. A vertical lamp-post at the mid point D of AC subtends an angle 30<sup>o</sup> at B. The height (in m) of the lamp-post is - | [{"identifier": "A", "content": "$$2\\sqrt {21} $$"}, {"identifier": "B", "content": "$${3 \\over 2}\\sqrt {21} $$"}, {"identifier": "C", "content": "$$7\\sqrt {3} $$"}, {"identifier": "D", "content": "$${2 \\over 3}\\sqrt {21} $$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267190/exam_images/txlbrnxh7uu1cyfihe9v.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Height and Distance Question 29 English Explanation">
<br>BD = hcot30<sup>o</sup> = h$$\sqrt 3 $$
<br><br>So, 7<sup>2</sup> + 5<sup>2</sup> = 2(h$$\sqrt 3 $$)<sup>2</sup> + 3<sup>2</sup>)
<br><br>$$ \Rightarrow $$ 37 = 3h<sup>2</sup> + 9
<br><br>$$ \Rightarrow $$ 3h<sup>2</sup> = 28
<br><br>$$ \Rightarrow $$ h = $$\sqrt {{{28} \over 3}} = {2 \over 3}\sqrt {21} $$ | mcq | jee-main-2019-online-10th-january-morning-slot |
FrbvgcqsNWYOAzdb0T3rsa0w2w9jxb41ua0 | maths | height-and-distance | height-and-distance | The angle of elevation of the top of a vertical tower standing on a horizontal plane is observed to be 45<sup>o</sup> from
a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the
top of the tower from B be 30<sup>o</sup>, then the distance (in m) of the foot of the tower from the point A is : | [{"identifier": "A", "content": "$$15\\left( {1 + \\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$15\\left( {3 - \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$15\\left( {3 + \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$15\\left( {5 - \\sqrt 3 } \\right)$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267678/exam_images/ssjeqbfwbhvdrbkypihv.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Evening Slot Mathematics - Height and Distance Question 24 English Explanation">
$${x \over d} = \tan {30^o} \Rightarrow {x \over d} = {1 \over {\sqrt 3 }} \Rightarrow d = \sqrt 3 x$$
<br><br>
and $${{x + 30} \over d} = \tan {45^o}$$ $$ \Rightarrow $$ d = x + 30<br><br>
$$d = {d \over {\sqrt 3 }} + 30 \Rightarrow \left( {1 - {1 \over {\sqrt 3 }}} \right)d = 30$$<br><br>
$$ \Rightarrow d = {{30\sqrt 3 } \over {\sqrt 3 - 1}} \Rightarrow d = {{30\sqrt 3 (\sqrt 3 + 1)} \over 2} = 15\sqrt 3 (\sqrt 3 + 1)$$<br><br>
$$ \Rightarrow $$ d = 15 ( 3 + $$\sqrt 3 $$) | mcq | jee-main-2019-online-12th-april-evening-slot |
TRgZU92si7ThZAHGJW3rsa0w2w9jwy1ds9o | maths | height-and-distance | height-and-distance | ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If
the angles of elevation of the top of the tower at A and B are cot<sup>–1</sup>
(3$$\sqrt 2 $$ ) and cosec<sup>–1</sup>
(2$$\sqrt 2 $$ ) respectively,
then the height of the tower (in metres) is : | [{"identifier": "A", "content": "$${{100} \\over {3\\sqrt 3 }}$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "10$$\\sqrt 5 $$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267486/exam_images/ucqctvtzjka0ninslfkd.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265297/exam_images/mga2pd8jfduhngdumk72.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264081/exam_images/seb5qh2vead30zbnfys2.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Height and Distance Question 25 English Explanation"></picture>
$$\Delta $$APM<br><br>
$${h \over {AM}} = {1 \over {3\sqrt 2 }}$$<br><br>
$$\Delta $$BPM<br><br>
$${h \over {BM}} = {1 \over {\sqrt 7 }}$$<br><br>
$$\Delta $$ABM<br><br>
AM<sup>2</sup> + MB<sup>2</sup> = (100)<sup>2</sup><br><br>'
$$ \Rightarrow $$ 18h<sup>2</sup> + 7h<sup>2</sup> = 100 × 100<br><br>
$$ \Rightarrow $$ h<sup>2</sup> = 4 × 100<br><br>
$$ \Rightarrow $$ h = 20 | mcq | jee-main-2019-online-10th-april-morning-slot |
pLTsdMGhjup9oQjoBN18hoxe66ijvwuynvi | maths | height-and-distance | height-and-distance | Two poles standing on a horizontal ground are of
heights 5m and 10 m respectively. The line joining
their tops makes an angle of 15º with ground. Then
the distance (in m) between the poles, is :- | [{"identifier": "A", "content": "$$5\\left( {2 + \\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$${5 \\over 2}\\left( {2 + \\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$10\\left( {\\sqrt3 - 1 } \\right)$$"}, {"identifier": "D", "content": "$$5\\left( {\\sqrt3 + 1 } \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264351/exam_images/b08oyn8k1lljfispvted.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Height and Distance Question 26 English Explanation">
<br>
$$\tan {15^o} = {5 \over d} \Rightarrow d = {5 \over {\tan {{15}^o}}}$$<br><br>
$$ \Rightarrow {{5\left( {\sqrt 3 + 1} \right)} \over {\sqrt 3 - 1}} \Rightarrow {{5\left( {4 + 2\sqrt 3 } \right)} \over 2}$$<br><br>
$$ = 5\left( {2 + \sqrt 3 } \right)$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
fNVBNnJtWYO5d0c1AjsBn | maths | height-and-distance | height-and-distance | Two vertical poles of heights, 20 m and 80 m stand
a part on a horizontal plane. The height (in meters)
of the point of intersection of the lines joining the
top of each pole to the foot of the other, from this
horizontal plane is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "18"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264407/exam_images/wgqikoy7kceictq8u4qg.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266318/exam_images/ksjej4im98rcph4nc2mb.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Height and Distance Question 27 English Explanation"></picture>
<br><br>From triangle BCD,
<br><br>tan $$\alpha $$ = $${{80} \over x}$$
<br><br>and from triangle BFE,
<br><br>tan $$\alpha $$ = $${{h} \over y}$$
<br><br>$$ \therefore $$ $${{80} \over x}$$ = $${{h} \over y}$$
<br><br>$$ \Rightarrow $$ $$y = {{hx} \over {80}}$$ ........ (1)
<br><br>From triangle ABC,
<br><br>tan $$\beta $$ = $${{20} \over x}$$
<br><br>and from triangle EFC,
<br><br>tan $$\beta $$ = $${{h} \over {x-y}}$$
<br><br>$$ \therefore $$ $${{20} \over x}$$ = $${{h} \over {x-y}}$$
<br><br>$$ \Rightarrow $$ $$x-y = {{hx} \over {20}}$$ ........ (2)
<br><br>By adding equation (1) and (2) we get,
<br><br>x = $${{hx} \over {80}}$$ + $${{hx} \over {20}}$$
<br><br>$$ \Rightarrow $$ 1 = $${{h} \over {80}}$$ + $${{h} \over {20}}$$
<br><br>$$ \Rightarrow $$ h = 16 m
<br><br>$$ \therefore $$ Height of intersection point is 16 m | mcq | jee-main-2019-online-8th-april-evening-slot |
THaZ2h7gD42exiDAfzKuX | maths | height-and-distance | height-and-distance | If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30<sup>o</sup> and the angle of depression of reflection of the cloud in the lake from P be 60<sup>o</sup>, then the height of the cloud (in meters) from
the surface of the lake is : | [{"identifier": "A", "content": "45"}, {"identifier": "B", "content": "42"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "60"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266845/exam_images/jc6rgs0t7o340wghq0oy.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Evening Slot Mathematics - Height and Distance Question 28 English Explanation">
<br>tan 30<sup>o</sup> = $${x \over y} \Rightarrow y = \sqrt 3 x\,\,\,\,....(i)$$
<br><br>tan 60<sup>o</sup> = $${{25 + x + 25} \over y}$$
<br><br>$$ \Rightarrow \,\,\sqrt 3 y = 50 + x$$
<br><br>$$ \Rightarrow $$ $$3x = 50 + x$$
<br><br>$$ \Rightarrow $$ x = 25 m
<br><br>$$ \therefore $$ Height of cloud from surface
<br><br>= 25 + 25 = 50m | mcq | jee-main-2019-online-12th-january-evening-slot |
pxH6cut2HzLmnMIUy0jgy2xukg0cyxoq | maths | height-and-distance | height-and-distance | The angle of elevation of the summit of a
mountain from a point on the ground is 45°.
After climbing up one km towards the summit
at an inclination of 30° from the ground, the
angle of elevation of the summit is found to be
60°. Then the height (in km) of the summit from
the ground is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 3 - 1}}$$"}, {"identifier": "B", "content": "$${{\\sqrt 3 + 1} \\over {\\sqrt 3 - 1}}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 3 + 1}}$$"}, {"identifier": "D", "content": "$${{\\sqrt 3 - 1} \\over {\\sqrt 3 + 1}}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265755/exam_images/vceiuoriishdtliybgle.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Mathematics - Height and Distance Question 20 English Explanation">
<br><br>In $$\Delta $$CDF
<br><br>sin 30<sup>o</sup> = $${z \over 1}$$
<br><br>$$ \Rightarrow $$ z = $${1 \over 2}$$ km
<br><br>cos 30<sup>o</sup> = $${Y \over 1}$$
<br><br>$$ \Rightarrow $$ Y = $${{\sqrt 3 } \over 2}$$ km
<br><br>Now in $$\Delta $$ABC,
<br><br>tan 45<sup>o</sup> = $${h \over {X + Y}}$$
<br><br>$$ \Rightarrow $$ h = X + Y
<br><br>$$ \Rightarrow $$ X = h - $${{\sqrt 3 } \over 2}$$
<br><br>Now in $$\Delta $$BDE
<br><br>tan 60<sup>o</sup> = $${{h - z} \over X}$$
<br><br>$$ \Rightarrow $$ $${\sqrt 3 }$$X = h - $${1 \over 2}$$
<br><br>$$ \Rightarrow $$ $$\sqrt 3 \left( {h - {{\sqrt 3 } \over 2}} \right)$$ = h - $${1 \over 2}$$
<br><br>$$ \Rightarrow $$ h = $${1 \over {\sqrt 3 - 1}}$$ km | mcq | jee-main-2020-online-6th-september-evening-slot |
g3Go8rysjUigdUl2cnjgy2xukfw17ig0 | maths | height-and-distance | height-and-distance | Let AD and BC be two vertical poles <br/>at A and B respectively on a horizontal ground. <br/>If
AD = 8 m, BC = 11 m and AB = 10 m; then the distance<br/> (in meters) of a point M on AB from the point
A such<br/> that MD<sup>2</sup> + MC<sup>2</sup> is minimum is ______. | [] | null | 5 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266427/exam_images/aq1jmc4l3u0dm4zx0orb.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Height and Distance Question 22 English Explanation">
<br><br>(MD)<sup>2</sup>
+ (MC)<sup>2</sup>
= h<sup>2</sup>
+ 64 + (h – 10)<sup>2</sup>
+ 121
<br><br>= 2h<sup>2</sup>
– 20h + 64 + 100 + 121
<br><br>= 2(h<sup>2</sup>
– 10h) + 285
<br><br>= 2(h – 5)<sup>2</sup>
+ 235
<br><br>It is minimum if h = 5. | integer | jee-main-2020-online-6th-september-morning-slot |
lewrHJiFB4GaqO5A3fjgy2xukfxgnp8b | maths | height-and-distance | height-and-distance | The angle of elevation of the top of a hill from a point on the horizontal plane passing through the
foot of the hill is found to be 45<sup>o</sup>. After walking a distance of 80 meters towards the top, up a slope
inclined at an angle of 30<sup>o</sup> to the horizontal plane, the angle of elevation of the top of the hill
becomes 75<sup>o</sup>. Then the height of the hill (in meters) is ____________. | [] | null | 80 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264592/exam_images/wpspq1xvs3grsccoimwk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Morning Slot Mathematics - Height and Distance Question 21 English Explanation">
<br><br>sin 30<sup>o</sup> = $${x \over {80}}$$ $$ \Rightarrow $$ x = 40
<br><br>cos 30<sup>o</sup> = $${y \over {80}}$$ $$ \Rightarrow $$ y = $$40\sqrt 3 $$
<br><br>Now, In $$\Delta $$AEF
<br><br>tan 75<sup>o</sup> = $${{h - x} \over {h - y}}$$
<br><br>$$ \Rightarrow $$ 2 + $$\sqrt 3 $$ = $${{h - 40} \over {h - 40\sqrt 3 }}$$
<br><br>$$ \Rightarrow $$ $$\left( {2 + \sqrt 3 } \right)\left( {h - 40\sqrt 3 } \right)$$ = h - 40
<br><br>$$ \Rightarrow $$ 2h - 80$${\sqrt 3 }$$ + $${\sqrt 3 }$$h - 120 = h - 40
<br><br>$$ \Rightarrow $$ h + $${\sqrt 3 }$$h = 80 + 80$${\sqrt 3 }$$
<br><br>$$ \Rightarrow $$ ($${\sqrt 3 }$$ + 1)h = 80($${\sqrt 3 }$$ + 1)
<br><br>$$ \Rightarrow $$ h = 80 m | integer | jee-main-2020-online-6th-september-morning-slot |
TYuBguLiKOQljYZ9Bbjgy2xukfakapsn | maths | height-and-distance | height-and-distance | The angle of elevation of a cloud C from a point P, 200 m above a still lake is 30°. If the angle of depression of the image of C in the lake from the point P is 60°,then PC (in m) is equal to :
| [{"identifier": "A", "content": "$$200\\sqrt 3 $$"}, {"identifier": "B", "content": "400"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "$$400\\sqrt 3 $$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267793/exam_images/nysg0p5njtrlfbcb1wam.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266112/exam_images/nyi5l1fouysd8mkjmgb3.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264022/exam_images/pke0hgjw8stat0i8us1y.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263473/exam_images/fkhdonbhbro4n6hcue7z.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266631/exam_images/b7eikgu2adhv0wlelap4.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264737/exam_images/gq1yc2t6g4c8b8dx5rhy.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266261/exam_images/mad9v6jss642a5iyxeks.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Evening Slot Mathematics - Height and Distance Question 23 English Explanation"></picture>
<br>Let PA = x<br><br>For $$\Delta $$APC<br><br>$$AC = {{PA} \over {\sqrt 3 }} = {x \over {\sqrt 3 }}$$<br><br>$$A{C^1} = AB + B{C^1}$$<br><br>$$A{C^1} = 200 + {x \over {\sqrt 3 }}$$<br><br>From $$\Delta {C^1}PA, $$
<br><br>$$A{C^1} = \sqrt 3 PA$$<br><br>$$ \Rightarrow \left( {200 + {x \over {\sqrt 3 }}} \right) = \sqrt 3 x $$
<br><br>$$\Rightarrow x = (200)(\sqrt 3 )$$<br><br>From $$\Delta APC, $$
<br><br>sin 30<sup>o</sup> = $${{{x \over {\sqrt 3 }}} \over {PC}}$$
<br><br>$$ \Rightarrow $$ $$PC = {{2x} \over {\sqrt 3 }} $$
<br><br>$$\Rightarrow PC = 400$$ | mcq | jee-main-2020-online-4th-september-evening-slot |
MxCfgJ8IDqtwtaRJnP1klrga9a8 | maths | height-and-distance | height-and-distance | Two vertical poles are 150 m apart and the height of one is three times that of the other. If
from the middle point of the line joining their feet, an observer finds the angles of elevation of
their tops to be complementary, then the height of the shorter pole (in meters) is : | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "20$$\\sqrt 3 $$"}, {"identifier": "D", "content": "25$$\\sqrt 3 $$"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266120/exam_images/yeugccdhox3noatlqnmc.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267418/exam_images/vespqs6qac89duryh7y9.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267255/exam_images/qmutzyssa7f1cgtddf35.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266649/exam_images/zqjilakeoz8ylsn177dr.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266319/exam_images/kzxpinvpy4sfwodbsfod.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264779/exam_images/ogqc9gd5vcautialjont.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267031/exam_images/aszqjhyd9ytrutgsg8cr.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Height and Distance Question 19 English Explanation"></picture>
<br>$$ \therefore $$ $$\tan (90 - \theta ) = {x \over {75}}$$<br><br>and $$\tan \theta = {{3x} \over {75}}$$<br><br>As $$\cot \theta .\tan \theta = 1$$<br><br>$$ \therefore $$ $${x \over {75}}.{{3x} \over {75}} = 1$$<br><br>$$ \Rightarrow x = 25\sqrt 3 $$ | mcq | jee-main-2021-online-24th-february-morning-slot |
cMNUUY1RfRfxsUTd461klrlwmtk | maths | height-and-distance | height-and-distance | The angle of elevation of a jet plane from a point A on the ground is 60$$^\circ$$. After a flight of 20 seconds at the speed of 432 km/hour, the angle of elevation changes to 30$$^\circ$$. If the jet plane is flying at a constant height, then its height is : | [{"identifier": "A", "content": "$$3600\\sqrt 3 $$ m"}, {"identifier": "B", "content": "$$1200\\sqrt 3 $$ m"}, {"identifier": "C", "content": "$$1800\\sqrt 3 $$ m"}, {"identifier": "D", "content": "$$2400\\sqrt 3 $$ m"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266806/exam_images/yei81ivy9h4mpf0cbjzv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 24th February Evening Shift Mathematics - Height and Distance Question 18 English Explanation">
<br>$$v = 432 \times {{1000} \over {60 \times 60}}$$ m/sec = 120 m/sec<br><br>Distance AB = v $$\times$$ 20 = 2400 meter<br><br>In $$\Delta$$PAC<br><br>$$\tan 60^\circ = {h \over {PC}} \Rightarrow PC = {h \over {\sqrt 3 }}$$<br><br>In $$\Delta$$PBD<br><br>$$\tan 30^\circ = {h \over {PD}} \Rightarrow PD = \sqrt 3 h$$<br><br>PD = PC + CD<br><br>$$\sqrt 3 h = {h \over {\sqrt 3 }} + 2400 \Rightarrow {{2h} \over {\sqrt 3 }} = 2400$$<br><br>$$h = 1200\sqrt 3 $$ meter
| mcq | jee-main-2021-online-24th-february-evening-slot |
7LjqYLBFSIq2nWK5cW1kls4jij6 | maths | height-and-distance | height-and-distance | A man is observing, from the top of a tower, a boat speeding towards the lower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is 30$$^\circ$$ (Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is 45$$^\circ$$. Then the time taken (in seconds) by the boat from B to reach the base of the tower is : | [{"identifier": "A", "content": "$$10(\\sqrt 3 + 1)$$"}, {"identifier": "B", "content": "$$10(\\sqrt 3 - 1)$$"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "$$10\\sqrt 3$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266282/exam_images/cyid3ux4etov3mikiruk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Morning Shift Mathematics - Height and Distance Question 17 English Explanation">
<br>$${h \over {x + y}} = \tan 30^\circ $$<br><br>$$x + y = \sqrt 3 h$$ ...... (1)<br><br>Also, <br><br>$${h \over y} = \tan 45^\circ $$<br><br>$$h = y$$ ..... (2)<br><br>put in (1)<br><br>$$x + y = \sqrt 3 y$$<br><br>$$x = \left( {\sqrt 3 - 1} \right)y$$<br><br>$${x \over {20}} = 'v'$$ speed<br><br>$$ \therefore $$ time taken to reach Foot from B<br><br>$$ = {y \over V}$$<br><br>$$ = {x \over {\left( {\sqrt 3 - 1} \right).x}} \times 20$$<br><br>$$ = 10\left( {\sqrt 3 + 1} \right)$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
y2vesUbzQ1pa4cPPZy1kmm2uv3f | maths | height-and-distance | height-and-distance | A pole stands vertically inside a triangular park ABC. Let the angle of elevation of the top of the pole from each corner of the park be $${\pi \over 3}$$. If the radius of the circumcircle of $$\Delta$$ABC is 2, then the height of the pole is equal to : | [{"identifier": "A", "content": "$${{1 \\over {\\sqrt 3 }}}$$"}, {"identifier": "B", "content": "2$${\\sqrt 3 }$$"}, {"identifier": "C", "content": "$${\\sqrt 3 }$$"}, {"identifier": "D", "content": "$${{{2\\sqrt 3 } \\over 3}}$$"}] | ["B"] | null | <picture><source media="(max-width: 1221px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263609/exam_images/jzim1jti8mbu5b7il5p0.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267155/exam_images/s0gaiujemitqjrle3a8g.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267050/exam_images/v7tveeyggmh6rhelsdju.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264646/exam_images/ml5k91ttxiiunmymt8mx.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267482/exam_images/ck5zbksuz8ke6he73fie.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264042/exam_images/c7wzo3oqwrimegfuybce.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - Height and Distance Question 16 English Explanation"></picture>
<br>Let PD = h, R = 2<br><br>As angle of elevation of top of pole from A, B, C are equal. So D must be circumcentre of $$\Delta$$ABC<br><br>$$\tan \left( {{\pi \over 3}} \right) = {{PD} \over R} = {h \over R}$$<br><br>$$h = R\tan \left( {{\pi \over 3}} \right) = 2\sqrt 3 $$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1krrso0hb | maths | height-and-distance | height-and-distance | Let in a right angled triangle, the smallest angle be $$\theta$$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then sin$$\theta$$ is equal to : | [{"identifier": "A", "content": "$${{\\sqrt 5 + 1} \\over 4}$$"}, {"identifier": "B", "content": "$${{\\sqrt 5 - 1} \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt 2 - 1} \\over 2}$$"}, {"identifier": "D", "content": "$${{\\sqrt 5 - 1} \\over 4}$$"}] | ["B"] | null | <br>Let a $$\Delta$$ABC having C = 90$$^\circ$$ and A = $$\theta$$<br><br>$${{\sin \theta } \over a} = {{\cos \theta } \over b} = {1 \over c}$$ ..... (i)<br><br>Also for triangle of reciprocals<br><br>$$\cos A = {{{{\left( {{1 \over c}} \right)}^2} + {{\left( {{1 \over b}} \right)}^2} - {{\left( {{1 \over a}} \right)}^2}} \over {2\left( {{1 \over c}} \right)\left( {{1 \over b}} \right)}}$$<br><br>$${1 \over {{c^2}}} + {1 \over {{{(c\cos \theta )}^2}}} = {1 \over {{{(c\sin \theta )}^2}}}$$<br><br>$$ \Rightarrow 1 + {\sec ^2}\theta = \cos e{c^2}\theta $$<br><br>$$ \Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {4{{\sin }^2}\theta {{\cos }^2}\theta }}$$<br><br>$$ \Rightarrow {1 \over 4} = {{{{\cos }^2}\theta } \over {{{\sin }^2}2\theta }}$$<br><br>$$ \Rightarrow 1 - {\cos ^2}2\theta = 4\cos 2\theta $$<br><br>$${\cos ^2}2\theta + 4\cos 2\theta - 1 = 0$$<br><br>$$\cos 2\theta = {{ - 4 \pm \sqrt {16 + 4} } \over 2}$$<br><br>$$\cos 2\theta = - 2 \pm \sqrt 5 $$<br><br>$$\cos 2\theta = \sqrt 5 - 2 = 1 - 2{\sin ^2}\theta $$<br><br>$$ \Rightarrow 2{\sin ^2}\theta = 3 - \sqrt 5 $$<br><br>$$ \Rightarrow {\sin ^2}\theta = {{3 - \sqrt 5 } \over 2}$$<br><br>$$ \Rightarrow \sin \theta = {{\sqrt 5 - 1} \over 2}$$<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwld6t6g/11cc28b4-5cf0-4c13-9cdb-f2eaaa5c2649/ca276b80-5174-11ec-b1ae-c1b029d25cc4/file-1kwld6t6h.jpeg" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Height and Distance Question 15 English Explanation"> | mcq | jee-main-2021-online-20th-july-evening-shift |
1krvrqs6v | maths | height-and-distance | height-and-distance | A spherical gas balloon of radius 16 meter subtends an angle 60$$^\circ$$ at the eye of the observer A while the angle of elevation of its center from the eye of A is 75$$^\circ$$. Then the height (in meter) of the top most point of the balloon from the level of the observer's eye is : | [{"identifier": "A", "content": "$$8(2 + 2\\sqrt 3 + \\sqrt 2 )$$"}, {"identifier": "B", "content": "$$8(\\sqrt 6 + \\sqrt 2 + 2)$$"}, {"identifier": "C", "content": "$$8(\\sqrt 2 + 2 + \\sqrt 3 )$$"}, {"identifier": "D", "content": "$$8(\\sqrt 6 - \\sqrt 2 + 2)$$"}] | ["B"] | null | <br>O $$\to$$ centre of sphere<br><br>P, Q $$\to$$ point of contact of tangents from A<br><br>Let T be top most point of balloon & R be foot of perpendicular from O to ground.<br><br>From triangle OAP, OA = 16cosec30$$^\circ$$ = 32<br><br>From triangle ABO, OR = OA sin75$$^\circ$$ = $$32{{\left( {\sqrt 3 + 1} \right)} \over {2\sqrt 2 }}$$<br><br>So level of top most point = OR + OT<br><br>$$ = 8\left( {\sqrt 6 + \sqrt 2 + 2} \right)$$<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwld2tjm/54d13fc9-0216-4da9-b4ee-1d4f67320097/5b332f20-5174-11ec-b1ae-c1b029d25cc4/file-1kwld2tjn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Height and Distance Question 14 English Explanation"> | mcq | jee-main-2021-online-25th-july-morning-shift |
1ktd24qoa | maths | height-and-distance | height-and-distance | A 10 inches long pencil AB with mid point C and a small eraser P are placed on the horizontal top of a table such that PC = $$\sqrt 5 $$ inches and $$\angle$$PCB = tan<sup>-1</sup>(2). The acute angle through which the pencil must be rotated about C so that the perpendicular distance between eraser and pencil becomes exactly 1 inch is :<br/><br/><img src="data:image/png;base64,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"/> | [{"identifier": "A", "content": "$${\\tan ^{ - 1}}\\left( {{3 \\over 4}} \\right)$$"}, {"identifier": "B", "content": "tan<sup>$$-$$1</sup>(1)"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}\\left( {{4 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}\\left( {{1 \\over 2}} \\right)$$"}] | ["A"] | null | <br>From figure,<br><br>$$\sin \beta = {1 \over {\sqrt 5 }}$$<br><br>$$\therefore$$ $$\tan \beta = {1 \over 2}$$<br><br>$$\tan (\alpha + \beta ) = 2$$<br><br>$${{\tan \alpha + \tan \beta )} \over {1 - \tan \alpha .\tan \beta }} = 2$$<br><br>$${{\tan \alpha + {1 \over 2}} \over {1 - \tan \alpha \left( {{1 \over 2}} \right)}} = 2$$<br><br>$$\tan \alpha = {3 \over 4}$$<br><br>$$\alpha = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$$<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwlcorx4/9da75a96-cc50-45bd-83d1-6ec5874ccc4f/d49e0490-5172-11ec-81c8-372e0b624874/file-1kwlcorx5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 26th August Evening Shift Mathematics - Height and Distance Question 13 English Explanation"> | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktg351uo | maths | height-and-distance | height-and-distance | Two poles, AB of length a metres and CD of length a + b (b $$\ne$$ a) metres are erected at the same horizontal level with bases at B and D. If BD = x and tan$$\angle$$ACB = $${1 \over 2}$$, then : | [{"identifier": "A", "content": "x<sup>2</sup> + 2(a + 2b)x $$-$$ b(a + b) = 0"}, {"identifier": "B", "content": "x<sup>2</sup> + 2(a + 2b)x + a(a + b) = 0"}, {"identifier": "C", "content": "x<sup>2</sup> $$-$$ 2ax + b(a + b) = 0"}, {"identifier": "D", "content": "x<sup>2</sup> $$-$$ 2ax + a(a + b) = 0"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264596/exam_images/ojvw2ffxtio4f1jffdyo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Height and Distance Question 12 English Explanation"> <br><br>$$\tan \theta = {1 \over 2}$$<br><br>$$\tan (\theta + \alpha ) = {x \over b},\tan \alpha = {x \over {a + b}}$$<br><br>$$ \Rightarrow {{{1 \over 2} + {x \over {a + b}}} \over {1 - {1 \over 2} \times {x \over {a + b}}}} = {x \over b}$$<br><br>$$ \Rightarrow {x^2} - 2ax + ab + {b^2} = 0$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1ktir6j4n | maths | height-and-distance | height-and-distance | A vertical pole fixed to the horizontal ground is divided in the ratio 3 : 7 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a point on the ground 18 m away from the base of the pole, then the height of the pole (in meters) is : | [{"identifier": "A", "content": "12$$\\sqrt {15} $$"}, {"identifier": "B", "content": "12$$\\sqrt {10} $$"}, {"identifier": "C", "content": "8$$\\sqrt {10} $$"}, {"identifier": "D", "content": "6$$\\sqrt {10} $$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266417/exam_images/igmotqewqjtulhutbk8z.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Mathematics - Height and Distance Question 11 English Explanation"><br><br>Let height of pole = 10$$l$$<br><br>$$\tan \alpha = {{3l} \over {18}} = {l \over 6}$$<br><br>$$\tan 2\alpha = {{10l} \over {18}}$$<br><br>$${{2\tan \alpha } \over {1 - {{\tan }^2}\alpha }} = {{10l} \over {18}}$$<br><br>use $$\tan \alpha = {l \over 6} \Rightarrow l = \sqrt {{{72} \over 5}} $$<br><br>height of pole = $$10l = 12\sqrt {10} $$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1l54tdyii | maths | height-and-distance | height-and-distance | <p>From the base of a pole of height 20 meter, the angle of elevation of the top of a tower is 60$$^\circ$$. The pole subtends an angle 30$$^\circ$$ at the top of the tower. Then the height of the tower is :</p> | [{"identifier": "A", "content": "$$15\\sqrt 3 $$"}, {"identifier": "B", "content": "$$20\\sqrt 3 $$"}, {"identifier": "C", "content": "20 + $$10\\sqrt 3 $$"}, {"identifier": "D", "content": "30"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5flljg6/71a3df8b-f1e7-4fa5-9624-d4e70f01a8df/eb523e50-0076-11ed-bd71-b57b399a7926/file-1l5flljg7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5flljg6/71a3df8b-f1e7-4fa5-9624-d4e70f01a8df/eb523e50-0076-11ed-bd71-b57b399a7926/file-1l5flljg7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Evening Shift Mathematics - Height and Distance Question 10 English Explanation"></p>
<p>Here AB is a tower and CD is a pole.</p>
<p>In triangle ABC,</p>
<p>$$\tan 60^\circ = {{AB} \over {AC}} = {{20 + h} \over x}$$ ...... (1)</p>
<p>In triangle BED,</p>
<p>$$\tan 30^\circ = {h \over x}$$ ...... (2)</p>
<p>Divide equation (1) by equation (2), we get</p>
<p>$${{\tan 60^\circ } \over {\tan 30^\circ }} = {{20 + h} \over x} \times {x \over h}$$</p>
<p>$$ \Rightarrow {{\sqrt 3 } \over {{1 \over {\sqrt 3 }}}} = {{20 + h} \over h}$$</p>
<p>$$ \Rightarrow 3 = {{20 + h} \over h}$$</p>
<p>$$ \Rightarrow 3h = 20 + h$$</p>
<p>$$ \Rightarrow h = 10\,m$$</p>
<p>$$\therefore$$ Height of tower $$ = 20 + 10 = 30\,m$$</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l567bggg | maths | height-and-distance | height-and-distance | <p>Let AB and PQ be two vertical poles, 160 m apart from each other. Let C be the middle point of B and Q, which are feet of these two poles. Let $${\pi \over 8}$$ and $$\theta$$ be the angles of elevation from C to P and A, respectively. If the height of pole PQ is twice the height of pole AB, then tan<sup>2</sup>$$\theta$$ is equal to</p> | [{"identifier": "A", "content": "$${{3 - 2\\sqrt 2 } \\over 2}$$"}, {"identifier": "B", "content": "$${{3 + \\sqrt 2 } \\over 2}$$"}, {"identifier": "C", "content": "$${{3 - 2\\sqrt 2 } \\over 4}$$"}, {"identifier": "D", "content": "$${{3 - \\sqrt 2 } \\over 4}$$"}] | ["C"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5oc0rd9/9c4c842e-b702-4002-a405-e2ab09551342/852123d0-0544-11ed-987f-3938cfc0f7f1/file-1l5oc0rda.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5oc0rd9/9c4c842e-b702-4002-a405-e2ab09551342/852123d0-0544-11ed-987f-3938cfc0f7f1/file-1l5oc0rda.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Height and Distance Question 9 English Explanation"></p>
<p>$${l \over {80}} = \tan \theta $$ ..... (i)</p>
<p>$${{2l} \over {80}} = \tan {\pi \over 8}$$ ..... (ii)</p>
<p>From (i) and (ii)</p>
<p>$${1 \over 2} = {{\tan \theta } \over {\tan {\pi \over 8}}} \Rightarrow {\tan ^2}\theta = {1 \over 4}{\tan ^2}{\pi \over 8}$$</p>
<p>$$ \Rightarrow {\tan ^2}\theta = {{\sqrt 2 - 1} \over {4(\sqrt 2 + 1)}} = {{3 - 2\sqrt 2 } \over 4}$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l6dwt2de | maths | height-and-distance | height-and-distance | <p>A tower PQ stands on a horizontal ground with base $$Q$$ on the ground. The point $$R$$ divides the tower in two parts such that $$Q R=15 \mathrm{~m}$$. If from a point $$A$$ on the ground the angle of elevation of $$R$$ is $$60^{\circ}$$ and the part $$P R$$ of the tower subtends an angle of $$15^{\circ}$$ at $$A$$, then the height of the tower is :
</p> | [{"identifier": "A", "content": "$$5(2 \\sqrt{3}+3) \\,\\mathrm{m}$$"}, {"identifier": "B", "content": "$$5(\\sqrt{3}+3) \\,\\mathrm{m}$$"}, {"identifier": "C", "content": "$$10(\\sqrt{3}+1) \\,\\mathrm{m}$$"}, {"identifier": "D", "content": "$$10(2 \\sqrt{3}+1) \\,\\mathrm{m}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79fj3e6/3edc2d44-ec0d-4c06-831f-a466d782744d/eb6330e0-24aa-11ed-8d2e-5f0df5271c2d/file-1l79fj3e7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79fj3e6/3edc2d44-ec0d-4c06-831f-a466d782744d/eb6330e0-24aa-11ed-8d2e-5f0df5271c2d/file-1l79fj3e7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Height and Distance Question 8 English Explanation"></p>
<p>For $$\Delta$$AQR,</p>
<p>$$\tan 60^\circ = {{15} \over x}$$ ....... (1)</p>
<p>From $$\Delta$$AQP,</p>
<p>$$\tan 75^\circ = {h \over x}$$</p>
<p>$$ \Rightarrow \left( {2 + \sqrt 3 } \right) = {h \over x}$$ [$$\because$$ $$\tan 75^\circ = 2 + \sqrt 3 $$]</p>
<p>$$ \Rightarrow h = \left( {2 + \sqrt 3 } \right)x$$</p>
<p>$$ = \left( {2 + \sqrt 3 } \right){{15} \over {\sqrt 3 }}$$ [From (1)]</p>
<p>$$ = \left( {2 + \sqrt 3 } \right) \times {{15\sqrt 3 } \over 3}$$</p>
<p>$$ = \left( {2 + \sqrt 3 } \right) \times 5\sqrt 3 $$</p>
<p>$$ = 5\left( {2\sqrt 3 + 3} \right)$$ m</p> | mcq | jee-main-2022-online-25th-july-morning-shift |
1l6jcbdiq | maths | height-and-distance | height-and-distance | <p>Let a vertical tower $$A B$$ of height $$2 h$$ stands on a horizontal ground. Let from a point $$P%$$ on the ground a man can see upto height $$h$$ of the tower with an angle of elevation $$2 \alpha$$. When from $$P$$, he moves a distance $$d$$ in the direction of $$\overrightarrow{A P}$$, he can see the top $$B$$ of the tower with an angle of elevation $$\alpha$$. If $$d=\sqrt{7} h$$, then $$\tan \alpha$$ is equal to</p> | [{"identifier": "A", "content": "$$\\sqrt{5}-2$$"}, {"identifier": "B", "content": "$$\\sqrt{3}-1$$"}, {"identifier": "C", "content": "$$ \\sqrt{7}-2$$"}, {"identifier": "D", "content": "$$\\sqrt{7}-\\sqrt{3}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7psamcq/756ef252-1665-43d7-bc8f-4dd4bf8b8c96/60d220a0-2da9-11ed-8542-f96181a425b5/file-1l7psamcr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7psamcq/756ef252-1665-43d7-bc8f-4dd4bf8b8c96/60d220a0-2da9-11ed-8542-f96181a425b5/file-1l7psamcr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Morning Shift Mathematics - Height and Distance Question 7 English Explanation"></p>
<p>$$\Delta{APM}$$ gives</p>
<p>$$\tan 2\alpha = {h \over x}$$ ..... (i)</p>
<p>$$\Delta{AQB}$$ gives</p>
<p>$$\tan 2\alpha = {{2h} \over {x + d}} = {{2h} \over {x + h\sqrt 7 }}$$ ...... (ii)</p>
<p>From (i) and (ii)</p>
<p>$$\tan 2\alpha = {{2\,.\,\tan 2\alpha } \over {1 + \sqrt 7 \,.\,\tan 2\alpha }}$$</p>
<p>Let $$t = \tan \alpha $$</p>
<p>$$ \Rightarrow t = {{2{{2t} \over {1 - {t^2}}}} \over {1 + \sqrt 7 \,.\,{{2t} \over {1 - {t^2}}}}}$$</p>
<p>$$ \Rightarrow {t^2} - 2\sqrt 7 t + 3 = 0$$</p>
<p>$$t = \sqrt 7 - 2$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6kkz98e | maths | height-and-distance | height-and-distance | <p>The angle of elevation of the top P of a vertical tower PQ of height 10 from a point A on the horizontal ground is $$45^{\circ}$$. Let R be a point on AQ and from a point B, vertically above $$\mathrm{R}$$, the angle of elevation of $$\mathrm{P}$$ is $$60^{\circ}$$. If $$\angle \mathrm{BAQ}=30^{\circ}, \mathrm{AB}=\mathrm{d}$$ and the area of the trapezium $$\mathrm{PQRB}$$ is $$\alpha$$, then the ordered pair $$(\mathrm{d}, \alpha)$$ is :</p> | [{"identifier": "A", "content": "$$(10(\\sqrt{3}-1), 25)$$"}, {"identifier": "B", "content": "$$\\left(10(\\sqrt{3}-1), \\frac{25}{2}\\right)$$"}, {"identifier": "C", "content": "$$(10(\\sqrt{3}+1), 25)$$"}, {"identifier": "D", "content": "$$\\left(10(\\sqrt{3}+1), \\frac{25}{2}\\right)$$"}] | ["A"] | null | <p>Let $$BR = x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qafj8i/a7a027fb-787f-4a20-838e-deb7a4fd2b6b/4dc88020-2df0-11ed-a744-1fb8f3709cfa/file-1l7qafj8j.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qafj8i/a7a027fb-787f-4a20-838e-deb7a4fd2b6b/4dc88020-2df0-11ed-a744-1fb8f3709cfa/file-1l7qafj8j.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Evening Shift Mathematics - Height and Distance Question 6 English Explanation"></p>
<p>$${x \over d} = {1 \over 2} \Rightarrow x = {d \over 2}$$</p>
<p>$${{10 - x} \over {10 - x\sqrt 3 }} = \sqrt 3 \Rightarrow 10 - x = 10\sqrt 3 - 3x$$</p>
<p>$$2x = 10(\sqrt 3 - 1)$$</p>
<p>$$x = 5(\sqrt 3 - 1)$$</p>
<p>$$d = 2x = 10(\sqrt 3 - 1)$$</p>
<p>$$\alpha = {1 \over 2}(x + 10)(10 - x\sqrt 3 )=$$ Area (PQRB)</p>
<p>$$ = {1 \over 2}\left( {5\sqrt 3 - 5 + 10} \right)\left( {10 - 5\sqrt 3 (\sqrt 3 - 1)} \right)$$</p>
<p>$$ = {1 \over 2}\left( {5\sqrt 3 + 5} \right)\left( {10 - 15 + 5\sqrt 3 } \right) - {1 \over 2}(75 - 25) = 25$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nnwt7m | maths | height-and-distance | height-and-distance | <p>A horizontal park is in the shape of a triangle $$\mathrm{OAB}$$ with $$\mathrm{AB}=16$$. A vertical lamp post $$\mathrm{OP}$$ is erected at the point $$\mathrm{O}$$ such that $$\angle \mathrm{PAO}=\angle \mathrm{PBO}=15^{\circ}$$ and $$\angle \mathrm{PCO}=45^{\circ}$$, where $$\mathrm{C}$$ is the midpoint of $$\mathrm{AB}$$. Then $$(\mathrm{OP})^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{32}{\\sqrt{3}}(\\sqrt{3}-1)$$"}, {"identifier": "B", "content": "$$\\frac{32}{\\sqrt{3}}(2-\\sqrt{3})$$"}, {"identifier": "C", "content": "$$\\frac{16}{\\sqrt{3}}(\\sqrt{3}-1)$$"}, {"identifier": "D", "content": "$$\\frac{16}{\\sqrt{3}}(2-\\sqrt{3})$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rri7rq/1c703de6-5ecf-4544-826f-2ec23ff8478d/dc32c760-2ebf-11ed-b92e-01f1dabc9173/file-1l7rri7rr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rri7rq/1c703de6-5ecf-4544-826f-2ec23ff8478d/dc32c760-2ebf-11ed-b92e-01f1dabc9173/file-1l7rri7rr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Evening Shift Mathematics - Height and Distance Question 5 English Explanation"></p>
<p>$$OP = OA\tan 15 = OB\tan 15$$ ...... (i)</p>
<p>$$OP = OC\tan 45 \Rightarrow OP = OC$$ ...... (ii)</p>
<p>$$OA = OB$$ ....... (iii)</p>
<p>$$O{C^2} + {8^2} = O{A^2}$$</p>
<p>$$O{P^2} + 64 = O{P^2}{\left( {{{\sqrt 3 + 1} \over {\sqrt 3 - 1}}} \right)^2}$$</p>
<p>$$64 = O{P^2}\left[ {{{{{\left( {\sqrt 3 + 1} \right)}^2} - {{\left( {\sqrt {3 - 1} } \right)}^2}} \over {{{\left( {\sqrt 3 - 1} \right)}^2}}}} \right]$$</p>
<p>$$ = O{P^2}\left( {{{4\sqrt 3 } \over {{{\left( {\sqrt 3 - 1} \right)}^2}}}} \right)$$</p>
<p>$$O{P^2} = {{64{{\left( {\sqrt 3 - 1} \right)}^2}} \over {4\sqrt 3 }} = {{32} \over {\sqrt 3 }}\left( {2 - \sqrt 3 } \right)$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6p25qly | maths | height-and-distance | height-and-distance | <p>The angle of elevation of the top of a tower from a point A due north of it is $$\alpha$$ and from a point B at a distance of 9 units due west of A is $$\cos ^{-1}\left(\frac{3}{\sqrt{13}}\right)$$. If the distance of the point B from the tower is 15 units, then $$\cot \alpha$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{6}{5}$$"}, {"identifier": "B", "content": "$$\\frac{9}{5}$$"}, {"identifier": "C", "content": "$$\\frac{4}{3}$$"}, {"identifier": "D", "content": "$$\\frac{7}{3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssh0oh/2868d3c3-c2c7-439f-b463-f42683b289b7/6cb0f920-2f50-11ed-88f3-17ddb055f60b/file-1l7ssh0oi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ssh0oh/2868d3c3-c2c7-439f-b463-f42683b289b7/6cb0f920-2f50-11ed-88f3-17ddb055f60b/file-1l7ssh0oi.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Morning Shift Mathematics - Height and Distance Question 4 English Explanation"></p>
<p>$$NA = \sqrt {{{15}^2} - {9^2}} = 12$$</p>
<p>$${h \over {15}} = \tan \theta = {2 \over 3}$$</p>
<p>$$h = 10$$ units</p>
<p>$$\cot \alpha = {{12} \over {10}} = {6 \over 5}$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1lgsvy5it | maths | height-and-distance | height-and-distance | <p>The angle of elevation of the top $$\mathrm{P}$$ of a tower from the feet of one person standing due South of the tower is $$45^{\circ}$$ and from the feet of another person standing due west of the tower is $$30^{\circ}$$. If the height of the tower is 5 meters, then the distance (in meters) between the two persons is equal to</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "$$\\frac{5}{2} \\sqrt{5}$$"}, {"identifier": "C", "content": "$$5 \\sqrt{5}$$"}, {"identifier": "D", "content": "5"}] | ["A"] | null | Let's denote the person standing due south as S and the one standing due west as W. Also, let the tower be at point T.
<br/><br/>From person S's perspective, we have a right triangle $\triangle SPT$. The height of the tower PT is given as 5m, which is the opposite side for angle S. The angle at S is $45^\circ$. From the tangent trigonometric ratio, we have :
<br/><br/>$
\tan{45^\circ} = \frac{PT}{ST} = \frac{5}{ST}
$
<br/><br/>Since $\tan{45^\circ}=1$, we get $ST = 5$m.
<br/><br/>Similarly, from person W's perspective, we have a right triangle $\triangle WPT$. The angle at W is $30^\circ$. From the tangent trigonometric ratio, we have:
<br/><br/>
$\tan{30^\circ} = \frac{PT}{WT} = \frac{5}{WT}$
<br/><br/>Since $\tan{30^\circ}=\frac{1}{\sqrt{3}}$, we get $WT = 5\sqrt{3}$m.
<br/><br/>Since S and W are perpendicular to each other (one is due south and the other is due west), $\triangle SWT$ forms a right triangle.
<br/><br/>We can find $SW$ (the distance between the two people) using the Pythagorean theorem:
<br/><br/>$
SW^2 = ST^2 + WT^2 = 5^2 + (5\sqrt{3})^2 = 25 + 75 = 100
$
<br/><br/>So, $SW = \sqrt{100} = 10$m.
<br/><br/>Hence, the correct answer is 10 meters, which corresponds to Option A. | mcq | jee-main-2023-online-11th-april-evening-shift |
1lh22i34s | maths | height-and-distance | height-and-distance | <p>From the top $$\mathrm{A}$$ of a vertical wall $$\mathrm{AB}$$ of height $$30 \mathrm{~m}$$, the angles of depression of the top $$\mathrm{P}$$ and bottom $$\mathrm{Q}$$ of a vertical tower $$\mathrm{PQ}$$ are $$15^{\circ}$$ and $$60^{\circ}$$ respectively, $$\mathrm{B}$$ and $$\mathrm{Q}$$ are on the same horizontal level. If $$\mathrm{C}$$ is a point on $$\mathrm{AB}$$ such that $$\mathrm{CB}=\mathrm{PQ}$$, then the area (in $$\mathrm{m}^{2}$$ ) of the quadrilateral $$\mathrm{BCPQ}$$ is equal to :</p> | [{"identifier": "A", "content": "$$200(3-\\sqrt{3})$$"}, {"identifier": "B", "content": "$$300(\\sqrt{3}-1)$$"}, {"identifier": "C", "content": "$$300(\\sqrt{3}+1)$$"}, {"identifier": "D", "content": "$$600(\\sqrt{3}-1)$$"}] | ["D"] | null | Given, $A B$ be a vertical wall of height $30 \mathrm{~m}$ and $P Q$ be a vertical tower.
<br><br>Such that $\angle B Q A=\angle T A Q=60^{\circ}$ (Alternate angles)
<br><br>and $\angle C P A=\angle T A P=15^{\circ}$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnoi86a2/300ca189-bd8f-4def-ab5a-3cd9c710ee8f/aee81b90-69b8-11ee-a65f-213955fd9d72/file-6y3zli1lnoi86a3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnoi86a2/300ca189-bd8f-4def-ab5a-3cd9c710ee8f/aee81b90-69b8-11ee-a65f-213955fd9d72/file-6y3zli1lnoi86a3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Morning Shift Mathematics - Height and Distance Question 1 English Explanation">
<br><br>Now, $\tan 60^{\circ}=\frac{A B}{B Q}=\frac{30}{B Q}$
<br><br>$$
\begin{array}{ll}
&\Rightarrow \sqrt{3}=\frac{30}{B Q} \\\\
&\Rightarrow B Q=\frac{30}{\sqrt{3}}=10 \sqrt{3}
\end{array}
$$
<br><br>$$
\begin{aligned}
& \text { and } \tan 15^{\circ}=\frac{A C}{P C} \\\\
& \Rightarrow 2-\sqrt{3}=\frac{A C}{10 \sqrt{3}} (\because P C=B Q) \\\\
& \Rightarrow A C=10 \sqrt{3}(2-\sqrt{3})=20 \sqrt{3}-30 \\\\
& \therefore B C=A B-A C=30-(20 \sqrt{3}-30)=60-20 \sqrt{3}
\end{aligned}
$$
<br><br>$\therefore$ Area of quadrilateral $B C P Q$
<br><br>$$
\begin{aligned}
& =B Q \times B C=10 \sqrt{3} \times(60-20 \sqrt{3}) \\\\
& =10(60 \sqrt{3}-60) \\\\
& =600(\sqrt{3}-1) \mathrm{m}^2
\end{aligned}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
ChoEdf9u4lZOoFejCdqYG | maths | hyperbola | common-tangent | Equation of a common tangent to the parabola y<sup>2</sup> = 4x and the hyperbola xy = 2 is : | [{"identifier": "A", "content": "x + y + 1 = 0"}, {"identifier": "B", "content": "4x + 2y + 1 = 0"}, {"identifier": "C", "content": "x \u2013 2y + 4 = 0"}, {"identifier": "D", "content": "x + 2y + 4 = 0"}] | ["D"] | null | Let the equation of tangent to parabola
<br><br>y<sup>2</sup> = 4x be y = mx + $${1 \over m}$$
<br><br>It is also a tangent to hyperbola xy = 2
<br><br>$$ \Rightarrow $$ x$$\left( {mx + {1 \over m}} \right)$$ = 2
<br><br>$$ \Rightarrow $$ x<sup>2</sup>m + $${x \over m}$$ $$-$$ 2 = 0
<br><br>D = 0 $$ \Rightarrow $$ m = $$-$$ $${1 \over 2}$$
<br><br>So tangent is 2y + x + 4 = 0 | mcq | jee-main-2019-online-11th-january-morning-slot |
oP6I4Fp5P7sBpVKC10jgy2xukfqdqhxe | maths | hyperbola | common-tangent | If the line y = mx + c is a common tangent to
the hyperbola
<br/>$${{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1$$ and the circle
x<sup>2</sup>
+ y<sup>2</sup>
= 36, then which one of the following is
true? | [{"identifier": "A", "content": "5m = 4"}, {"identifier": "B", "content": "8m + 5 = 0"}, {"identifier": "C", "content": "c<sup>2</sup> = 369"}, {"identifier": "D", "content": "4c<sup>2</sup> = 369"}] | ["D"] | null | $${{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1$$
<br><br>$$ \therefore $$ c = $$ \pm $$ $$\sqrt {{a^2}{m^2} - {b^2}} $$
<br><br>$$ \Rightarrow $$ c = $$ \pm $$ $$\sqrt {100{m^2} - 64} $$
<br><br>General tangent to hyperbola in slope form is
<br><br>y = mx $$ \pm $$ $$\sqrt {100{m^2} - 64} $$
<br><br>This tangent is also tangent to the circle x<sup>2</sup>
+ y<sup>2</sup>
= 36, whose center (0, 0) and radius = 6.
<br><br>Distance of the tangent from the center is
<br><br>$$\left| {{{\sqrt {100{m^2} - 64} } \over {\sqrt {{m^2} + 1} }}} \right|$$ = 6
<br><br>$$ \Rightarrow $$ 100m<sup>2</sup>
— 64 = 36m<sup>2</sup>
+ 36
<br><br>$$ \Rightarrow $$ 64m<sup>2</sup>
= 100
<br><br>$$ \Rightarrow $$ m = $${{10} \over 8}$$
<br><br>$$ \therefore $$ c<sup>2</sup> = 100 $$ \times $$ $${{100} \over {64}}$$ - 64
<br><br>$$ \Rightarrow $$ c<sup>2</sup> = $${{{{100}^2} - {{64}^2}} \over {64}}$$
<br><br>$$ \Rightarrow $$ c<sup>2</sup> = $${{164 \times 36} \over {64}}$$
<br><br>$$ \Rightarrow $$ 4c<sup>2</sup> = 369 | mcq | jee-main-2020-online-5th-september-evening-slot |
1l6klz0xz | maths | hyperbola | common-tangent | <p>A common tangent $$\mathrm{T}$$ to the curves $$\mathrm{C}_{1}: \frac{x^{2}}{4}+\frac{y^{2}}{9}=1$$ and $$C_{2}: \frac{x^{2}}{42}-\frac{y^{2}}{143}=1$$ does not pass through the fourth quadrant. If $$\mathrm{T}$$ touches $$\mathrm{C}_{1}$$ at $$\left(x_{1}, y_{1}\right)$$ and $$\mathrm{C}_{2}$$ at $$\left(x_{2}, y_{2}\right)$$, then $$\left|2 x_{1}+x_{2}\right|$$ is equal to ______________.</p> | [] | null | 20 | <p>Equation of tangent to ellipse $${{{x^2}} \over 4} + {{{y^2}} \over 9} = 1$$ and given slope m is : $$y = mx + \sqrt {4{m^2} + 9} $$ ..... (i)</p>
<p>For slope m equation of tangent to hyperbola is :</p>
<p>$$y = mx + \sqrt {42{m^2} - 143} $$ ....... (ii)</p>
<p>Tangents from (i) and (ii) are identical then</p>
<p>$$4{m^2} + 9 = 42{m^2} - 143$$</p>
<p>$$\therefore$$ $$m = \, \pm \,2$$ (+2 is not acceptable)</p>
<p>$$\therefore$$ $$m = - 2$$.</p>
<p>Hence, $${x_1} = {8 \over 5}$$ and $${x_2} = {{84} \over 5}$$</p>
<p>$$\therefore$$ $$|2{x_1} + {x_2}| = \left| {{{16} \over 5} + {{84} \over 5}} \right| = 20$$</p> | integer | jee-main-2022-online-27th-july-evening-shift |
1l6m6ue5c | maths | hyperbola | common-tangent | <p>For the hyperbola $$\mathrm{H}: x^{2}-y^{2}=1$$ and the ellipse $$\mathrm{E}: \frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1$$, a $$>\mathrm{b}>0$$, let the</p>
<p>(1) eccentricity of $$\mathrm{E}$$ be reciprocal of the eccentricity of $$\mathrm{H}$$, and</p>
<p>(2) the line $$y=\sqrt{\frac{5}{2}} x+\mathrm{K}$$ be a common tangent of $$\mathrm{E}$$ and $$\mathrm{H}$$.</p>
<p>Then $$4\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)$$ is equal to _____________.</p> | [] | null | 3 | <p>The equation of tangent to hyperbola $${x^2} - {y^2} = 1$$ within slope $$m$$ is equal to $$y = mx\, \pm \,\sqrt {{m^2} - 1} $$ ...... (i)</p>
<p>And for same slope $$m$$, equation of tangent to ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ is $$y = mx\, \pm \,\sqrt {{a^2}{m^2} + {b^2}} $$ ...... (ii)</p>
<p>$$\because$$ Equation (i) and (ii) are identical</p>
<p>$$\therefore$$ $${a^2}{m^2} + {b^2} = {m^2} - 1$$</p>
<p>$$\therefore$$ $${m^2} = {{1 + {b^2}} \over {1 - {a^2}}}$$</p>
<p>But equation of common tangent is $$y = \sqrt {{5 \over 2}} x + k$$</p>
<p>$$\therefore$$ $$m = \sqrt {{5 \over 2}} \Rightarrow {5 \over 2} = {{1 + {b^2}} \over {1 - {a^2}}}$$</p>
<p>$$\therefore$$ $$5{a^2} + 2{b^2} = 3$$ ....... (i)</p>
<p>eccentricity of ellipse $$ = {1 \over {\sqrt 2 }}$$</p>
<p>$$\therefore$$ $$1 - {{{b^2}} \over {{a^2}}} = {1 \over 2}$$</p>
<p>$$ \Rightarrow {a^2} = 2{b^2}$$ ....... (ii)</p>
<p>From equation (i) and (ii) : $${a^2} = {1 \over 2},\,{b^2} = {1 \over 4}$$</p>
<p>$$\therefore$$ $$4({a^2} + {b^2}) = 3$$</p> | integer | jee-main-2022-online-28th-july-morning-shift |
Z3V5G1McOharWAOJoJmAb | maths | hyperbola | locus | The locus of the point of intersection of the straight lines,
<br/><br/>tx $$-$$ 2y $$-$$ 3t = 0
<br/><br/>x $$-$$ 2ty + 3 = 0 <b>(t $$ \in $$ R)</b>, is : | [{"identifier": "A", "content": "an ellipse with eccentricity $${2 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "an ellipse with the length of major axis 6 "}, {"identifier": "C", "content": "a hyperbola with eccentricity $$\\sqrt 5 $$ "}, {"identifier": "D", "content": "a hyperbola with the length of conjugate axis 3 "}] | ["D"] | null | Here, tx $$-$$ 2y $$-$$ 3t = 0 & x $$-$$ 2ty + 3 = 0
<br><br>On solving, we get;
<br><br>y = $${{6t} \over {2{t^2} - 2}}$$ = $${{3t} \over {{t^2} - 1}}$$ & x = $${{3{t^2} + 3} \over {{t^2} - 1}}$$
<br><br>Put t = tan$$\theta $$
<br><br>$$ \therefore $$ x = $$-$$ 3 sec 2$$\theta $$ & 2y = 3 ($$-$$ tan 2$$\theta $$)
<br><br>$$ \because $$ sec<sup>2</sup>2$$\theta $$ $$-$$ tan<sup>2</sup>2$$\theta $$ = 1
<br><br>$$ \Rightarrow $$ $${{{x^2}} \over 9}$$ $$-$$ $${{{y^2}} \over {9/4}}$$ = 1
<br><br>which represents at hyperbola
<br><br>$$ \therefore $$ a<sup>2</sup> = 9 & b<sup>2</sup> = 9/4
<br><br>$$\lambda $$(T.A.) = 6; e<sup>2</sup> = 1 + $${{9/4} \over 9}$$ = 1 + $${1 \over 4}$$ $$ \Rightarrow $$ e = $${{\sqrt 5 } \over 2}$$ | mcq | jee-main-2017-online-8th-april-morning-slot |
Q6AKcTjpG375yPdeFFUqx | maths | hyperbola | locus | If the tangents drawn to the hyperbola 4y<sup>2</sup> = x<sup>2</sup> + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is : | [{"identifier": "A", "content": "x<sup>2</sup> $$-$$ 4y<sup>2</sup> + 16x<sup>2</sup>y<sup>2</sup> = 0"}, {"identifier": "B", "content": "x<sup>2</sup> $$-$$ 4y<sup>2</sup> $$-$$ 16x<sup>2</sup>y<sup>2</sup> = 0"}, {"identifier": "C", "content": "4x<sup>2</sup> $$-$$ y<sup>2</sup> + 16x<sup>2</sup>y<sup>2</sup> = 0"}, {"identifier": "D", "content": "4x<sup>2</sup> $$-$$ y<sup>2</sup> $$-$$ 16x<sup>2</sup>y<sup>2</sup> = 0"}] | ["B"] | null | Equation of hyperbola is :
<br><br>4y<sup>2</sup> = x<sup>2</sup> + 1 $$ \Rightarrow $$ $$-$$ x<sup>2</sup> + 4y<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ $$-$$ $${{{x^2}} \over {{1^2}}}$$ + $${{{y^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}}$$ = 1
<br><br>$$ \therefore $$ a = 1, b = $${1 \over 2}$$
<br><br>Now, tangent to the curve at point (x<sub>1</sub>, y<sub>1</sub>) is given by
<br><br>4 $$ \times $$ 2y<sub>1</sub>$${{dy} \over {dx}}$$ = 2x<sub>1</sub>
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{2{x_1}} \over {8{y_1}}}$$ = $${{{x_1}} \over {4{y_1}}}$$
<br><br>Equation of tangent at (x<sub>1</sub>, y<sub>1</sub>) is
<br><br>y = mx + c
<br><br>$$ \Rightarrow $$ y = $${{{x_1}} \over {4{y_1}}}$$ . x + c
<br><br>As tangent passes through (x<sub>1</sub>, y<sub>1</sub>)
<br><br>$$ \therefore $$ y<sub>1</sub> = $${{{x_1}{x_1}} \over {4{y_1}}} + c$$
<br><br>$$ \Rightarrow $$ C = $${{4y_1^2 - x_1^2} \over {4{y_1}}}$$ = $${1 \over {4{y_1}}}$$
<br><br>Therefore, y = $${{{x_1}} \over {4{y_1}}}x + {1 \over {4{y_1}}}$$
<br><br>$$ \Rightarrow $$ 4y<sub>1</sub>y = x<sub>1</sub>x + 1
<br><br>which intersects x axis at A $$\left( {{{ - 1} \over {{x_1}}},0} \right)$$ and y axis at $$B\left( {0,{1 \over {4{y_1}}}} \right)$$
<br><br>Let midpoint of AB is (h, k)
<br><br>$$ \therefore $$ h = $${{ - 1} \over {2{x_1}}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x<sub>1</sub> = $${{ - 1} \over {2h}}$$ & y<sub>1</sub> = $${1 \over {8k}}$$
<br><br>Thus, 4$${\left( {{1 \over {8k}}} \right)^2}$$ = $${\left( {{{ - 1} \over {2h}}} \right)^2}$$ + 1
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${1 \over {16{k^2}}}$$ = $${1 \over {4{h^2}}}$$ + 1
<br><br>$$ \Rightarrow $$ $$\,\,\,$$ 1 = $${{16{k^2}} \over {4{h^2}}}$$ + 16k<sup>2</sup>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ h<sup>2</sup> = 4k<sup>2</sup> + 16h<sup>2</sup> k.
<br><br>So, required equation is
<br><br>x<sup>2</sup> $$-$$ 4y<sup>2</sup> $$-$$ 16x<sup>2</sup> y<sup>2</sup> = 0 | mcq | jee-main-2018-online-15th-april-morning-slot |
eAGN1CCOclwn4x0jehdMT | maths | hyperbola | locus | The locus of the point of intersection of the lines, $$\sqrt 2 x - y + 4\sqrt 2 k = 0$$ and $$\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0$$ (k is any non-zero real parameter), is : | [{"identifier": "A", "content": "an ellipse whose eccentricity is $${1 \\over {\\sqrt 3 }}.$$"}, {"identifier": "B", "content": "an ellipse with length of its major axis $$8\\sqrt 2 .$$"}, {"identifier": "C", "content": "a hyperbola whose eccentricity is $$\\sqrt 3 .$$"}, {"identifier": "D", "content": "a hyperbola with length of its transverse axis $$8\\sqrt 2 .$$"}] | ["D"] | null | Here, lines are :
<br><br>$$\sqrt 2 x$$ $$-$$ y + 4$$\sqrt 2 k$$ = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....$$(i)
<br><br>and $$\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)$$
<br><br>Put the value of y from (i) in (ii) we get;
<br><br>$$ \Rightarrow $$2$$\sqrt 2 $$kx + 4$$\sqrt 2 $$(k<sup>2</sup> $$-$$ 1) = 0
<br><br>$$ \Rightarrow $$ x = $${{2\left( {1 - {k^2}} \right)} \over k}$$, y = $${{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}$$
<br><br>$$\therefore\,\,\,$$ $${\left( {{y \over {4\sqrt 2 }}} \right)^2} - {\left( {{x \over 4}} \right)^2} = 1$$
<br><br>$$\therefore\,\,\,$$ length of transverse axis
<br><br>2a = 2 $$ \times $$ 4$${\sqrt 2 }$$ = 8$${\sqrt 2 }$$
<br><br>Hence, the locus is a hyperbola with length of its transverse axis equal to 8$${\sqrt 2 }$$ | mcq | jee-main-2018-online-16th-april-morning-slot |
MiEqnyGdBW0sqgoDHLsL2 | maths | hyperbola | locus | A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is : | [{"identifier": "A", "content": "an ellipse"}, {"identifier": "B", "content": "a parabola"}, {"identifier": "C", "content": "a hyperbola"}, {"identifier": "D", "content": "a straight line"}] | ["B"] | null | Let equation of circle is
<br><br>x<sup>2</sup><sup></sup> + y<sup>2</sup> + 2fx + 2fy + e = 0, it passes through (0, 2b)
<br><br>$$ \Rightarrow $$ 0 + 4b<sup>2</sup> + 2g $$ \times $$ 0 + 4f + c = 0
<br><br>$$ \Rightarrow $$ 4b<sup>2</sup> + 4f + c = 0 . . . (i)
<br><br>$$2\sqrt {{g^2} - c} = 4a$$ . . . (ii)
<br><br>g<sup>2</sup> $$-$$ c = 4a<sup>2</sup> $$ \Rightarrow $$ c = $$\left( {{g^2} - 4{a^2}} \right)$$
<br><br>Putting in equation (1)
<br><br>$$ \Rightarrow $$ 4b<sup>2</sup> + 4f + g<sup>2</sup> $$-$$ 4a<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + 4y + 4(b<sup>2</sup> $$-$$ a<sup>2</sup>) = 0, it represent a hyperbola. | mcq | jee-main-2019-online-11th-january-evening-slot |
mhXzw5ULbGC8X7A7vQ1kls5f8bf | maths | hyperbola | locus | The locus of the point of intersection of the lines $$\left( {\sqrt 3 } \right)kx + ky - 4\sqrt 3 = 0$$ and $$\sqrt 3 x - y - 4\left( {\sqrt 3 } \right)k = 0$$ is a conic, whose eccentricity is _________. | [] | null | 2 | $$\sqrt 3 kx + ky = 4\sqrt 3 $$ ........(1)<br><br>$$\sqrt 3 kx - ky = 4\sqrt 3 {k^2}$$ ....... (2)<br><br>Adding equation (1) & (2)<br><br>$$2\sqrt 3 kx = 4\sqrt 3 ({k^2} + 1)$$<br><br>$$x = 2\left( {k + {1 \over k}} \right)$$ ......... (3)<br><br>Substracting equation (1) & (2)<br><br>$$y = 2\sqrt 3 \left( {{1 \over k} - k} \right)$$ ........(4)<br><br>$$\therefore$$ $${{{x^2}} \over 4} - {{{y^2}} \over {12}} = 4$$<br><br>$${{{x^2}} \over {16}} - {{{y^2}} \over {48}} = 1$$ (Hyperbola)<br><br>$$ \therefore $$ $${e^2} = 1 + {{48} \over {16}}$$<br><br>$$ \Rightarrow $$ $$e = 2$$ | integer | jee-main-2021-online-25th-february-morning-slot |
AvQIY4RyhtIXgia6Mp1kmhx9otb | maths | hyperbola | locus | The locus of the midpoints of the chord of the circle, x<sup>2</sup> + y<sup>2</sup> = 25 which is tangent to the hyperbola, $${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$ is : | [{"identifier": "A", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $$-$$ 9x<sup>2</sup> + 16y<sup>2</sup> = 0"}, {"identifier": "B", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $$-$$ 9x<sup>2</sup> + 144y<sup>2</sup> = 0"}, {"identifier": "C", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $$-$$ 16x<sup>2</sup> + 9y<sup>2</sup> = 0"}, {"identifier": "D", "content": "(x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> $$-$$ 9x<sup>2</sup> $$-$$ 16y<sup>2</sup> = 0"}] | ["A"] | null | tangent of hyperbola<br><br>$$y = mx \pm \sqrt {9{m^2} - 16} $$ ..... (i)<br><br>which is a chord of circle with mid-point (h, k)<br><br>so equation of chord T = S<sub>1</sub><br><br>hx + ky = h<sup>2</sup> + k<sup>2</sup><br><br>$$y = - {{hx} \over k} + {{{h^2} + {k^2}} \over k}$$ ..... (ii)<br><br>by (i) and (ii)<br><br>$$m = - {h \over k}$$ and $$\sqrt {9{m^2} - 16} $$ = $${{{h^2} + {k^2}} \over k}$$<br><br>$$9{{{h^2}} \over {{k^2}}} - 16 = {{{{\left( {{h^2} + {k^2}} \right)}^2}} \over {{k^2}}}$$<br><br>$$ \therefore $$ Locus : 9x<sup>2</sup> $$-$$ 16y<sup>2</sup> = (x<sup>2</sup> + y<sup>2</sup>)<sup>2</sup> | mcq | jee-main-2021-online-16th-march-morning-shift |
1krvscm6d | maths | hyperbola | locus | The locus of the centroid of the triangle formed by any point P on the hyperbola $$16{x^2} - 9{y^2} + 32x + 36y - 164 = 0$$, and its foci is : | [{"identifier": "A", "content": "$$16{x^2} - 9{y^2} + 32x + 36y - 36 = 0$$"}, {"identifier": "B", "content": "$$9{x^2} - 16{y^2} + 36x + 32y - 144 = 0$$"}, {"identifier": "C", "content": "$$16{x^2} - 9{y^2} + 32x + 36y - 144 = 0$$"}, {"identifier": "D", "content": "$$9{x^2} - 16{y^2} + 36x + 32y - 36 = 0$$"}] | ["A"] | null | Given hyperbola is <br><br>$$16{(x + 1)^2} - 9{(y - 2)^2} = 164 + 16 - 36 = 144$$<br><br>$$ \Rightarrow {{{{(x + 1)}^2}} \over 9} - {{{{(y - 2)}^2}} \over {16}} = 1$$<br><br>Eccentricity, $$e = \sqrt {1 + {{16} \over 9}} = {5 \over 3}$$<br><br>$$\Rightarrow$$ foci are (4, 2) and ($$-$$6, 2)<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267464/exam_images/h22mhjtns4c9oikrrw1r.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Hyperbola Question 39 English Explanation"><br>Let the centroid be (h, k) & A($$\alpha$$, $$\beta$$) be point on hyperbola.<br><br>So, $$h = {{\alpha - 6 + 4} \over 3},k = {{\beta + 2 + 2} \over 3}$$<br><br>$$ \Rightarrow \alpha = 3h + 2,\beta = 3k - 4$$<br><br>($$\alpha$$, $$\beta$$) lies on hyperbola so<br><br>$$16{(3h + 2 + 1)^2} - 9{(3k - 4 - 2)^2} = 144$$<br><br>$$ \Rightarrow 144{(h + 1)^2} - 81{(k - 2)^2} = 144$$<br><br>$$ \Rightarrow 16({h^2} + 2h + 1) - 9({k^2} - 4k + 4) = 16$$<br><br>$$ \Rightarrow 16{x^2} - 9{y^2} + 32x + 36y - 36 = 0$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1ktd0j2lu | maths | hyperbola | locus | The locus of the mid points of the chords of the hyperbola x<sup>2</sup> $$-$$ y<sup>2</sup> = 4, which touch the parabola y<sup>2</sup> = 8x, is : | [{"identifier": "A", "content": "y<sup>3</sup>(x $$-$$ 2) = x<sup>2</sup>"}, {"identifier": "B", "content": "x<sup>3</sup>(x $$-$$ 2) = y<sup>2</sup>"}, {"identifier": "C", "content": "y<sup>2</sup>(x $$-$$ 2) = x<sup>3</sup>"}, {"identifier": "D", "content": "x<sup>2</sup>(x $$-$$ 2) = y<sup>3</sup>"}] | ["C"] | null | T = S<sub>1</sub><br><br>xh $$-$$ yk = h<sup>2</sup> $$-$$ k<sup>2</sup><br><br>$$y = {{xh} \over k} - {{({h^2} - {k^2})} \over k}$$<br><br>this touches y<sup>2</sup> = 8x then $$c = {a \over m}$$<br><br>$$\left( {{{{k^2} - {h^2}} \over k}} \right) = {{2k} \over h}$$<br><br>2y<sup>2</sup> = x(y<sup>2</sup> $$-$$ x<sup>2</sup>)<br><br>y<sup>2</sup>(x $$-$$ 2) = x<sup>3</sup> | mcq | jee-main-2021-online-26th-august-evening-shift |
1ldu5514a | maths | hyperbola | locus | <p>Let T and C respectively be the transverse and conjugate axes of the hyperbola $$16{x^2} - {y^2} + 64x + 4y + 44 = 0$$. Then the area of the region above the parabola $${x^2} = y + 4$$, below the transverse axis T and on the right of the conjugate axis C is :</p> | [{"identifier": "A", "content": "$$4\\sqrt 6 - {{28} \\over 3}$$"}, {"identifier": "B", "content": "$$4\\sqrt 6 - {{44} \\over 3}$$"}, {"identifier": "C", "content": "$$4\\sqrt 6 + {{28} \\over 3}$$"}, {"identifier": "D", "content": "$$4\\sqrt 6 + {{44} \\over 3}$$"}] | ["C"] | null | $16(x+2)^{2}-(y-2)^{2}=16$
<br><br>
$$
\frac{(x+2)^{2}}{1}-\frac{(y-2)^{2}}{16}=1
$$
<br><br>
TA $: y=2$
<br><br>
$\mathrm{CA}: x=-2$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef56ked/6c51cd9e-287f-4e86-804d-923bd55624d6/ab431f50-b264-11ed-a126-5dfa1a9d5fb8/file-1lef56kee.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lef56ked/6c51cd9e-287f-4e86-804d-923bd55624d6/ab431f50-b264-11ed-a126-5dfa1a9d5fb8/file-1lef56kee.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Evening Shift Mathematics - Hyperbola Question 18 English Explanation"><br>
$$
A=\left|\int_{-2}^{\sqrt{6}}\left(2-\left(x^{2}-4\right)\right) d x\right|
$$
<br><br>
$$
\begin{aligned}
& =6 x-\left.\frac{x^{3}}{3}\right|_{-2} ^{\sqrt{6}} \\\\
& =\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)
\end{aligned}
$$
<br><br>
$$
=\frac{12 \sqrt{6}}{3}+\frac{28}{3}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
1lgutulh7 | maths | hyperbola | locus | <p>Let R be a rectangle given by the lines $$x=0, x=2, y=0$$ and $$y=5$$. Let A$$(\alpha,0)$$ and B$$(0,\beta),\alpha\in[0,2]$$ and $$\beta\in[0,5]$$, be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the mid-point of AB lies on a :</p> | [{"identifier": "A", "content": "hyperbola"}, {"identifier": "B", "content": "straight line"}, {"identifier": "C", "content": "parabola"}, {"identifier": "D", "content": "circle"}] | ["A"] | null | We have, $R$ be a rectangle formed by the lines $x=0$, $x=2, y=0$ and $y=5$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln228t36/fd6163aa-9150-4285-8e02-935227613d64/f02c4220-5d60-11ee-8969-dbcde4c067b7/file-6y3zli1ln228t37.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ln228t36/fd6163aa-9150-4285-8e02-935227613d64/f02c4220-5d60-11ee-8969-dbcde4c067b7/file-6y3zli1ln228t37.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Hyperbola Question 14 English Explanation">
<br><br>Let $P Q R S$ be the rectangle such that
<br><br>$$
P(0,0), Q(2,0), R(2,5) \text { and } S(0,5)
$$
<br><br>$P Q=S R=2$ units and $P S=Q R=5$ units
<br><br>$\therefore$ Area of rectangle $P Q R S=(2 \times 5) \mathrm{sq}$ units $=10 \mathrm{sq}$ units
<br><br>Area of $\triangle P A B=\frac{1}{5} \times$ Area of rectangle
<br><br>$$
\begin{aligned}
\frac{1}{2} \times \alpha \times \beta & =\frac{1}{5} \times 10 \text { sq units } \\\\
& =2 \text { sq units } \\\\
\frac{1}{2} \times \alpha \times \beta & =2 \\\\
\Rightarrow \alpha \beta & =4 \text { sq units } .........(i)
\end{aligned}
$$
<br><br>Let $M(h, k)$ be the mid-point of $A B$.
<br><br>Here, $h=\frac{\alpha}{2}, k=\frac{\beta}{2}$
<br><br>or $\alpha=2 h$ and $\beta=2 k$
<br><br>On substitute values of $\alpha$ and $\beta$ in Eq. (i), we get
<br><br>$$
\begin{aligned}
\quad(2 h)(2 k) & =4 \\\\
\Rightarrow h k & =1 \\\\
\therefore \text { Locus of } M \text { is } x y & =1 \text {, which is a hyperbola. }
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-morning-shift |
ZI8EaNj1ctW4fj8P | maths | hyperbola | normal-to-hyperbola | The normal to a curve at $$P(x,y)$$ meets the $$x$$-axis at $$G$$. If the distance of $$G$$ from the origin is twice the abscissa of $$P$$, then the curve is a : | [{"identifier": "A", "content": "circle "}, {"identifier": "B", "content": "hyperbola "}, {"identifier": "C", "content": "ellipse "}, {"identifier": "D", "content": "parabola"}] | ["B"] | null | Equation of normal at $$P\left( {x,y} \right)$$ is $$Y - y = - {{dx} \over {dy}}\left( {x - x} \right)$$
<br><br>Coordinate of $$G$$ at $$X$$ axis is $$\left( {X,0} \right)$$ (let)
<br><br>$$\therefore$$ $$0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x$$
<br><br>$$ \Rightarrow X = x + y{{dy} \over {dx}}$$ $$\therefore$$ Co-ordinate of $$G\left( {x + y{{dy} \over {dx}},0} \right)$$
<br><br>Given distance of $$G$$ from origin $$=$$ twice of the abscissa of $$P.$$
<br><br>as distance cannot be $$-ve,$$ therefore abscissa $$x$$ should be $$+ve$$
<br><br>$$\therefore$$ $$x + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx$$
<br><br>On Integrating $$ \Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}$$
<br><br>$$\therefore$$ the curve is a hyperbola | mcq | aieee-2007 |
2IuA3TuSpLnwUxoO8Vx2H | maths | hyperbola | normal-to-hyperbola | A normal to the hyperbola, 4x<sup>2</sup> $$-$$ 9y<sup>2</sup> = 36 meets the co-ordinate axes $$x$$ and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the ocus of P is : | [{"identifier": "A", "content": "4x<sup>2</sup> + 9y<sup>2</sup> = 121"}, {"identifier": "B", "content": "9x<sup>2</sup> + 4y<sup>2</sup> = 169"}, {"identifier": "C", "content": "4x<sup>2</sup> $$-$$ 9y<sup>2</sup> = 121"}, {"identifier": "D", "content": "9x<sup>2</sup> $$-$$ 4y<sup>2</sup> = 169"}] | ["D"] | null | Given, 4x<sup>2</sup> $$-$$ 9y<sup>2</sup> = 36
<br><br>After differentiating w.r.t.x, we get
<br><br>4.2x $$-$$ 9.2.y.$${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ Slope of tangent = $${{dy} \over {dx}}$$ = $${{4x} \over {9y}}$$
<br><br>So, slope of normal = $${{ - 9y} \over {4x}}$$
<br><br>Now, equation of normal at point (x<sub>0</sub>, y<sub>0</sub>) is given by
<br><br>y $$-$$ y<sub>0</sub> = $${{ - 9{y_0}} \over {4{x_0}}}$$ (x $$-$$ x<sub>0</sub>)
<br><br>As normal intersects X axis at A, Then
<br><br>A = $$\left( {{{13{x_0}} \over 9},0} \right)$$ and B $$ \equiv $$ $$\left( {0,{{13{y_0}} \over 4}} \right)$$
<br><br>As OABP is parallelogram
<br><br>$$ \therefore $$ midpoint of OB $$ \equiv $$ $$\left( {0,{{13{y_0}} \over 8}} \right)$$ $$ \equiv $$ Midpoint of AP
<br><br>So, P(x, y) $$ \equiv $$ $$\left( {{{ - 13{x_0}} \over 9},{{13{y_0}} \over 4}} \right)$$ ...(i)
<br><br>$$ \because $$ (x<sub>0</sub>, y<sub>0</sub>) lies on hyperbola, therefore
<br><br>4(x<sub>0</sub>)<sup>2</sup> $$-$$ 9(y<sub>0</sub>)<sup>2</sup> = 36
<br><br>From equation (i) : x<sub>0</sub> = $${{ - 9x} \over {13}}$$ and y<sub>0</sub> = $${{4y} \over {13}}$$
<br><br>From equation (ii), we get
<br><br>9x<sup>2</sup> $$-$$ 4y<sup>2</sup> = 169
<br><br>Hence, locus of point P is : 9x<sup>2</sup> $$-$$ 4y<sup>2</sup> = 169 | mcq | jee-main-2018-online-15th-april-evening-slot |
0oVM7UCfTdYqh9poTc18hoxe66ijvwpdyef | maths | hyperbola | normal-to-hyperbola | If the line y = mx + 7$$\sqrt 3 $$ is normal to the
hyperbola
$${{{x^2}} \over {24}} - {{{y^2}} \over {18}} = 1$$ , then a value of m is : | [{"identifier": "A", "content": "$${3 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${{\\sqrt 15 } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt 5 } \\over 2}$$"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 5 }}$$"}] | ["D"] | null | Given line y = mx + 7$$\sqrt 3 $$ .....(1)
<br><br>Given hyperbola
<br><br>$${{{x^2}} \over {24}} - {{{y^2}} \over {18}} = 1$$
<br><br>Here $${a^2} = 24$$ and $${b^2} = 18$$
<br><br>We know the equation of normal to the hyperbola is
<br><br>$$y = mx \pm {{m\left( {{a^2} + {b^2}} \right)} \over {\sqrt {{a^2} - {b^2}{m^2}} }}$$
<br><br>$$ \Rightarrow $$ $$y = mx \pm {{m\left( {42} \right)} \over {\sqrt {24 - 18{m^2}} }}$$ .....(2)
<br><br>Comparing (1) and (2), we get
<br><br>$${{m\left( {42} \right)} \over {\sqrt {24 - 18{m^2}} }}$$ = $$7\sqrt 3 $$
<br><br>$$ \Rightarrow $$ 36m<sup>2</sup> = 72 - 54m<sup>2</sup>
<br><br>$$ \Rightarrow $$ 90m<sup>2</sup> = 72
<br><br>$$ \Rightarrow $$ m<sup>2</sup> = $${{72} \over {90}}$$
<br><br>$$ \Rightarrow $$ m = $${2 \over {\sqrt 5 }}$$ | mcq | jee-main-2019-online-9th-april-morning-slot |
tTDEM7hQQr5lkaZQzQ7k9k2k5hjxu5z | maths | hyperbola | normal-to-hyperbola | If a hyperbola passes through the point
P(10, 16) and it has vertices at (± 6, 0), then the
equation of the normal to it at P is : | [{"identifier": "A", "content": "2x + 5y = 100"}, {"identifier": "B", "content": "x + 3y = 58"}, {"identifier": "C", "content": "x + 2y = 42"}, {"identifier": "D", "content": "3x + 4y = 94"}] | ["A"] | null | Let hyperbola is $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
<br><br>vertices ($$ \pm $$a, 0 ) = ($$ \pm $$6, 0) $$ \Rightarrow $$ a = 6
<br><br>Hyperbola passes through p(10, 16)
<br><br>$$ \therefore $$ $${{{{10}^2}} \over {{6^2}}} - {{{{16}^2}} \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ b = 12
<br><br>$$ \therefore $$ Required hyperbola is $${{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1$$
<br><br>Equation of normal will be
<br><br>$${{{a^2}x} \over {{x_1}}} + {{{b^2}y} \over {{y_1}}} = {a^2} + {b^2}$$
<br><br>At P(10,16) normal is
<br><br>$${{36x} \over {10}} + {{144y} \over {16}} = 36 + 144$$
<br><br>$$ \Rightarrow $$ 18x + 45y = 900
<br><br>$$ \Rightarrow $$ 2x + 5y = 100 | mcq | jee-main-2020-online-8th-january-evening-slot |
qmELQCfVJYJt8ZkIAejgy2xukf7fotkp | maths | hyperbola | normal-to-hyperbola | Let P(3, 3) be a point on the hyperbola, <br/>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. If the normal to it at P intersects the x-axis
at (9, 0) and e is its eccentricity, then the ordered pair (a<sup>2</sup>, e<sup>2</sup>) is equal to : | [{"identifier": "A", "content": "$$\\left( {{9 \\over 2},2} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{3 \\over 2},2} \\right)$$"}, {"identifier": "C", "content": "(9,3) "}, {"identifier": "D", "content": "$$\\left( {{9 \\over 2},3} \\right)$$"}] | ["D"] | null | Given hyperbola, $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$<br><br>Point P (3, 3) is on the parabola<br><br>$$ \therefore $$ $${9 \over {{a^2}}} - {9 \over {{b^2}}} = 1$$ ...(1)<br><br>Equation of normal at (x<sub>1</sub>, y<sub>1</sub>),<br><br>$${{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2}{e^2}$$<br><br>Normal at p(3, 3),<br><br>$${{{a^2}x} \over 3} - {{{b^2}y} \over 3} = {a^2}{e^2}$$ ... (2)<br><br>It intersect x-axis at (9, 0),<br><br>Putting in equation (2),<br><br>$${{9{a^2}} \over 3} - 0 = {a^2}{e^2}$$<br><br>$$ \Rightarrow 3{a^2} = {a^2}{e^2}$$<br><br>$$ \Rightarrow {e^2} = 3$$<br><br>Also, $${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$<br><br>$$ \Rightarrow 3 = 1 + {{{b^2}} \over {{a^2}}}$$<br><br>$$ \Rightarrow {b^2} = 2{a^2}$$<br><br>Putting the value of b<sup>2</sup> in equation (1),<br><br>$${9 \over {{a^2}}} - {9 \over {2{a^2}}} = 1$$<br><br>$$ \Rightarrow {a^2} = {9 \over 2}$$<br><br>$$ \therefore $$ $$({a^2},{e^2}) = \left( {{9 \over 2},3} \right)$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
1ktcyu7x8 | maths | hyperbola | normal-to-hyperbola | The point $$P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$$ lies on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ having eccentricity $${{\sqrt 5 } \over 2}$$. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to : | [{"identifier": "A", "content": "$$4\\sqrt 3 $$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$$6\\sqrt 3 $$"}, {"identifier": "D", "content": "$$3\\sqrt 6 $$"}] | ["C"] | null | $$P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)$$ lies on hyperbola <br><br>$$ \Rightarrow {{24} \over {{a^2}}} - {3 \over {{b^2}}} = 1$$ ...... (i)<br><br>$$e = {{\sqrt 5 } \over 2} \Rightarrow {b^2} = {a^2}\left( {{5 \over 4} - 1} \right) \Rightarrow 4{b^2} = {a^2}$$<br><br>Put in (i) $$ \Rightarrow {6 \over {{b^2}}} - {3 \over {{b^2}}} = 1 \Rightarrow b = \sqrt 3 $$<br><br>$$ \Rightarrow a = \sqrt {12} $$<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265061/exam_images/w1lkvrlkm2a0iclvpvya.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Evening Shift Mathematics - Hyperbola Question 38 English Explanation"><br>Tangent at P :<br><br>$${{ - x} \over {\sqrt 6 }} - {y \over {\sqrt 3 }} = 1 \Rightarrow Q(0,\sqrt 3 )$$<br><br>Slope of $$T = - {1 \over {\sqrt 2 }}$$<br><br>Normal at P : <br><br>$$y - \sqrt 3 = \sqrt 2 (x + 2\sqrt 6 )$$<br><br>$$ \Rightarrow R = (0,5\sqrt 3 )$$<br><br>$$QR = 6\sqrt 3 $$ | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktgpapht | maths | hyperbola | normal-to-hyperbola | Let A (sec$$\theta$$, 2tan$$\theta$$) and B (sec$$\phi$$, 2tan$$\phi$$), where $$\theta$$ + $$\phi$$ = $$\pi$$/2, be two points on the hyperbola 2x<sup>2</sup> $$-$$ y<sup>2</sup> = 2. If ($$\alpha$$, $$\beta$$) is the point of the intersection of the normals to the hyperbola at A and B, then (2$$\beta$$)<sup>2</sup> is equal to ____________. | [] | null | Bonus | Since, point A (sec$$\theta$$, 2tan$$\theta$$) lies on the hyperbola 2x<sup>2</sup> $$-$$ y<sup>2</sup> = 2<br><br>Therefore, 2sec<sup>2</sup>$$\theta$$ $$-$$ 4tan<sup>2</sup>$$\theta$$ = 2<br><br>$$\Rightarrow$$ 2 + 2tan<sup>2</sup>$$\theta$$ $$-$$ 4tan<sup>2</sup>$$\theta$$ = 2<br><br>$$\Rightarrow$$ tan$$\theta$$ = 0 $$\Rightarrow$$ $$\theta$$ = 0<br><br>Similarly, for point B, we will get $$\phi$$ = 0.<br><br>but according to question $$\theta$$ + $$\phi$$ = $${\pi \over 2}$$ which is not possible.<br><br>Hence, it must be a 'BONUS'. | integer | jee-main-2021-online-27th-august-evening-shift |
1l58g2o6j | maths | hyperbola | normal-to-hyperbola | <p>The normal to the hyperbola <br/><br/>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$$ at the point $$\left( {8,3\sqrt 3 } \right)$$ on it passes through the point :</p> | [{"identifier": "A", "content": "$$\\left( {15, - 2\\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {9,2\\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 1,9\\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - 1,6\\sqrt 3 } \\right)$$"}] | ["C"] | null | <p>Given hyperbola : $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over 9} = 1$$</p>
<p>$$\because$$ It passes through $$(8,3\sqrt 3 )$$</p>
<p>$$\because$$ $${{64} \over {{a^2}}} - {{27} \over 9} = 1 \Rightarrow {a^2} = 16$$</p>
<p>Now, equation of normal to hyperbola</p>
<p>$${{16x} \over 8} + {{9y} \over {3\sqrt 3 }} = 16 + 9$$</p>
<p>$$ \Rightarrow 2x + \sqrt 3 y = 25$$ ...... (i)</p>
<p>$$\left( { - 1,9\sqrt 3 } \right)$$ satisfies (i)</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59lgdx4 | maths | hyperbola | normal-to-hyperbola | <p>Let the eccentricity of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be $${5 \over 4}$$. If the equation of the normal at the point $$\left( {{8 \over {\sqrt {5} }},{{12} \over {5}}} \right)$$ on the hyperbola is $$8\sqrt 5 x + \beta y = \lambda $$, then $$\lambda$$ $$-$$ $$\beta$$ is equal to ___________.</p> | [] | null | 85 | <p>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1\left( {e = {5 \over 4}} \right)$$</p>
<p>So, $${b^2} = {a^2}\left( {{{25} \over {16}} - 1} \right) \Rightarrow b = {3 \over 4}a$$</p>
<p>Also $$\left( {{8 \over {\sqrt 5 }},{{12} \over 5}} \right)$$ lies on the given hyperbola</p>
<p>So, $${{64} \over {5{a^2}}} - {{144} \over {25\left( {{{9{a^2}} \over {16}}} \right)}} = 1 \Rightarrow a = {8 \over 5}$$ and $$b = {6 \over 5}$$</p>
<p>Equation of normal</p>
<p>$${{64} \over {25}}\left( {{x \over {{8 \over {\sqrt 5 }}}}} \right) + {{36} \over {25}}\left( {{y \over {{{12} \over 5}}}} \right) = 4$$</p>
<p>$$ \Rightarrow {8 \over {5\sqrt 5 }}x + {3 \over 5}y = 4$$</p>
<p>$$ \Rightarrow 8\sqrt 5 x + 15y = 100$$</p>
<p>So, $$\beta$$ = 15 and $$\lambda$$ = 100</p>
<p>Gives $$\lambda$$ $$-$$ $$\beta$$ = 85</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l6gim5lt | maths | hyperbola | normal-to-hyperbola | <p>Let the tangent drawn to the parabola $$y^{2}=24 x$$ at the point $$(\alpha, \beta)$$ is perpendicular to the line $$2 x+2 y=5$$. Then the normal to the hyperbola $$\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$$ at the point $$(\alpha+4, \beta+4)$$ does NOT pass through the point :</p> | [{"identifier": "A", "content": "(25, 10)"}, {"identifier": "B", "content": "(20, 12)"}, {"identifier": "C", "content": "(30, 8)"}, {"identifier": "D", "content": "(15, 13)"}] | ["D"] | null | <p>Any tangent to $${y^2} = 24x$$ at ($$\alpha$$, $$\beta$$)</p>
<p>$$\beta y = 12(x + \alpha )$$</p>
<p>Slope $$ = {{12} \over \beta }$$ and perpendicular to $$2x + 2y = 5$$</p>
<p>$$ \Rightarrow {{12} \over \beta } = 1 \Rightarrow \beta = 12,\,\alpha = 6$$</p>
<p>Hence hyperbola is $${{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1$$ and normal is drawn at (10, 16)</p>
<p>Equation of normal $${{36\,.\,x} \over {10}} + {{144\,.\,y} \over {16}} = 36 + 144$$</p>
<p>$$ \Rightarrow {x \over {50}} + {y \over {20}} = 1$$</p>
<p>This does not pass though (15, 13) out of given option.</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1ldv2wg2b | maths | hyperbola | normal-to-hyperbola | <p>The vertices of a hyperbola H are ($$\pm$$ 6, 0) and its eccentricity is $${{\sqrt 5 } \over 2}$$. Let N be the normal to H at a point in the first quadrant and parallel to the line $$\sqrt 2 x + y = 2\sqrt 2 $$. If d is the length of the line segment of N between H and the y-axis then d$$^2$$ is equal to _____________.</p> | [] | null | 216 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lebzl8r1/919fa76e-b803-49d8-86c9-4e9df9241622/72ada3d0-b0a8-11ed-98af-57b0199700ba/file-1lebzl8r2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lebzl8r1/919fa76e-b803-49d8-86c9-4e9df9241622/72ada3d0-b0a8-11ed-98af-57b0199700ba/file-1lebzl8r2.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 25th January Morning Shift Mathematics - Hyperbola Question 17 English Explanation"><br>
$$
H: \frac{x^2}{36}-\frac{y^2}{9}=1
$$<br><br>
equation of normal is $6 x \cos \theta+3 y \cot \theta=45$<br><br>
$$
\begin{aligned}
& \text { slope }=-2 \sin \theta=-\sqrt{2} \\\\
& \Rightarrow \theta=\frac{\pi}{4}
\end{aligned}
$$<br><br>
Equation of normal is $\sqrt{2} x+y=15$<br><br>
$P:(a \sec \theta, b \tan \theta)$<br><br>
$\Rightarrow \mathrm{P}(6 \sqrt{2}, 3)$ and $\mathrm{K}(0,15)$<br><br>
$$
d^2=216
$$ | integer | jee-main-2023-online-25th-january-morning-shift |
jaoe38c1lscn7o5t | maths | hyperbola | position-of-point-and-chord-joining-of-two-points | <p>Let $$e_1$$ be the eccentricity of the hyperbola $$\frac{x^2}{16}-\frac{y^2}{9}=1$$ and $$e_2$$ be the eccentricity of the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \mathrm{a} > \mathrm{b}$$, which passes through the foci of the hyperbola. If $$\mathrm{e}_1 \mathrm{e}_2=1$$, then the length of the chord of the ellipse parallel to the $$x$$-axis and passing through $$(0,2)$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{8 \\sqrt{5}}{3}$$\n"}, {"identifier": "B", "content": "$$3 \\sqrt{5}$$\n"}, {"identifier": "C", "content": "$$4 \\sqrt{5}$$\n"}, {"identifier": "D", "content": "$$\\frac{10 \\sqrt{5}}{3}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& H: \frac{x^2}{16}-\frac{y^2}{9}=1 \qquad e_1=\frac{5}{4} \\
& \therefore e_1 e_2=1 \Rightarrow e_2=\frac{4}{5}
\end{aligned}$$</p>
<p>Also, ellipse is passing through $$( \pm 5,0)$$</p>
<p>$$\begin{aligned}
& \therefore a=5 \text { and } b=3 \\
& E: \frac{x^2}{25}+\frac{y^2}{9}=1
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1vjyru/5d14555c-4b38-417d-b959-37b3238917a8/12d5a2a0-d410-11ee-b9d5-0585032231f0/file-1lt1vjyrv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1vjyru/5d14555c-4b38-417d-b959-37b3238917a8/12d5a2a0-d410-11ee-b9d5-0585032231f0/file-1lt1vjyrv.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Hyperbola Question 10 English Explanation"></p>
<p>End point of chord are $$\left( \pm \frac{5 \sqrt{5}}{3}, 2\right)$$</p>
<p>$$\therefore \mathrm{L}_{\mathrm{PQ}}=\frac{10 \sqrt{5}}{3}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
p6VqpFyaPpNleFFg | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | The foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1$$ and the hyperbola $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$ coincide. Then the value of $${b^2}$$ is : | [{"identifier": "A", "content": "$$9$$"}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$5$$ "}, {"identifier": "D", "content": "$$7$$ "}] | ["D"] | null | $${{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}$$
<br><br>$$a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,$$
<br><br>$$e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}$$
<br><br>$$\therefore$$ Foci $$ = \left( { \pm 3,0} \right)$$
<br><br>$$\therefore$$ foci of ellipse $$=$$ foci of hyperbola
<br><br>$$\therefore$$ for ellipse $$ae=3$$ but $$a=4,$$
<br><br>$$\therefore$$ $$e = {3 \over 4}$$
<br><br>Then $${b^2} = {a^2}\left( {1 - {e^2}} \right)$$
<br><br>$$ \Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7$$ | mcq | aieee-2003 |
Yyy5G3cAbrt43UGL | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | For the Hyperbola $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$ , which of the following remains constant when $$\alpha $$ varies$$=$$? | [{"identifier": "A", "content": "abscissae of vertices "}, {"identifier": "B", "content": "abscissae of foci "}, {"identifier": "C", "content": "eccentricity "}, {"identifier": "D", "content": "directrix."}] | ["B"] | null | Given, equation of hyperbola is $${{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1$$
<br><br>We know that the equation of hyperbola is
<br><br>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ Here, $${a^2} = {\cos ^2}\alpha $$ and $${b^2} = {\sin ^2}\alpha $$
<br><br>We know that, $${b^2} = {a^2}\left( {{e^2} - 1} \right)$$
<br><br>$$ \Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)$$
<br><br>$$ \Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}$$
<br><br>$$ \Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha $$
<br><br>$$\therefore$$ $$ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1$$
<br><br>Co-ordinate of foci are $$\left( { \pm \alpha e,0} \right)\,\,$$ i.e. $$\left( { \pm 1,0} \right)$$
<br><br>Hence, abscissae of foci remain constant when $$\alpha $$ varies. | mcq | aieee-2007 |
LPJEFjsEwH1i9kZv | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | The eccentricity of the hyperbola whose length of the latus rectum is equal to $$8$$ and the length of its conjugate axis is equal to half of the distance between its foci, is : | [{"identifier": "A", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${\\sqrt 3 }$$ "}, {"identifier": "C", "content": "$${{4 \\over 3}}$$"}, {"identifier": "D", "content": "$${4 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | $${{2{b^2}} \over a} = 8$$ and $$2b = {1 \over 2}\left( {2ae} \right)$$
<br><br>$$ \Rightarrow 4{b^2} = {a^2}{e^2}$$
<br><br>$$ \Rightarrow 4{a^2}\left( {{e^2} - 1} \right) = {a^2}{e^2}$$
<br><br>$$ \Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}$$ | mcq | jee-main-2016-offline |
8CXP96he2JPl6kqXEpCMi | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | Let a and b respectively be the semitransverse and semi-conjugate axes of a
hyperbola whose eccentricity satisfies the equation 9e<sup>2</sup> − 18e + 5 = 0. If S(5, 0) is a focus and 5x = 9 is the corresponding directrix of this hyperbola, then a<sup>2</sup> − b<sup>2</sup> is equal to :
| [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "$$-$$ 7"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$-$$ 5"}] | ["B"] | null | As S(5, 0) is the focus.
<br><br>$$ \therefore $$ ae = 5 . . . (1)
<br><br>As 5x = 9
<br><br>$$ \therefore $$ x = $${9 \over 5}$$ is the directrix
<br><br>As we know directrix = $${a \over e}$$
<br><br>$$ \therefore $$ $${a \over e} = {9 \over 5}$$ . . . .(2)
<br><br>Solving (1) and (2), we get
<br><br>a = 3 and e = $${5 \over 3}$$
<br><br>As we know,
<br><br>b<sup>2</sup> = a<sup>2</sup> (e<sup>2</sup> $$-$$ 1) = 9$$\left( {{{25} \over 9} - 1} \right)$$ = 16
<br><br>$$ \therefore $$ a<sup>2</sup> $$-$$ b<sup>2</sup> = 9 $$-$$ 16 $$=$$ $$-$$ 7 | mcq | jee-main-2016-online-9th-april-morning-slot |
wTrhrAus7jXGW0OMPsaAA | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | A hyperbola whose transverse axis is along the major axis of the conic, $${{{x^2}} \over 3} + {{{y^2}} \over 4} = 4$$ and has vertices at the foci of this conic. If the eccentricity of the hyperbola is $${3 \over 2},$$ then which of the following points does <b>NOT</b> lie on it? | [{"identifier": "A", "content": "(0, 2)"}, {"identifier": "B", "content": "$$\\left( {\\sqrt 5 ,2\\sqrt 2 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {\\sqrt {10} ,2\\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {5,2\\sqrt 3 } \\right)$$ "}] | ["D"] | null | $${{{x^2}} \over {12}} + {{{y^2}} \over {16}}$$ = 1
<br><br>e = $$\sqrt {1 - {{12} \over {16}}} $$ = $${1 \over 2}$$
<br><br>Foci (0, 2) & (0, $$-$$ 2)
<br><br>So, transverse axis of hyperbola
<br><br>= 2b = 4 <br><br>$$ \Rightarrow $$ b = 2 & a<sup>2</sup> = 1<sup>2</sup> (e<sup>2</sup> $$-$$ 1)
<br><br>$$ \Rightarrow $$ a<sup>2</sup> = 4$$\left( {{9 \over 4} - 1} \right)$$
<br><br>$$ \Rightarrow $$ a<sup>2</sup> = 5
<br><br>$$ \therefore $$ It's equation is $${{{x^2}} \over 5} - {{{y^2}} \over 4}$$ = $$-$$ 1
<br><br>The point (5, 2$$\sqrt 3 $$) does not satisfy the above equation. | mcq | jee-main-2016-online-10th-april-morning-slot |
c5COgvW6rHjsd2dmukCP2 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | Let $$0 < \theta < {\pi \over 2}$$. If the eccentricity of the
<br/><br/> hyperbola $${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}$$ = 1 is greater
<br/><br/>than 2, then the length of
its latus rectum lies in the interval :
| [{"identifier": "A", "content": "(3, $$\\infty $$) "}, {"identifier": "B", "content": "$$\\left( {{3 \\over 2},2} \\right]$$"}, {"identifier": "C", "content": "$$\\left( {1,{3 \\over 2}} \\right]$$"}, {"identifier": "D", "content": "$$\\left( {2,3} \\right]$$"}] | ["A"] | null | Given hyperbola,
<br><br>$${{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1$$
<br><br>here a = cos$$\theta $$
<br><br>and b = sin$$\theta $$
<br><br>We know, eccentricity of the hyperbola is,
<br><br>$$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$
<br><br>$$ \therefore $$ Here eccentricity
<br><br>(e) = $$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} $$
<br><br>Given that,
<br><br>$$\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2$$
<br><br>$$ \Rightarrow $$ $$\sqrt {1 + {{\tan }^2}\theta } > 2$$
<br><br>$$ \Rightarrow $$ 1 + tan<sup>2</sup>$$\theta $$ > 4
<br><br>$$ \Rightarrow $$ tan<sup>2</sup>$$\theta $$ > 3
<br><br>$$ \Rightarrow $$ tan$$\theta $$ > $$ \pm \sqrt 3 $$
<br><br>As given $$\theta $$ $$ \in $$ $$\left( {0,{\pi \over 2}} \right)$$
<br><br>possible value of tan$$\theta $$ > $$\sqrt 3 $$
<br><br>So, $$\theta $$ can be in the range $${\pi \over 3} < \theta < {\pi \over 2}$$
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264616/exam_images/y5muchtddz9gnxydcpim.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Hyperbola Question 63 English Explanation">
<br><br>We know latus ractum (LR) = $${{2{b^2}} \over a}$$
<br><br>$$ \therefore $$ LR = $${{2{{\sin }^2}\theta } \over {\cos \theta }}$$
<br><br>= 2 tan$$\theta $$ sin$$\theta $$
<br><br>We know in the range $${\pi \over 3} < \theta < {\pi \over 2}$$ tan$$\theta $$ and sin$$\theta $$ both are increasing function.
<br><br>So, at $${\pi \over 3}$$ value of LR will be minimum and at $${\pi \over 2}$$ value of LR will be maximum.
<br><br>$$ \therefore $$ Minimum value of LR = 2tan$${\pi \over 3}$$ sin$${\pi \over 3}$$
<br><br>= 2 $$ \times $$ $$\sqrt 3 \times {{\sqrt 3 } \over 2}$$
<br>= 3
<br><br>Maximum value of LR = 2tan$${\pi \over 2}$$ sin$${\pi \over 2}$$
<br><br>= 2$$\left( \infty \right) \times 1$$
<br><br>= $$\infty $$
<br><br>$$ \therefore $$ Interval of LR = (3, $$\infty $$)
| mcq | jee-main-2019-online-9th-january-morning-slot |
kh3oFVUOUj5Bwo7LP4i2O | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is : | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$$\\sqrt 3 $$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$${2 \\over {\\sqrt 3 }}$$"}] | ["D"] | null | Let the equation of hyperbola
<br><br>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}}$$ = 1
<br><br>Given 2a = 4
<br><br>$$ \Rightarrow $$ $$a$$ = 2
<br><br>It passes through (4, 2)
<br><br>$$ \therefore $$ $${{16} \over 4} - {4 \over {{b^2}}}$$ = 1
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = $${4 \over 3}$$
<br><br>e = $$\sqrt {1 + {{{b^2}} \over {{a^2}}}} $$ = $$\sqrt {1 + {{4/3} \over 4}} $$
<br><br>= $$\sqrt {1 + {1 \over 3}} $$ = $${2 \over {\sqrt 3 }}$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
udGNmB5egVI3TF1a5s5Vu | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the
eccentricity of the hyperbola is : | [{"identifier": "A", "content": "$${{13} \\over 6}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${{13} \\over 12}$$"}, {"identifier": "D", "content": "$${{13} \\over 8}$$"}] | ["C"] | null | 2b = 5 and 2ae = 13
<br><br>b<sup>2</sup> = a<sup>2</sup>(e<sup>2</sup> $$-$$ 1) $$ \Rightarrow $$ $${{25} \over 4}$$ = $${{169} \over 4}$$ $$-$$ a<sup>2</sup>
<br><br>$$ \Rightarrow $$ a $$=$$ 6 $$ \Rightarrow $$ e $$=$$ $${{13} \over {12}}$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
7NBZ9LCV3YKefkvN211Fe | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If the vertices of a hyperbola be at (–2, 0) and (2, 0) and one of its foci be at (–3, 0), then which one of the following points does not lie on this hyperbola? | [{"identifier": "A", "content": "$$\\left( {6,5\\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {2\\sqrt 6 ,5} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - 6,2\\sqrt {10} } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {4,\\sqrt {15} } \\right)$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266288/exam_images/sqc6djvwoxgp7yny26un.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Hyperbola Question 57 English Explanation">
<br>ae = 3, e = $${3 \over 2}$$, b<sup>2</sup> = 4$$\left( {{9 \over 4} - 1} \right)$$, b<sup>2</sup> = 5
<br><br>$${{{x^2}} \over 4} - {{{y^2}} \over 5} = 1$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
sCkkVj9Lsm6OEcTclf3rsa0w2w9jwxs6pli | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If a directrix of a hyperbola centred at the origin and passing through the point (4, –2$$\sqrt 3 $$ ) is 5x = 4$$\sqrt 5 $$ and
its eccentricity is e, then : | [{"identifier": "A", "content": "4e<sup>4</sup> \u2013 24e<sup>2</sup> + 27 = 0"}, {"identifier": "B", "content": "4e<sup>4</sup> \u2013 24e<sup>2</sup> + 35 = 0"}, {"identifier": "C", "content": "4e<sup>4</sup> \u2013 12e<sup>2</sup> - 27 = 0"}, {"identifier": "D", "content": "4e<sup>4</sup> + 8e<sup>2</sup> - 35 = 0"}] | ["B"] | null | 5x = 4$$\sqrt 5 $$
<br><br>$$ \Rightarrow $$ $$x = {4 \over {\sqrt 5 }}$$<br><br>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267388/exam_images/fqdlbbxdjjchifyrvnwu.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265977/exam_images/ch6ei3b6khax7yryqclp.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266673/exam_images/nphwuscqpaxgnfxexsnx.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265768/exam_images/pzouu2nlusw3fsyzoegr.webp"><source media="(max-width: 1040px)" srcset="https://imagex.cdn.examgoal.net/jf7697gjy9md3qt/27986382-f5a0-4da5-994f-
/e8a83650-a9df-11e9-a277-6507805b1ccf/file-jf7697gjy9md3qu-1040w.jpg"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267427/exam_images/s7uodmyv0nur1udqxtay.webp"><source media="(max-width: 1400px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264897/exam_images/xowsgrwqtlaiauuz99z9.webp"><source media="(max-width: 1580px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263302/exam_images/kbwjk5emkktzwcdp2myt.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266897/exam_images/bkhe3kwsqinvpy72llmd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Morning Slot Mathematics - Hyperbola Question 54 English Explanation"></picture>
<br>
Equation of hyperbola<br><br>
$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ it passes through $$\left( {4, - 2\sqrt 3 } \right)$$<br><br>
$$ \therefore {e^2} = 1 + {{{b^2}} \over {{a^2}}}$$ <br><br>
$$ \Rightarrow {a^2}{e^2} - {a^2} = {b^2}$$<br><br>
$$ \Rightarrow {{16} \over {{a^2}}} - {{12} \over {{a^2}{e^2} - {a^2}}} = 1$$<br><br>
$$ \Rightarrow {4 \over {{a^2}}}\left[ {{4 \over 1} - {3 \over {{e^2} - 1}}} \right] = 1$$<br><br>
$$ \Rightarrow 4{e^2} - 4 - 3 = ({e^2} - 1)\left( {{{{a^2}} \over 4}} \right)$$<br><br>
$$ \Rightarrow 4(4{e^3} - 7) = ({e^2} - 1){\left( {{{4e} \over {\sqrt 5 }}} \right)^2}$$<br><br>
$$ \Rightarrow 4{e^4} - 24{e^2} + 35 = 0$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
jnsfPBe0KmkT0ZCwSQ3rsa0w2w9jx24zm31 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If 5x + 9 = 0 is the directrix of the hyperbola 16x<sup>2</sup>
– 9y<sup>2</sup>
= 144, then its corresponding focus is : | [{"identifier": "A", "content": "$$\\left( {{5 \\over 3},0} \\right)$$"}, {"identifier": "B", "content": "(5, 0)"}, {"identifier": "C", "content": "(- 5, 0)"}, {"identifier": "D", "content": "$$\\left( { - {5 \\over 3},0} \\right)$$"}] | ["C"] | null | $${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$<br><br>
$$ \therefore $$ a = 3 and b = 4<br><br>
$${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$<br><br>
$$ \Rightarrow {e^2} = 1 + {{16} \over 9}$$<br><br>
$$ \Rightarrow $$ e = $$5 \over 3$$<br><br>
$$ \therefore $$ focus is (–ae, 0) = (–5, 0) | mcq | jee-main-2019-online-10th-april-evening-slot |
GmH9VlfoYhBOK87JlK7k9k2k5it6e9b | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | If e<sub>1</sub> and e<sub>2</sub> are the eccentricities of the ellipse,
$${{{x^2}} \over {18}} + {{{y^2}} \over 4} = 1$$ and the hyperbola, $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ respectively and (e<sub>1</sub>, e<sub>2</sub>) is a point on the ellipse,
15x<sup>2</sup> + 3y<sup>2</sup> = k, then k is equal to : | [{"identifier": "A", "content": "17"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "15"}, {"identifier": "D", "content": "14"}] | ["B"] | null | e<sub>1</sub> = $$\sqrt {1 - {4 \over {18}}} $$ = $${{\sqrt 7 } \over 3}$$
<br><br>e<sub>1</sub> = $$\sqrt {1 + {4 \over 9}} $$ = $${{\sqrt {13} } \over 3}$$
<br><br>$$ \because $$ (e<sub>1</sub>, e<sub>2</sub> ) lies on 15x<sup>2</sup>
+ 3y<sup>2</sup>
= k
<br><br>$$ \therefore $$ $$15\left( {{7 \over 9}} \right) + 3\left( {{{13} \over 9}} \right)$$ = k
<br><br>$$ \Rightarrow $$ k = 16 | mcq | jee-main-2020-online-9th-january-morning-slot |
cKUyQMOGtgxolaNvzyjgy2xukezb2lny | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | For some $$\theta \in \left( {0,{\pi \over 2}} \right)$$, if the eccentricity of the
<br/>hyperbola, x<sup>2</sup>–y<sup>2</sup>sec<sup>2</sup>$$\theta $$ = 10 is
$$\sqrt 5 $$ times the
<br/>eccentricity of the ellipse, x<sup>2</sup>sec<sup>2</sup>$$\theta $$ + y<sup>2</sup> = 5, then
the length of the latus rectum of the ellipse, is : | [{"identifier": "A", "content": "$$\\sqrt {30} $$"}, {"identifier": "B", "content": "$$2\\sqrt 6 $$"}, {"identifier": "C", "content": "$${{4\\sqrt 5 } \\over 3}$$"}, {"identifier": "D", "content": "$${{2\\sqrt 5 } \\over 3}$$"}] | ["C"] | null | Given equation of hyperbola $$ \Rightarrow {x^2} - {y^2}{\sec ^2}\theta = 10$$<br><br>
$$ \Rightarrow {{{x^2}} \over {10}} - {{{y^2}} \over {10{{\cos }^2}\theta }} = 1$$<br><br>
Hence eccentricity of hyperbola<br><br>
$$\left( {{e_H}} \right) = \sqrt {1 + {{10{{\cos }^2}\theta } \over {10}}} $$ ...(i)<br><br>
$$\left\{ { \because \,\, e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} } \right\}$$<br><br>
Now equation of ellipse $$ \Rightarrow {x^2}{\sec ^2}\theta + {y^2} = 5$$<br><br>
$$ \Rightarrow {{{x^2}} \over {5{{\cos }^2}}} + {{{y^2}} \over 5} = 1\,$$ $$\,\left\{ {e = 1 - {{{a^2}} \over {{b^2}}}} \right\}$$<br><br>
Hence eccenticity of ellipse<br><br>
$$\left( {{e_E}} \right) = \sqrt {1 - {{5{{\cos }^2}\theta } \over 5}} $$<br><br>
$$\left( {{e_E}} \right) = \sqrt {1 - {{\cos }^2}\theta } $$ ...(ii)<br><br>
given $$ {e_H} = \sqrt 5 {e_e}$$<br><br>
Hence $$\sqrt {1 + {{\cos }^2}\theta } = \sqrt 5 \times \left( {\sqrt {1 - {{\cos }^2}\theta } } \right)$$<br><br>
Squaring both sides<br><br>
$$1 + {\cos ^2}\theta = 5\left( {1 - {{\cos }^2}\theta } \right)$$<br><br>
$$1 + {\cos ^2}\theta = 5 - 5{\cos ^2}\theta $$<br><br>
$$6{\cos ^2}\theta = 4$$<br><br>
$${\cos ^2}\theta = {2 \over 3}$$ ...(iii)<br><br>
Now length of latus rectum of ellipse = $$ = {{2{a^2}} \over b} = {{10{{\cos }^2}\theta } \over {\sqrt 5 }} = {{20} \over {3\sqrt 5 }} = {{4\sqrt 5 } \over 3}$$
| mcq | jee-main-2020-online-2nd-september-evening-slot |
SAyayGIQQef05hoXTRjgy2xukezmfbtw | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | A hyperbola having the transverse axis of
length
$$\sqrt 2 $$ has the same foci as that of the ellipse
3x<sup>2</sup> + 4y<sup>2</sup> = 12, then this hyperbola does not
pass through which of the following points? | [{"identifier": "A", "content": "$$\\left( {1, - {1 \\over {\\sqrt 2 }}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {\\sqrt {{3 \\over 2}} ,{1 \\over {\\sqrt 2 }}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\sqrt {{3 \\over 2}} ,1} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over {\\sqrt 2 }},0} \\right)$$"}] | ["B"] | null | Ellipse : $${{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$<br><br>eccentricity = $$\sqrt {1 - {3 \over 4}} = {1 \over 2}$$<br><br>$$ \therefore $$ foci = ($$ \pm $$ 1, 0)<br><br>for hyperbola, given $$2a = \sqrt 2 \Rightarrow a = {1 \over {\sqrt 2 }}$$<br><br>$$ \therefore $$ hyperbola will be<br><br>$${{{x^2}} \over {1/2}} - {{{y^2}} \over {{b^2}}} = 1$$<br><br>eccentricity = $$\sqrt {1 + 2{b^2}} $$<br><br>$$ \therefore $$ foci $$ = \left( { \pm \sqrt {{{1 + 2{b^2}} \over 2}} ,0} \right)$$<br><br>$$ \because $$ Ellipse and hyperbola have same foci<br><br>$$ \Rightarrow \sqrt {{{1 + 2{b^2}} \over 2}} = 1$$<br><br>$$ \Rightarrow {b^2} = {1 \over 2}$$<br><br>$$ \therefore $$ Equation of hyperbola : $${{{x^2}} \over {1/2}} - {{{y^2}} \over {1/2}} = 1$$<br><br>$$ \Rightarrow $$ $${x^2} - {y^2} = {1 \over 2}$$<br><br>Clearly $$\left( {\sqrt {{3 \over 2}} ,{1 \over {\sqrt 2 }}} \right)$$ does not lie on it. | mcq | jee-main-2020-online-3rd-september-morning-slot |
yxkRQU9lhUVkEZFF71jgy2xukf418vtf | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | Let e<sub>1</sub>
and e<sub>2</sub>
be the eccentricities of the
ellipse, <br/>$${{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1$$(b < 5) and the hyperbola,
<br/>$${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$$ respectively satisfying e<sub>1</sub>e<sub>2</sub>
= 1. If $$\alpha $$
<br/>and $$\beta $$ are the distances between the foci of the
<br/>ellipse and the foci of the hyperbola
<br/>respectively, then the ordered pair ($$\alpha $$, $$\beta $$) is
equal to :
| [{"identifier": "A", "content": "(8, 10)"}, {"identifier": "B", "content": "(8, 12)"}, {"identifier": "C", "content": "$$\\left( {{{24} \\over 5},10} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{{20} \\over 3},12} \\right)$$"}] | ["A"] | null | For ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1\,\,\,(b < 5)$$<br><br>Let e<sub>1</sub> is eccentricity of ellipse <br><br>$$ \therefore $$ b<sup>2</sup> = 25 (1 $$ - $$ $${e_1}^2$$) ........(1)<br><br>Again for hyperbola<br><br>$${{{x^2}} \over {16}} - {{{y^2}} \over {{b^2}}} = 1$$<br><br>Let e<sub>2</sub> is eccentricity of hyperbola.<br><br>$$ \therefore $$ $${b^2} = 16({e_1}^2 - 1)$$ ......(2)<br><br>by (1) & (2)<br><br>$$25(1 - {e_1}^2)\, = \,16({e_1}^2 - 1)$$<br><br>Now e<sub>1</sub> . e<sub>2</sub> = 1 (given)<br><br>$$ \therefore $$ $$25(1 - {e_1}^2)\, = \,16\left( {{{1 - {e_1}^2} \over {{e_1}^2}}} \right)$$<br><br>or e<sub>1</sub> = $${4 \over 5}$$ $$ \therefore $$ e<sub>2</sub> = $${5 \over 4}$$<br><br>Now distance between foci is 2ae<br><br>$$ \therefore $$ Distance for ellipse = $$2 \times 5 \times {4 \over 5} = 8 = \alpha $$<br><br>Distance for hyperbola = $$2 \times 4 \times {5 \over 4} = 10 = \beta $$<br><br>$$ \therefore $$ ($$\alpha $$, $$\beta $$) $$ \equiv $$ (8, 10) | mcq | jee-main-2020-online-3rd-september-evening-slot |
iii96ZkKaJJmJnFmw61klt9jx19 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | A hyperbola passes through the foci of the ellipse $${{{x^2}} \over {25}} + {{{y^2}} \over {16}} = 1$$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is : | [{"identifier": "A", "content": "$${{{x^2}} \\over 9} - {{{y^2}} \\over 4} = 1$$"}, {"identifier": "B", "content": "$${{{x^2}} \\over 9} - {{{y^2}} \\over 16} = 1$$"}, {"identifier": "C", "content": "$${{{x^2}} \\over 9} - {{{y^2}} \\over 25} = 1$$"}, {"identifier": "D", "content": "x<sup>2</sup> $$-$$ y<sup>2</sup> = 9"}] | ["B"] | null | $${e_1} = \sqrt {1 - {{16} \over {25}}} = {3 \over 5}$$ foci ($$ \pm $$ae, 0)<br><br>Foci = ($$ \pm $$3, 0)<br><br>Let equation of hyperbola be $${{{x^2}} \over {{A^2}}} - {{{y^2}} \over {{B^2}}} = 1$$<br><br>Passes through ($$ \pm $$3, 0)<br><br>A<sup>2</sup> = 9, A = 3, $${e_2} = {5 \over 3}$$<br><br>$${e_2}^2 = 1 + {{{B^2}} \over {{A^2}}}$$<br><br>$${{25} \over 9} = 1 + {{{B^2}} \over 9} \Rightarrow {B^2} = 16$$
<br><br>Equation of the hyperbola
<br><br> $${{{x^2}} \over 9} - {{{y^2}} \over {16}} = 1$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
1l546bltp | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, a > 0, b > 0, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $$4(2\sqrt 2 + \sqrt {14} )$$. If the eccentricity H is $${{\sqrt {11} } \over 2}$$, then the value of a<sup>2</sup> + b<sup>2</sup> is equal to __________.</p> | [] | null | 88 | <p>$$2a + 2b = 4\left( {2\sqrt 2 + \sqrt {14} } \right)$$ ...... (1)</p>
<p>$$1 + {{{b^2}} \over {{a^2}}} = {{11} \over {14}}$$ ....... (2)</p>
<p>$$ \Rightarrow {{{b^2}} \over {{a^2}}} = {7 \over 4}$$ ....... (3)</p>
<p>and $$a + b = 4\sqrt 2 + 2\sqrt {14} $$ ...... (4)</p>
<p>By (3) and (4)</p>
<p>$$ \Rightarrow a + {{\sqrt 7 } \over 2}a = 4\sqrt 2 + 2\sqrt {14} $$</p>
<p>$$ \Rightarrow {{a\left( {2 + \sqrt 7 } \right)} \over 2} = 2\sqrt 2 \left( {2 + \sqrt 7 } \right)$$</p>
<p>$$ \Rightarrow a = 4\sqrt 2 \Rightarrow {a^2} = 32$$ and $${b^2} = 56$$</p>
<p>$$ \Rightarrow {a^2} + {b^2} = 32 + 56 = 88$$</p> | integer | jee-main-2022-online-29th-june-morning-shift |
1l55hrsqt | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. Let e' and l' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If $${e^2} = {{11} \over {14}}l$$ and $${\left( {e'} \right)^2} = {{11} \over 8}l'$$, then the value of $$77a + 44b$$ is equal to :</p> | [{"identifier": "A", "content": "100"}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "120"}, {"identifier": "D", "content": "130"}] | ["D"] | null | <p>$$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$, then</p>
<p>$${e^2} = {{11} \over {14}}l$$ (l be the length of LR)</p>
<p>$$ \Rightarrow {a^2} + {b^2} = {{11} \over 7}{b^2}a$$ ..... (i)</p>
<p>and $$e{'^2} = {{11} \over 8}l'$$ (l' be the length of LR of conjugate hyperbola)</p>
<p>$$ \Rightarrow {a^2} + {b^2} = {{11} \over 4}{a^2}b$$ ....... (ii)</p>
<p>By (i) and (ii)</p>
<p>$$7a = 4b$$</p>
<p>then by (i)</p>
<p>$${{16} \over {49}}{b^2} + {b^2} = {{11} \over 7}{b^2}\,.\,{{4b} \over 7}$$</p>
<p>$$ \Rightarrow 44b = 65$$ and $$77a = 65$$</p>
<p>$$\therefore$$ $$77a + 44b = 130$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l5bbaufk | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the hyperbola $$H:{{{x^2}} \over {{a^2}}} - {y^2} = 1$$ and the ellipse $$E:3{x^2} + 4{y^2} = 12$$ be such that the length of latus rectum of H is equal to the length of latus rectum of E. If $${e_H}$$ and $${e_E}$$ are the eccentricities of H and E respectively, then the value of $$12\left( {e_H^2 + e_E^2} \right)$$ is equal to ___________.</p> | [] | null | 42 | <p>$$\because$$ $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over 1} = 1$$</p>
<p>$$\therefore$$ Length of latus rectum $$ = {2 \over a}$$</p>
<p>$$E:{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1$$</p>
<p>Length of latus rectum $$ = {6 \over 2} = 3$$</p>
<p>$$\because$$ $${2 \over a} = 3 \Rightarrow a = {2 \over 3}$$</p>
<p>$$\therefore$$ $$12\left( {e_H^2 + e_E^2} \right) = 12\left( {1 + {9 \over 4}} \right) + \left( {1 - {3 \over 4}} \right) = 42$$</p> | integer | jee-main-2022-online-24th-june-evening-shift |
1l6f2jnb6 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the foci of the ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$$ and the hyperbola $$\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$$ coincide. Then the length of the latus rectum of the hyperbola is :</p> | [{"identifier": "A", "content": "$$\\frac{32}{9}$$"}, {"identifier": "B", "content": "$$\\frac{18}{5}$$"}, {"identifier": "C", "content": "$$\\frac{27}{4}$$"}, {"identifier": "D", "content": "$$\\frac{27}{10}$$"}] | ["D"] | null | <p>Ellipse : $${{{x^2}} \over {16}} + {{{y^2}} \over 7} = 1$$</p>
<p>Eccentricity $$ = \sqrt {1 - {7 \over {16}}} = {3 \over 4}$$</p>
<p>Foci $$ \equiv ( \pm \,a\,e,0) \equiv ( \pm \,3,0)$$</p>
<p>Hyperbola : $${{{x^2}} \over {\left( {{{144} \over {25}}} \right)}} - {{{y^2}} \over {\left( {{\alpha \over {25}}} \right)}} = 1$$</p>
<p>Eccentricity $$ = \sqrt {1 + {\alpha \over {144}}} = {1 \over {12}}\sqrt {144 + \alpha } $$</p>
<p>Foci $$ \equiv ( \pm \,a\,e,0) \equiv \left( { \pm \,{{12} \over 5}\,.\,{1 \over {12}}\sqrt {144 + \alpha } ,\,0} \right)$$</p>
<p>If foci coincide then $$3 = {1 \over 5}\sqrt {144 + \alpha } \Rightarrow \alpha = 81$$</p>
<p>Hence, hyperbola is $${{{x^2}} \over {{{\left( {{{12} \over 5}} \right)}^2}}} - {{{y^2}} \over {{{\left( {{9 \over 5}} \right)}^2}}} = 1$$</p>
<p>Length of latus rectum $$ = 2\,.\,{{{{81} \over {25}}} \over {{{12} \over 5}}} = {{27} \over {10}}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6hyxly9 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>If the line $$x-1=0$$ is a directrix of the hyperbola $$k x^{2}-y^{2}=6$$, then the hyperbola passes through the point :</p> | [{"identifier": "A", "content": "$$(-2 \\sqrt{5}, 6)$$"}, {"identifier": "B", "content": "$$(-\\sqrt{5}, 3)$$"}, {"identifier": "C", "content": "$$(\\sqrt{5},-2)$$"}, {"identifier": "D", "content": "$$(2 \\sqrt{5}, 3 \\sqrt{6})$$"}] | ["C"] | null | <p>Given hyperbola : $${{{x^2}} \over {6/k}} - {{{y^2}} \over 6} = 1$$</p>
<p>Eccentricity $$ = e = \sqrt {1 + {6 \over {6/k}}} = \sqrt {1 + k} $$</p>
<p>Directrices : $$x = \, \pm \,{a \over e} \Rightarrow x = \, \pm \,{{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }}$$</p>
<p>As given : $${{\sqrt 6 } \over {\sqrt k \sqrt {k + 1} }} = 1$$</p>
<p>$$ \Rightarrow k = 2$$</p>
<p>Here hyperbola is $${{{x^2}} \over 3} - {{{y^2}} \over 6} = 1$$</p>
<p>Checking the option gives $$\left( {\sqrt 5 , - 2} \right)$$ satisfies it.</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6je3gl6 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>An ellipse $$E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ passes through the vertices of the hyperbola $$H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1$$. Let the major and minor axes of the ellipse $$E$$ coincide with the transverse and conjugate axes of the hyperbola $$H$$, respectively. Let the product of the eccentricities of $$E$$ and $$H$$ be $$\frac{1}{2}$$. If $$l$$ is the length of the latus rectum of the ellipse $$E$$, then the value of $$113 l$$ is equal to _____________.</p> | [] | null | 1552 | <p>Vertices of hyperbola $$ = (0,\, \pm \,8)$$</p>
<p>As ellipse pass through it i.e.,</p>
<p>$$0 + {{64} \over {{b^2}}} = 1 \Rightarrow {b^2} = 64$$ ...... (1)</p>
<p>As major axis of ellipse coincide with transverse axis of hyperbola we have b > a i.e.</p>
<p>$${e_E} = \sqrt {1 - {{{a^2}} \over {64}}} = {{\sqrt {64 - {a^2}} } \over 8}$$</p>
<p>and $${e_H} = \sqrt {1 + {{49} \over {64}}} = {{\sqrt {113} } \over 8}$$</p>
<p>$$\therefore$$ $${e_E}\,.\,{e_H} = {1 \over 2} = {{\sqrt {64 - {a^2}} \sqrt {113} } \over {64}}$$</p>
<p>$$ \Rightarrow (64 - {a^2})(113) = {32^2}$$</p>
<p>$$ \Rightarrow {a^2} = 64 - {{1024} \over {113}}$$</p>
<p>L.R of ellipse $$ = {{2{a^2}} \over b} = {2 \over 8}\left( {{{113 \times 64 - 1024} \over {113}}} \right)$$</p>
<p>$$ = l = {{1552} \over {113}}$$</p>
<p>$$\therefore$$ $$113l = 1552$$</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6nni0or | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the hyperbola $$H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$$ pass through the point $$(2 \sqrt{2},-2 \sqrt{2})$$. A parabola is drawn whose focus is same as the focus of $$\mathrm{H}$$ with positive abscissa and the directrix of the parabola passes through the other focus of $$\mathrm{H}$$. If the length of the latus rectum of the parabola is e times the length of the latus rectum of $$\mathrm{H}$$, where e is the eccentricity of H, then which of the following points lies on the parabola?</p> | [{"identifier": "A", "content": "$$(2 \\sqrt{3}, 3 \\sqrt{2})$$"}, {"identifier": "B", "content": "$$\\mathbf(3 \\sqrt{3},-6 \\sqrt{2})$$"}, {"identifier": "C", "content": "$$(\\sqrt{3},-\\sqrt{6})$$"}, {"identifier": "D", "content": "$$(3 \\sqrt{6}, 6 \\sqrt{2})$$"}] | ["B"] | null | <p>$$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$</p>
<p>Focus of parabola : $$(ae,\,0)$$</p>
<p>Directrix : $$x = - ae$$.</p>
<p>Equation of parabola $$ \equiv {y^2} = 4aex$$</p>
<p>Length of latus rectum of parabola $$ = 4ae$$</p>
<p>Length of latus rectum of hyperbola $$ = {{2.{b^2}} \over a}$$</p>
<p>as given, $$4ae = {{2{b^2}} \over a}\,.\,e$$</p>
<p>$$2 = {{{b^2}} \over {{a^2}}}$$ ...... (i)</p>
<p>$$\because$$ H passes through $$\left( {2\sqrt 2 , - 2\sqrt 2 } \right) \Rightarrow {8 \over {{a^2}}} - {8 \over {{b^2}}} = 1$$ ........ (ii)</p>
<p>From (i) and (ii) $${a^2} = 4$$ and $${b^2} = 8 \Rightarrow e = \sqrt 3 $$</p>
<p>$$\Rightarrow$$ Equation of parabola is $${y^2} = 8\sqrt 3 x$$.</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
ldo9yx82 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | Let $\mathrm{H}$ be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is : | [{"identifier": "A", "content": "$\\frac{5}{2}$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$\\frac{3}{2}$"}] | ["C"] | null | $ 2 \mathrm{ae}=|(1+\sqrt{2})-(1+\sqrt{2})|=2 \sqrt{2}$
<br/><br/>$$ \Rightarrow $$ $\mathrm{ae}=\sqrt{2}$
<br/><br/>$$ \Rightarrow $$ $\mathrm{a}=1$
<br/><br/>$\Rightarrow \mathrm{b}=1 \because \mathrm{e}=\sqrt{2} \Rightarrow$ Hyperbola is rectangular
<br/><br/>$\Rightarrow \mathrm{L} . \mathrm{R}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1lguwg2ap | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$\mathrm{H}_{\mathrm{n}}: \frac{x^{2}}{1+n}-\frac{y^{2}}{3+n}=1, n \in N$$. Let $$\mathrm{k}$$ be the smallest even value of $$\mathrm{n}$$ such that the eccentricity of $$\mathrm{H}_{\mathrm{k}}$$ is a rational number. If $$l$$ is the length of the latus rectum of $$\mathrm{H}_{\mathrm{k}}$$, then $$21 l$$ is equal to ____________.</p> | [] | null | 306 | We have,
<br/><br/>$$
H_n \Rightarrow \frac{x^2}{1+n}-\frac{y^2}{3+n}=1, n \in N
$$
<br/><br/>Here, $a^2=1+n$ and $b^2=3+n$
<br/><br/>$$
\begin{aligned}
\operatorname{Eccentricity}(e) & =\sqrt{1+\frac{b^2}{a^2}} \\\\
& =\sqrt{1+\left(\frac{3+n}{1+n}\right)}=\sqrt{\frac{2 n+4}{n+1}}=\sqrt{\frac{2(n+2)}{n+1}}
\end{aligned}
$$
<br/><br/>The smallest even value for which $e \in Q$ is 48 .
<br/><br/>$$
\begin{aligned}
\therefore n & =48 \\\\
\therefore e & =\sqrt{\frac{2(48+2)}{48+1}}=\frac{10}{7}
\end{aligned}
$$
<br/><br/>$$
\begin{array}{ll}
\Rightarrow a^2=n+1=49, b^2=n+3=51 \\\\
\therefore 21 l=21 \times\left(\frac{2 b^2}{a}\right)=21 \times 2 \times \frac{51}{7}=306
\end{array}
$$ | integer | jee-main-2023-online-11th-april-morning-shift |
1lh2yqxxv | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the eccentricity of an ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ is reciprocal to that of the hyperbola $$2 x^{2}-2 y^{2}=1$$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is ___________.</p> | [] | null | 2 | Equation of hyperbola is $2 x^2-2 y^2=1$
<br/><br/>$\Rightarrow \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$
<br/><br/>Here, $a=b$
<br/><br/>$\therefore$ Eccentricity of rectangular hyperbola is $\sqrt{2}$
<br/><br/>$\therefore$ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$
<br/><br/>Since, ellipse intersects the hyperbola at right angles
<br/><br/>$\therefore$ Ellipse and the hyperbola are confocal
<br/><br/>$\therefore$ Foci of hyperbola $=(1,0)$
<br/><br/>$\Rightarrow$ Foci of ellipse, $a e=1$
<br/><br/>$$
\begin{array}{ll}
&\Rightarrow a\left(\frac{1}{\sqrt{2}}\right)=1 \\\\
&\Rightarrow a=\sqrt{2} \\\\
&\therefore e^2=1-\frac{b^2}{a^2} \\\\
&\Rightarrow \frac{1}{2}=1-\frac{b^2}{2} \\\\
&\Rightarrow \frac{b^2}{2}=\frac{1}{2} \\\\
&\Rightarrow b^2=1
\end{array}
$$
<br/><br/>$\therefore$ Length of the latus rectum $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$
<br/><br/>$\therefore$ Square of length of the latus rectum $=2$ | integer | jee-main-2023-online-6th-april-evening-shift |
lsaosyr4 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | For $0<\theta<\pi / 2$, if the eccentricity of the hyperbola
<br/><br/>$x^2-y^2 \operatorname{cosec}^2 \theta=5$ is $\sqrt{7}$ times eccentricity of the<br/><br/> ellipse $x^2 \operatorname{cosec}^2 \theta+y^2=5$, then the value of $\theta$ is : | [{"identifier": "A", "content": "$\\frac{\\pi}{6}$"}, {"identifier": "B", "content": "$\\frac{5 \\pi}{12}$"}, {"identifier": "C", "content": "$\\frac{\\pi}{3}$"}, {"identifier": "D", "content": "$\\frac{\\pi}{4}$"}] | ["C"] | null | <p>To find the value of $\theta$, we need to determine the relationship between the eccentricities of the given hyperbola and ellipse. Let's start by writing down the standard forms of ellipse and hyperbola and then relate them to the given equations.</p>
<p>The standard form of an ellipse is:
<p>$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$</p>
<p>The eccentricity $e$ of an ellipse is given by:</p>
<p>$e = \sqrt{1 - \frac{b^2}{a^2}}\ \text{for}\ a > b$</p>
<p>For the given ellipse:</p>
<p>$x^2 \operatorname{cosec}^2 \theta + y^2 = 5$</p>
<p>We can compare it with the standard form by writing it as:</p>
<p>$\frac{x^2}{\frac{5}{\operatorname{cosec}^2 \theta}} + \frac{y^2}{5} = 1$</p>
<p>From this we have $a^2 = \frac{5}{\operatorname{cosec}^2 \theta}$ and $b^2 = 5$. The eccentricity of the ellipse (let's call it $e_1$) is:</p>
<p>$e_1 = \sqrt{1 - \frac{5}{a^2}} = \sqrt{1 - \frac{5}{\frac{5}{\operatorname{cosec}^2 \theta}}} = \sqrt{1 - \operatorname{sin}^2 \theta} = \operatorname{cos} \theta$</p>
<p>The standard form of a hyperbola is:
<p>$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$</p>
<p>The eccentricity $e$ of a hyperbola is given by:</p>
<p>$e = \sqrt{1 + \frac{b^2}{a^2}}$</p>
<p>For the given hyperbola:</p>
<p>$x^2 - y^2 \operatorname{cosec}^2 \theta = 5$</p>
<p>We can rewrite it as:</p>
<p>$\frac{x^2}{5} - \frac{y^2}{\frac{5}{\operatorname{cosec}^2 \theta}} = 1$</p>
<p>From this we have $a^2 = 5$ and $b^2 = \frac{5}{\operatorname{cosec}^2 \theta}$. The eccentricity of the hyperbola (let's call it $e_2$) is:</p>
<p>$e_2 = \sqrt{1 + \frac{\frac{5}{\operatorname{cosec}^2 \theta}}{5}} = \sqrt{1 + \operatorname{sin}^2 \theta}$</p>
<p>According to the question the eccentricity of the hyperbola is $\sqrt{7}$ times the eccentricity of the ellipse, so we can write:
<p>$e_2 = \sqrt{7} \cdot e_1$</p>
<p>Substitute $e_1$ and $e_2$ from the above:</p>
<p>$\sqrt{1 + \operatorname{sin}^2 \theta} = \sqrt{7} \cdot \operatorname{cos} \theta$</p>
<p>Square both sides to eliminate the square root:</p>
<p>$1 + \operatorname{sin}^2 \theta = 7 \cdot \operatorname{cos}^2 \theta$</p>
<p>$1 + \operatorname{sin}^2 \theta = 7 \cdot (1 - \operatorname{sin}^2 \theta)$</p>
<p>$1 + \operatorname{sin}^2 \theta = 7 - 7 \cdot \operatorname{sin}^2 \theta$</p>
<p>$8 \cdot \operatorname{sin}^2 \theta = 6$</p>
<p>$\operatorname{sin}^2 \theta = \frac{6}{8} = \frac{3}{4}$</p>
<p>$\operatorname{sin} \theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$</p>
<p>Looking at the options provided and knowing the sine values for common angles, we can see that $\operatorname{sin} \theta = \frac{\sqrt{3}}{2}$ corresponds to $\theta = \frac{\pi}{3}$.
<p>The correct answer is:</p>
<p>Option C.</p>
<p>$\frac{\pi}{3}$</p>
| mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lse5pw17 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>If the foci of a hyperbola are same as that of the ellipse $$\frac{x^2}{9}+\frac{y^2}{25}=1$$ and the eccentricity of the hyperbola is $$\frac{15}{8}$$ times the eccentricity of the ellipse, then the smaller focal distance of the point $$\left(\sqrt{2}, \frac{14}{3} \sqrt{\frac{2}{5}}\right)$$ on the hyperbola, is equal to</p> | [{"identifier": "A", "content": "$$14 \\sqrt{\\frac{2}{5}}-\\frac{4}{3}$$\n"}, {"identifier": "B", "content": "$$7 \\sqrt{\\frac{2}{5}}+\\frac{8}{3}$$\n"}, {"identifier": "C", "content": "$$7 \\sqrt{\\frac{2}{5}}-\\frac{8}{3}$$\n"}, {"identifier": "D", "content": "$$14 \\sqrt{\\frac{2}{5}}-\\frac{16}{3}$$"}] | ["C"] | null | <p>$$\begin{aligned}
& \frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1 \\
& \mathrm{a}=3, \mathrm{~b}=5 \\
& \mathrm{e}=\sqrt{1-\frac{9}{25}}=\frac{4}{5} \therefore \text { foci }=(0, \pm \mathrm{be})=(0, \pm 4) \\
& \quad \therefore \mathrm{e}_{\mathrm{H}}=\frac{4}{5} \times \frac{15}{8}=\frac{3}{2}
\end{aligned}$$</p>
<p>Let equation hyperbola</p>
<p>$$\begin{aligned}
& \frac{\mathrm{x}^2}{\mathrm{~A}^2}-\frac{\mathrm{y}^2}{\mathrm{~B}^2}=-1 \\
& \therefore \mathrm{B} \cdot \mathrm{e}_{\mathrm{H}}=4 \quad \therefore \mathrm{B}=\frac{8}{3} \\
& \therefore \mathrm{A}^2=\mathrm{B}^2\left(\mathrm{e}_{\mathrm{H}}^2-1\right)=\frac{64}{9}\left(\frac{9}{4}-1\right) \therefore \mathrm{A}^2=\frac{80}{9} \\
& \therefore \frac{\mathrm{x}^2}{\frac{80}{9}}-\frac{\mathrm{y}^2}{\frac{64}{9}}=-1 \\
& \text { Directrix }: \mathrm{y}= \pm \frac{\mathrm{B}}{\mathrm{e}_{\mathrm{H}}}= \pm \frac{16}{9} \\
& \mathrm{PS}=\mathrm{e} \cdot \mathrm{PM}=\frac{3}{2}\left|\frac{14}{3} \cdot \sqrt{\frac{2}{5}}-\frac{16}{9}\right| \\
& =7 \sqrt{\frac{2}{5}}-\frac{8}{3}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lse5xrza | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the foci and length of the latus rectum of an ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b b e( \pm 5,0)$$ and $$\sqrt{50}$$, respectively. Then, the square of the eccentricity of the hyperbola $$\frac{x^2}{b^2}-\frac{y^2}{a^2 b^2}=1$$ equals</p> | [] | null | 51 | <p>$$\begin{aligned}
& \text { focii } \equiv( \pm 5,0) ; \frac{2 b^2}{a}=\sqrt{50} \\
& a=5 \quad b^2=\frac{5 \sqrt{2} a}{2} \\
& b^2=a^2\left(1-e^2\right)=\frac{5 \sqrt{2} a}{2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{a}\left(1-\mathrm{e}^2\right)=\frac{5 \sqrt{2}}{2} \\
& \Rightarrow \frac{5}{\mathrm{e}}\left(1-\mathrm{e}^2\right)=\frac{5 \sqrt{2}}{2} \\
& \Rightarrow \sqrt{2}-\sqrt{2} \mathrm{e}^2=\mathrm{e} \\
& \Rightarrow \sqrt{2} \mathrm{e}^2+\mathrm{e}-\sqrt{2}=0 \\
& \Rightarrow \sqrt{2} \mathrm{e}^2+2 \mathrm{e}-\mathrm{e}-\sqrt{2}=0 \\
& \Rightarrow \sqrt{2} \mathrm{e}(\mathrm{e}+\sqrt{2})-1(1+\sqrt{2})=0 \\
& \Rightarrow(\mathrm{e}+\sqrt{2})(\sqrt{2} \mathrm{e}-1)=0 \\
& \therefore \mathrm{e} \neq-\sqrt{2} ; \mathrm{e}=\frac{1}{\sqrt{2}}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{\mathrm{x}^2}{\mathrm{~b}^2}-\frac{\mathrm{y}^2}{\mathrm{a}^2 \mathrm{~b}^2}=1 \quad \mathrm{a}=5 \sqrt{2} \\
& b=5 \\
& a^2 b^2=b^2\left(e_1^2-1\right) \Rightarrow e_1^2=51 \\
&
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-morning-shift |
1lsg4ewgx | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$P$$ be a point on the hyperbola $$H: \frac{x^2}{9}-\frac{y^2}{4}=1$$, in the first quadrant such that the area of triangle formed by $$P$$ and the two foci of $$H$$ is $$2 \sqrt{13}$$. Then, the square of the distance of $$P$$ from the origin is</p> | [{"identifier": "A", "content": "26"}, {"identifier": "B", "content": "22"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "18"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxrj67/57cc4f4f-ed29-40fc-903d-a6afb2548230/871275f0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxrj68.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxrj67/57cc4f4f-ed29-40fc-903d-a6afb2548230/871275f0-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxrj68.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Hyperbola Question 7 English Explanation"></p>
<p>$$\begin{aligned}
& \frac{x^2}{9}-\frac{y^2}{4}=1 \\
& a^2=9, b^2=4 \\
& b^2=a^2\left(e^2-1\right) \Rightarrow e^2=1+\frac{b^2}{a^2} \\
& e^2=1+\frac{4}{9}=\frac{13}{9} \\
& e=\frac{\sqrt{13}}{3} \Rightarrow s_1 s_2=2 \mathrm{ae}=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Area of } \triangle \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{2} \times \beta \times \mathrm{S}_1 \mathrm{~S}_2=2 \sqrt{13} \\
& \Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 \\
& \begin{aligned}
& \frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2} \\
& \text { Distance of P from origin }=\sqrt{\alpha^2+\beta^2} \\
&=\sqrt{18+4}=\sqrt{22}
\end{aligned}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsgckxsg | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the latus rectum of the hyperbola $$\frac{x^2}{9}-\frac{y^2}{b^2}=1$$ subtend an angle of $$\frac{\pi}{3}$$ at the centre of the hyperbola. If $$\mathrm{b}^2$$ is equal to $$\frac{l}{\mathrm{~m}}(1+\sqrt{\mathrm{n}})$$, where $$l$$ and $$\mathrm{m}$$ are co-prime numbers, then $$\mathrm{l}^2+\mathrm{m}^2+\mathrm{n}^2$$ is equal to ________.</p> | [] | null | 182 | <p>LR subtends $$60^{\circ}$$ at centre</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnnba9/95b74993-a660-473c-9d3c-c4fef0db6ac2/87c00510-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnnbaa.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnnba9/95b74993-a660-473c-9d3c-c4fef0db6ac2/87c00510-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnnbaa.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Morning Shift Mathematics - Hyperbola Question 6 English Explanation"></p>
<p>$$\begin{aligned}
& \Rightarrow \tan 30^{\circ}=\frac{\mathrm{b}^2 / \mathrm{a}}{\mathrm{ae}}=\frac{\mathrm{b}^2}{\mathrm{a}^2 \mathrm{e}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \mathrm{e}=\frac{\sqrt{3} \mathrm{~b}^2}{9}
\end{aligned}$$</p>
<p>Also, $$\mathrm{e}^2=1+\frac{\mathrm{b}^2}{9} \Rightarrow 1+\frac{\mathrm{b}^2}{9}=\frac{3 \mathrm{~b}^4}{81}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{b}^4=3 \mathrm{~b}^2+27 \\
& \Rightarrow \mathrm{b}^4-3 \mathrm{~b}^2-27=0 \\
& \Rightarrow \mathrm{b}^2=\frac{3}{2}(1+\sqrt{13}) \\
& \Rightarrow \ell=3, \mathrm{~m}=2, \mathrm{n}=13 \\
& \Rightarrow \ell^2+\mathrm{m}^2+\mathrm{n}^2=182
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-morning-shift |
luxwe7n5 | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let the foci of a hyperbola $$H$$ coincide with the foci of the ellipse $$E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$$ and the eccentricity of the hyperbola $$H$$ be the reciprocal of the eccentricity of the ellipse $$E$$. If the length of the transverse axis of $$H$$ is $$\alpha$$ and the length of its conjugate axis is $$\beta$$, then $$3 \alpha^2+2 \beta^2$$ is equal to</p> | [{"identifier": "A", "content": "225"}, {"identifier": "B", "content": "237"}, {"identifier": "C", "content": "242"}, {"identifier": "D", "content": "205"}] | ["A"] | null | <p>$$E: \frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$$</p>
<p>$$\begin{aligned}
\text { Eccentricity of ellipse, } e_E & =\sqrt{1-\frac{b^2}{a^2}} \\
& =\sqrt{1-\frac{75}{100}} \\
& e_E=\frac{1}{2}
\end{aligned}$$</p>
<p>$$\therefore e_H=2$$ [ as eccentricity of hyperbola is reciprocal of eccentricity of ellipse]</p>
<p>Transverse axis of hyperbola $$=\alpha$$</p>
<p>Conjugate axis of hyperbola $$=\beta$$</p>
<p>Also, foci of ellipse $$(1 \pm a e, 1)$$</p>
<p>$$\begin{aligned}
& =\left(1 \pm\left(10 \times \frac{1}{2}\right), 1\right) \\
& =(1 \pm 5,1) \\
& =(6,1) \text { and }(-4,1)
\end{aligned}$$</p>
<p>Distance between foci $$=10$$</p>
<p>$$\begin{aligned}
& 2 a e=10 \\
& \Rightarrow a=\frac{5}{2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { also, } e^2=1+\frac{b^2}{a^2} \\
& \begin{aligned}
4 & =1+\frac{4 b^2}{25} \\
b^2 & =\frac{75}{4} \\
b & =\frac{\sqrt{75}}{2}
\end{aligned} \\
& \Rightarrow \quad \alpha=5 \\
& \text { and } \beta=\sqrt{75} \\
& 3 \alpha^2+2 \beta^2=3(5)^2+2(75)=225
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
lv3vef1a | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$\mathrm{S}$$ be the focus of the hyperbola $$\frac{x^2}{3}-\frac{y^2}{5}=1$$, on the positive $$x$$-axis. Let $$\mathrm{C}$$ be the circle with its centre at $$\mathrm{A}(\sqrt{6}, \sqrt{5})$$ and passing through the point $$\mathrm{S}$$. If $$\mathrm{O}$$ is the origin and $$\mathrm{SAB}$$ is a diameter of $$\mathrm{C}$$, then the square of the area of the triangle OSB is equal to __________.</p> | [] | null | 40 | <p>$$\begin{aligned}
& \frac{x^2}{3}-\frac{y^2}{5}=1 \\
& 5=3\left(e^2-1\right) \Rightarrow e=\sqrt{\frac{8}{3}} \\
& S \equiv(2 \sqrt{2}, 0)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4m7nur/89a66068-e09d-4eb2-bb5b-6e5807946115/8a9cdd30-10f6-11ef-8553-fdfc6347789d/file-1lw4m7nus.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4m7nur/89a66068-e09d-4eb2-bb5b-6e5807946115/8a9cdd30-10f6-11ef-8553-fdfc6347789d/file-1lw4m7nus.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Hyperbola Question 3 English Explanation"></p>
<p>$$A$$ is mid-point of $$B S$$</p>
<p>$$\Rightarrow \quad B(2 \sqrt{6}-2 \sqrt{2}, 2 \sqrt{5})$$</p>
<p>$$\Delta (OSB) = \left| {{1 \over 2}\left| {\matrix{
0 & 0 & 1 \cr
{2\sqrt 2 } & 0 & 1 \cr
{2\sqrt 6 - 2\sqrt 2 } & {2\sqrt 5 } & 1 \cr
} } \right|} \right| = 2\sqrt {10} $$<./p>
</p><p>$$(\Delta(O S B))^2=40$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv5grwkp | maths | hyperbola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4 \sqrt{3}$$. Suppose the point $$(\alpha, 6), \alpha>0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha, 6)$$, then $$\alpha^2+\beta$$ is equal to</p> | [{"identifier": "A", "content": "170"}, {"identifier": "B", "content": "171"}, {"identifier": "C", "content": "169"}, {"identifier": "D", "content": "172"}] | ["B"] | null | <p>$$\begin{aligned}
& H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=-1 \\
& e=\sqrt{1+\frac{a^2}{b^2}}=\sqrt{3} \\
& \Rightarrow 1+\frac{a^2}{b^2}=3 \\
& \Rightarrow \frac{a^2}{b^2}=2 \quad \text{.... (1)}\\
& \frac{2 a^2}{b}=4 \sqrt{3}
\end{aligned}$$</p>
<p>Using equation (1)</p>
<p>$$\begin{aligned}
& \frac{4 b^2}{b}=4 \sqrt{3} \\
& \Rightarrow b=\sqrt{3} \\
& a=\sqrt{6} \\
& H: \frac{x^2}{6}-\frac{y^2}{3}=-1 \\
& \frac{\alpha^2}{6}-12=-1 \\
& \frac{\alpha^2}{6}=11 \\
& \begin{array}{c}
\alpha^2=66 \\
\text { Focus }:(0, b c)(0,-b c) \\
(0,3),(0,-3)
\end{array} \\
& \beta=\sqrt{\alpha^2+9} \times \sqrt{\alpha^2+81} \\
& \beta=105 \\
& \alpha^2+\beta=66+105 \\
& =171
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
urhXbbS4Mwr9Upfs | maths | hyperbola | tangent-to-hyperbola | The locus of a point $$P\left( {\alpha ,\beta } \right)$$ moving under the condition that the line $$y = \alpha x + \beta $$ is tangent to the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ is : | [{"identifier": "A", "content": "an ellipse "}, {"identifier": "B", "content": "a circle "}, {"identifier": "C", "content": "a parabola "}, {"identifier": "D", "content": "a hyperbola "}] | ["D"] | null | Tangent to the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ is
<br><br>$$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $$
<br><br>Given that $$y = \alpha x + \beta $$ is the tangent of hyperbola
<br><br>$$ \Rightarrow m = \alpha $$ and $${a^2}{m^2} - {b^2} = {\beta ^2}$$
<br><br>$$\therefore$$ $${a^2}{\alpha ^2} - {b^2} = {\beta ^2}$$
<br><br>Locus is $${a^2}{x^2} - {y^2} = {b^2}$$ which is hyperbola. | mcq | aieee-2005 |
frFcYTdloe0SbBmg | maths | hyperbola | tangent-to-hyperbola | A hyperbola passes through the point P$$\left( {\sqrt 2 ,\sqrt 3 } \right)$$ and has foci at $$\left( { \pm 2,0} \right)$$. Then the tangent to this hyperbola at P also passes through the point : | [{"identifier": "A", "content": "$$\\left( {2\\sqrt 2 ,3\\sqrt 3 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {\\sqrt 3 ,\\sqrt 2 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\sqrt 2 , - \\sqrt 3 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {3\\sqrt 2 ,2\\sqrt 3 } \\right)$$"}] | ["A"] | null | Equation of hyperbola is $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
<br><br>foci is (±2, 0)
<br><br>$$ \Rightarrow $$ ae = 2
<br><br>$$ \Rightarrow $$ a<sup>2</sup>e<sup>2</sup> = 4
<br><br>Since b<sup>2</sup> = a<sup>2</sup> (e<sup>2</sup> – 1)
<br><br>b<sup>2</sup> = a<sup>2</sup> e<sup>2</sup> – a<sup>2</sup>
<br><br>$$ \therefore $$ a<sup>2</sup> + b<sup>2</sup> = 4 .....(1)
<br><br>Also Hyperbola passes through $$\left( {\sqrt 2 ,\sqrt 3 } \right)$$
<br><br>$$ \therefore $$ $${2 \over {{a^2}}} - {3 \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ $${2 \over {4 - {b^2}}} - {3 \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ (b<sup>2</sup> – 3) (b<sup>2</sup> + 4) = 0
<br><br>$$ \therefore $$ b<sup>2</sup> = 3 or b<sup>2</sup> = -4
<br><br> For b<sup>2</sup> = 3
<br><br>$$ \Rightarrow $$ a<sup>2</sup> = 1
<br><br>$${{{x^2}} \over 1} - {{{y^2}} \over 3} = 1$$
<br><br>Equation of tangent is $${{\sqrt 2 x} \over 1} - {{\sqrt 3 y} \over 3} = 1$$
<br><br>It satisfy point $$\left( {2\sqrt 2 ,3\sqrt 3 } \right)$$. | mcq | jee-main-2017-offline |
grnK93LwSd8r9LVV | maths | hyperbola | tangent-to-hyperbola | Tangents are drawn to the hyperbola 4x<sup>2</sup> - y<sup>2</sup> = 36 at the points P and Q.
<br/><br/>If these tangents intersect at the
point T(0, 3) then the area (in sq. units) of $$\Delta $$PTQ is : | [{"identifier": "A", "content": "$$36\\sqrt 5 $$"}, {"identifier": "B", "content": "$$45\\sqrt 5 $$"}, {"identifier": "C", "content": "$$54\\sqrt 3 $$"}, {"identifier": "D", "content": "$$60\\sqrt 3 $$"}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265692/exam_images/dqhksg6rhe7itnkgtxn3.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Hyperbola Question 71 English Explanation">
<br><br>Here PQ is the chord of contact.
<br><br>Equation of PQ is
<br><br>x.0 $$-$$ y.3 = 36
<br><br>$$ \Rightarrow \,\,\,\,y = - 12$$
<br><br>Putting value of y = $$-$$ 12 in the equation
<br><br>4x<sup>2</sup> $$-$$ y<sup>2</sup> = 36
<br><br>we get , 4x<sup>2</sup> $$-$$ 144 = 36
<br><br>$$ \Rightarrow \,\,\,\,4{x^2}\, = \,180$$
<br><br>$$ \Rightarrow \,\,\,\,\,{x^2}\,\, = \,\,45$$
<br><br>$$ \Rightarrow \,\,\,\,x = \pm \,3\sqrt 5 $$
<br><br>$$\therefore\,\,\,$$ Coordinate of $$P = \left( { - 3\sqrt 5 , - 12} \right)$$ and
<br><br> Coordinate of $$Q = \left( {3\sqrt 5 , - 12} \right)$$
<br><br>$$\therefore\,\,\,$$ Length of PQ $$ = 3\sqrt 5 + 3\sqrt 5 $$
<br><br>$$ = 6\sqrt 5 .$$
<br><br>TM is the height of the triangle Length of TM = 12 + 3 = 15
<br><br>$$\therefore\,\,\,$$ Area of $$\Delta PQT$$ = $${1 \over 2} \times 6\sqrt 5 \times 15$$
<br><br>$$ = 45\sqrt 5 $$ sq. units | mcq | jee-main-2018-offline |
V1oEoOwIdxchnOnBaDwIG | maths | hyperbola | tangent-to-hyperbola | The equation of a tangent to the hyperbola 4x<sup>2</sup> – 5y<sup>2</sup> = 20 parallel to the line x – y = 2 is : | [{"identifier": "A", "content": "x $$-$$ y + 9 = 0"}, {"identifier": "B", "content": "x $$-$$ y $$-$$ 3 = 0"}, {"identifier": "C", "content": "x $$-$$ y + 1 = 0"}, {"identifier": "D", "content": "x $$-$$ y + 7 = 0"}] | ["C"] | null | Hyperbola $${{{x^2}} \over 5} - {{{y^2}} \over 4} = 1$$
<br><br>slope of tangent = 1
<br><br>equation of tangent y = x $$ \pm $$ $$\sqrt {5 - 4} $$
<br><br>$$ \Rightarrow $$ y = x $$ \pm $$ 1
<br><br>$$ \Rightarrow $$ y = x + 1
<br><br>or
<br><br>$$ \Rightarrow $$ y = x $$-$$ 1 | mcq | jee-main-2019-online-10th-january-morning-slot |
serhntUUiJrREZf1pHxE5 | maths | hyperbola | tangent-to-hyperbola | If the eccentricity of the standard hyperbola
passing through the point (4,6) is 2, then the
equation of the tangent to the hyperbola at (4,6)
is : | [{"identifier": "A", "content": "2x \u2013 y \u2013 2 = 0"}, {"identifier": "B", "content": "3x \u2013 2y = 0"}, {"identifier": "C", "content": "2x \u2013 3y + 10 = 0"}, {"identifier": "D", "content": "x \u2013 2y + 8 = 0"}] | ["A"] | null | Formula for standard hyperbola :
<br><br>$${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$
<br><br>It passes through (4, 6)
<br><br>$$ \therefore $$ $${{16} \over {{a^2}}} - {{36} \over {{b^2}}} = 1$$ ........(1)
<br><br>We know, $${e^2} = 1 + {{{b^2}} \over {{a^2}}}$$
<br><br>$$ \Rightarrow $$ 4 = $$1 + {{{b^2}} \over {{a^2}}}$$ [ as e = 2 ]
<br><br>$$ \Rightarrow $$ $${{b^2} = 3{a^2}}$$ ....(2)
<br><br>Putting $${{b^2} = 3{a^2}}$$ in equation (1)
<br><br>$${{16} \over {{a^2}}} - {{36} \over {3{a^2}}} = 1$$
<br><br>$$ \Rightarrow $$ $${{a^2} = 4}$$
<br><br>$$ \therefore $$ , $${{b^2} = 12}$$
<br><br>So Equation of hyperbola is
<br><br>$${{{x^2}} \over 4} - {{{y^2}} \over {12}} = 1$$
<br><br>Equation of tangent to the hyperbola at (4, 6) is
<br><br>$${{4x} \over 4} - {{6y} \over {12}} = 1$$
<br><br>$$ \Rightarrow $$ $$x - {y \over 2} = 1$$
<br><br>$$ \Rightarrow $$ 2x – y – 2 = 0 | mcq | jee-main-2019-online-8th-april-evening-slot |
MyQ9Y4KXRg1ZCSnoHLjgy2xukews984x | maths | hyperbola | tangent-to-hyperbola | A line parallel to the straight line 2x – y = 0 is
tangent to the hyperbola
<br/>$${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$$ at the point
$$\left( {{x_1},{y_1}} \right)$$. Then $$x_1^2 + 5y_1^2$$ is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "8"}] | ["B"] | null | Tangent of hyperbola $${{{x^2}} \over 4} - {{{y^2}} \over 2} = 1$$ at the point (x<sub>1</sub>, y<sub>1</sub>) is
<br><br>$${{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1$$ which is parallel to 2x – y = 0
<br><br>$$ \therefore $$ Slope of tangent $${{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1$$ = Slope of 2x – y = 0
<br><br>$$ \Rightarrow $$ $${{{x_1}} \over {2{y_1}}}$$ = 2
<br><br>$$ \Rightarrow $$ x<sub>1</sub> = 4y<sub>1</sub> ....(1)
<br><br>(x<sub>1</sub>, y<sub>1</sub>) lies on hyperbola
<br><br>$$ \therefore $$ $${{x_1^2} \over 4} - {{y_1^2} \over 2} = 1$$ ....(2)
<br><br>From (1) & (2)
<br><br>$${{{{\left( {4{y_1}} \right)}^2}} \over 4} - {{y_1^2} \over 2} = 1$$
<br><br>$$ \Rightarrow $$ $$4y_1^2 - {{y_1^2} \over 2} = 1$$
<br><br>$$ \Rightarrow $$ $$7y_1^2 = 2$$
<br><br>$$ \Rightarrow $$ $$y_1^2 = {2 \over 7}$$
<br><br>Now $$x_1^2 + 5y_1^2$$
<br><br>= $${\left( {4{y_1}} \right)^2} + {\left( {5{y_1}} \right)^2}$$
<br><br>= 21$$y_1^2$$
<br><br>= 21$$ \times {2 \over 7}$$
<br><br>= 6 | mcq | jee-main-2020-online-2nd-september-morning-slot |
jH56590JjDMSL9PSa01kmm3kmwv | maths | hyperbola | tangent-to-hyperbola | Consider a hyperbola H : x<sup>2</sup> $$-$$ 2y<sup>2</sup> = 4. Let the tangent at a <br/>point P(4, $${\sqrt 6 }$$) meet the x-axis at Q and latus rectum at R(x<sub>1</sub>, y<sub>1</sub>), x<sub>1</sub> > 0. If F is a focus of H which is nearer to the point P, then the area of $$\Delta$$QFR is equal to : | [{"identifier": "A", "content": "$${\\sqrt 6 }$$ $$-$$ 1"}, {"identifier": "B", "content": "$${7 \\over {\\sqrt 6 }}$$ $$-$$ 2"}, {"identifier": "C", "content": "$${4\\sqrt 6 }$$ $$-$$ 1"}, {"identifier": "D", "content": "$${4\\sqrt 6 }$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3b0ez3v/5fa06fb6-2e73-4b95-aa16-dacb000be7c4/83af90b0-d658-11ec-9a06-bd4ec5b93eb4/file-1l3b0ez3w.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3b0ez3v/5fa06fb6-2e73-4b95-aa16-dacb000be7c4/83af90b0-d658-11ec-9a06-bd4ec5b93eb4/file-1l3b0ez3w.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - Hyperbola Question 41 English Explanation"></p>
<p>Given,</p>
<p>$${x^2} - 2{y^2} = 4$$</p>
<p>$$ \Rightarrow {{{x^2}} \over 4} - {{{y^2}} \over 2} = 1 \Rightarrow {{{x^2}} \over {{{(2)}^2}}} - {{{y^2}} \over {{{(\sqrt 2 )}^2}}} = 1$$</p>
<p>Here, a = 2, $$b = \sqrt 2 $$</p>
<p>$$\therefore$$ $$e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {2 \over 4}} = \sqrt {1 + {1 \over 2}} = \sqrt {{3 \over 2}} $$</p>
<p>So, Focus (F) = ($$\pm$$ a e, 0) = ($$\pm$$ $$\sqrt 6 $$, 0)</p>
<p>Now, equation of tangent at $$P(4,\sqrt 6 )$$ is</p>
<p>$$x{x_1} - 2y{y_1} = 4$$</p>
<p>$$ \Rightarrow x\,.\,4 - 2y\,.\,\sqrt 6 = 4$$</p>
<p>$$ \Rightarrow 4x - 2\sqrt 6 y = 4$$</p>
<p>$$ \Rightarrow 2x - \sqrt 6 y = 2$$ ....... (i)</p>
<p>Putting y = 0 in Eq. (i), we get x-intercept of tangent i.e. x = 1</p>
<p>$$\therefore$$ Q $$\equiv$$ (1, 0)</p>
<p>Hence, equation of corresponding latus rectum is $$x = \sqrt 6 $$</p>
<p>$$\therefore$$ $$R \equiv \left( {\sqrt 6 ,{{2(\sqrt 6 - 1)} \over {\sqrt 6 }}} \right)$$ [putting $$x = \sqrt 6 $$ in Eq. (i), we get $$y = {{2(\sqrt 6 - 1)} \over {\sqrt 6 }}$$]</p>
<p>$$\therefore$$ Area of $$\Delta QFR = {1 \over 2} \times (QF) \times (RF)$$</p>
<p>$$ = {1 \over 2}(\sqrt 6 - 1) \times {{2(\sqrt 6 - 1)} \over {\sqrt 6 }} = {{{{(\sqrt 6 - 1)}^2}} \over {\sqrt 6 }} = \left( {{7 \over {\sqrt 6 }} - 2} \right)$$</p> | mcq | jee-main-2021-online-18th-march-evening-shift |
1krua3nf6 | maths | hyperbola | tangent-to-hyperbola | Let a line L : 2x + y = k, k > 0 be a tangent to the hyperbola x<sup>2</sup> $$-$$ y<sup>2</sup> = 3. If L is also a tangent to the parabola y<sup>2</sup> = $$\alpha$$x, then $$\alpha$$ is equal to : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "$$-$$12"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "$$-$$24"}] | ["D"] | null | Tangent to hyperbola of <br><br>Slope m = $$-$$2 (given)<br><br>y = $$-$$2x $$\pm$$ $$\sqrt {3(3)} $$<br><br>$$\left( {y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} } \right)$$<br><br>$$\Rightarrow$$ y + 2x = $$\pm$$ 3 $$\Rightarrow$$ 2x + y = 3 (k > 0)<br><br>For parabola y<sup>2</sup> = $$\alpha$$x<br><br>$$y = mx + {\alpha \over {4m}}$$<br><br>$$ \Rightarrow y = - 2x + {\alpha \over { - 8}}$$<br><br>$$ \Rightarrow {\alpha \over { - 8}} = 3$$<br><br>$$ \Rightarrow \alpha = - 24$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1l566s5yb | maths | hyperbola | tangent-to-hyperbola | <p>Let the eccentricity of the hyperbola $$H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$ be $$\sqrt {{5 \over 2}} $$ and length of its latus rectum be $$6\sqrt 2 $$. If $$y = 2x + c$$ is a tangent to the hyperbola H, then the value of c<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "32"}] | ["B"] | null | <p>$$1 + {{{b^2}} \over {{a^2}}} = {5 \over 2} \Rightarrow {{{b^2}} \over {{a^2}}} = {3 \over 2}$$</p>
<p>$${{2{b^2}} \over a} = 6\sqrt 2 \Rightarrow 2.\,{3 \over 2}.\,a = 6\sqrt 2 $$</p>
<p>$$ \Rightarrow a = 2\sqrt 2 ,\,{b^2} = 12$$</p>
<p>$${c^2} = {a^2}{m^2} - {b^2} = 8.4 - 12 = 20$$</p> | mcq | jee-main-2022-online-28th-june-morning-shift |
1l58h85nt | maths | hyperbola | tangent-to-hyperbola | <p>Let a line L<sub>1</sub> be tangent to the hyperbola $${{{x^2}} \over {16}} - {{{y^2}} \over 4} = 1$$ and let L<sub>2</sub> be the line passing through the origin and perpendicular to L<sub>1</sub>. If the locus of the point of intersection of L<sub>1</sub> and L<sub>2</sub> is $${({x^2} + {y^2})^2} = \alpha {x^2} + \beta {y^2}$$, then $$\alpha$$ + $$\beta$$ is equal to _____________.</p> | [] | null | 12 | <p>Equation of L<sub>1</sub> is</p>
<p>$${{x\sec \theta } \over 4} - {{y\tan \theta } \over 2} = 1$$ ..... (i)</p>
<p>Equation of line L<sub>2</sub> is</p>
<p>$${{x\tan \theta } \over 2} + {{y\sec \theta } \over 4} = 0$$ ..... (ii)</p>
<p>$$\because$$ Required point of intersection of L<sub>1</sub> and L<sub>2</sub> is (x<sub>1</sub>, y<sub>1</sub>) then</p>
<p>$${{{x_1}\sec \theta } \over 4} - {{{y_1}\tan \theta } \over 2} - 1 = 0$$ ...... (iii)</p>
<p>and $${{{y_1}\sec \theta } \over 4} - {{{x_1}\tan \theta } \over 2} = 0$$ ...... (iv)</p>
<p>From equations (iii) and (iv)</p>
<p>$$\sec \theta = {{4{x_1}} \over {x_1^2 + y_1^2}}$$ and $$\tan \theta = {{ - 2{y_1}} \over {x_1^2 + y_1^2}}$$</p>
<p>$$\therefore$$ Required locus of (x<sub>1</sub>, y<sub>1</sub>) is</p>
<p>$${({x^2} + {y^2})^2} = 16{x^2} - 4{y^2}$$</p>
<p>$$\therefore$$ $$\alpha$$ = 16, $$\beta$$ = $$-$$4</p>
<p>$$\therefore$$ $$\alpha$$ + $$\beta$$ = 12</p> | integer | jee-main-2022-online-26th-june-evening-shift |
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