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luxwdx79 | maths | parabola | position-of-point-and-chord-joining-of-two-points | <p>Consider the circle $$C: x^2+y^2=4$$ and the parabola $$P: y^2=8 x$$. If the set of all values of $$\alpha$$, for which three chords of the circle $$C$$ on three distinct lines passing through the point $$(\alpha, 0)$$ are bisected by the parabola $$P$$ is the interval $$(p, q)$$, then $$(2 q-p)^2$$ is equal to __________.</p> | [] | null | 80 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1nqf2u/3f5eb39b-71fd-4a3c-a3b6-2db1d6eda7bc/23637760-0f56-11ef-91cd-f19f7dc20f18/file-1lw1nqf2v.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1nqf2u/3f5eb39b-71fd-4a3c-a3b6-2db1d6eda7bc/23637760-0f56-11ef-91cd-f19f7dc20f18/file-1lw1nqf2v.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) 9th April Evening Shift Mathematics - Parabola Question 8 English Explanation"></p>
<p>Chord with the middle point $$(\alpha, 0)$$</p>
<p>$$\begin{aligned}
& \Rightarrow T=S_1 \\
& \Rightarrow y y_1-4\left(x+x_1\right)=y_1^2-8 x_1 \\
& \Rightarrow-4(x+\alpha)=0-8 \alpha \\
& \Rightarrow x+\alpha=2 \alpha \Rightarrow x=\alpha
\end{aligned}$$</p>
<p>For circle chord with $$(2 t^2, 4 t)$$ as mid point</p>
<p>$$\begin{aligned}
& \Rightarrow \quad T=S_1 \\
& \Rightarrow \quad x x_1+y y_1-4=x_1^2+y_1^2-4 \\
& \Rightarrow \quad 2 t^2 x+4 t y=4 t^4+16 t^2
\end{aligned}$$</p>
<p>Passes through $$(\alpha, 0)$$</p>
<p>$$\begin{aligned}
\Rightarrow & 2 t^2 \alpha=4 t^4+16 t^2 \\
\Rightarrow & 2 \alpha=4 t^2+16 \Rightarrow \alpha=2 t^2+8=x_0+8 \\
& x^2+y^2=4 \text { and } y^2=8 x \\
\Rightarrow & x^2+8 x-4=0 \Rightarrow x_0=\frac{-8+\sqrt{80}}{2} \\
\Rightarrow & p=8 \text { and } q=4+\frac{\sqrt{80}}{2} \Rightarrow(2 q-p)^2=80
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-evening-shift |
9lMAXEX3SRWedqek | maths | parabola | question-based-on-basic-definition-and-parametric-representation | The locus of the vertices of the family of parabolas
<br/>$$y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a$$ is : | [{"identifier": "A", "content": "$$xy = {{105} \\over {64}}$$ "}, {"identifier": "B", "content": "$$xy = {{3} \\over {4}}$$"}, {"identifier": "C", "content": "$$xy = {{35} \\over {16}}$$"}, {"identifier": "D", "content": "$$xy = {{64} \\over {105}}$$"}] | ["A"] | null | Given parabola is $$y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a$$
<br><br>$$ \Rightarrow y = {{{a^3}} \over 3}\left( {{x^3} + {3 \over {2a}} + x + {9 \over {16{a^2}}}} \right) - {{3a} \over {16}} - 2a$$
<br><br>$$ \Rightarrow y + {{35a} \over {16}} = {{{a^3}} \over 3}{\left( {x + {3 \over {4a}}} \right)^2}$$
<br><br>$$\therefore$$ Vertex of parabola is $$\left( {{{ - 3} \over {4a}},{{ - 35a} \over {16}}} \right)$$
<br><br>To find locus of this vertex,
<br><br>$$x = {{ - 3} \over {4a}}\,\,$$ and $$\,\,y = {{ - 35a} \over {16}}$$
<br><br>$$ \Rightarrow a = {{ - 3} \over {4x}}\,\,$$ and $$a = - {{16y} \over {35}}$$
<br><br>$$ \Rightarrow {{ - 3} \over {4x}} = {{ - 16y} \over {35}} \Rightarrow 64xy = 105$$
<br><br>$$ \Rightarrow xy = {{105} \over {64}}$$ which is the required locus. | mcq | aieee-2006 |
xfL4pvtMgdFvKaeW | maths | parabola | question-based-on-basic-definition-and-parametric-representation | A parabola has the origin as its focus and the line $$x=2$$ as the directrix. Then the vertex of the parabola is at : | [{"identifier": "A", "content": "$$(0,2)$$ "}, {"identifier": "B", "content": "$$(1,0)$$ "}, {"identifier": "C", "content": "$$(0,1)$$ "}, {"identifier": "D", "content": "$$(2,0)$$ "}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266932/exam_images/otkuicrgj5mighgdx4s8.webp" loading="lazy" alt="AIEEE 2008 Mathematics - Parabola Question 114 English Explanation">
<br><br>Vertex of a parabola is the mid point of focus and the point
<br><br>where directrix meets the axis of the parabola.
<br><br>Here focus is $$O\left( {0,0} \right)$$ and directrix meets the axis at $$B\left( {2,0} \right)$$
<br><br>$$\therefore$$ Vertex of the parabola is $$(1,0)$$ | mcq | aieee-2008 |
ae6Mbgijd5ZpQ7Tgz9vBJ | maths | parabola | question-based-on-basic-definition-and-parametric-representation | Axis of a parabola lies along x-axis. If its vertex and focus are at distances 2 and 4 respectively from the
origin, on the positive x-axis then which of the following points does not lie on it? | [{"identifier": "A", "content": "(5, 2$$\\sqrt 6$$) "}, {"identifier": "B", "content": "(6, 4$$\\sqrt 2$$) "}, {"identifier": "C", "content": "(8, 6)"}, {"identifier": "D", "content": "(4, -4)"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263284/exam_images/idjgwu7hbabliruceroz.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Parabola Question 101 English Explanation">
<br><br>So the equation of the parabola,
<br><br>$${\left( {y - 0} \right)^2} = 4.a\left( {x - 2} \right)$$
<br><br>$$ \Rightarrow $$ y<sup>2</sup> = 4.2 (x $$-$$ 2)
<br><br>$$ \Rightarrow $$ y<sup>2</sup> = 8 (x $$-$$ 2)
<br><br>By checking each options you can see. point (8, 6) does not lie on the parabola. | mcq | jee-main-2019-online-9th-january-morning-slot |
F5sELRdLShJaMlbMoRhND | maths | parabola | question-based-on-basic-definition-and-parametric-representation | If $$\theta $$ denotes the acute angle between the curves, y = 10 – x<sup>2</sup> and y = 2 + x<sup>2</sup> at a point of their intersection, the |tan $$\theta $$| is equal to : | [{"identifier": "A", "content": "$$8 \\over 15$$"}, {"identifier": "B", "content": "$$4 \\over 9$$"}, {"identifier": "C", "content": "$$7 \\over 17$$"}, {"identifier": "D", "content": "$$8 \\over 17$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267822/exam_images/urqnnuedt8c4vkagbrve.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Morning Slot Mathematics - Parabola Question 100 English Explanation">
<br><br>Angle between the curves is the acute angle between the tangents at the point of intersection.
<br><br>y = 10 $$-$$ x<sup>2</sup> (for curve 1)
<br><br>and y = 2 + x<sup>2</sup> (for curve 2)
<br><br>$$ \therefore $$ 10 $$-$$ x<sup>2</sup> = 2 + x<sup>2</sup>
<br><br>$$ \Rightarrow $$ 2x<sup>2</sup> = 8
<br><br>$$ \Rightarrow $$ x<sup>2</sup> = 4
<br><br>$$ \Rightarrow $$ x = 2, $$-$$ 2
<br><br>$$ \therefore $$ points of intersection (2, 6) and ($$-$$ 2, 6)
<br><br>$${{dy} \over {dx}}$$ for curve 1 = $$-$$ 2x
<br><br>$$ \therefore $$ Slope(m<sub>1</sub>) of curve 1 is = $$-$$ 2(2) = $$-$$ 4
<br><br>$${{dy} \over {dx}}$$ for curve 2 = 2x
<br><br>$$ \therefore $$ slope (m<sub>2</sub>) of curve 2 = 2 $$ \times $$ 2 = 4
<br><br>$$ \therefore $$ tan$$\theta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$
<br><br>= $$\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|$$
<br><br>= $${8 \over {15}}$$ | mcq | jee-main-2019-online-9th-january-morning-slot |
1ktiqgayf | maths | parabola | question-based-on-basic-definition-and-parametric-representation | The length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (> R) respectively from the origin, is : | [{"identifier": "A", "content": "4(S + R)"}, {"identifier": "B", "content": "2(S $$-$$ R)"}, {"identifier": "C", "content": "4(S $$-$$ R)"}, {"identifier": "D", "content": "2(S + R)"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264985/exam_images/rrktibvdvq5qvsqqrcjk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Mathematics - Parabola Question 57 English Explanation"><br><br>V $$\to$$ Vertex<br><br>F $$\to$$ focus<br><br>VF = S $$-$$ R<br><br>So, latus rectum = 4(S $$-$$ R) | mcq | jee-main-2021-online-31st-august-morning-shift |
1l55ilnlq | maths | parabola | question-based-on-basic-definition-and-parametric-representation | <p>If vertex of a parabola is (2, $$-$$1) and the equation of its directrix is 4x $$-$$ 3y = 21, then the length of its latus rectum is :</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "16"}] | ["B"] | null | <p>Vertex of Parabola : (2, $$-$$1)</p>
<p>and directrix : 4x $$-$$ 3y = 21</p>
<p>Distance of vertex from the directrix</p>
<p>$$a = \left| {{{8 + 3 - 21} \over {\sqrt {25} }}} \right| = 2$$</p>
<p>$$\therefore$$ length of latus rectum = 4a = 8</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l56r6xzv | maths | parabola | question-based-on-basic-definition-and-parametric-representation | <p>If the equation of the parabola, whose vertex is at (5, 4) and the directrix is $$3x + y - 29 = 0$$, is $${x^2} + a{y^2} + bxy + cx + dy + k = 0$$, then $$a + b + c + d + k$$ is equal to :</p> | [{"identifier": "A", "content": "575"}, {"identifier": "B", "content": "$$-$$575"}, {"identifier": "C", "content": "576"}, {"identifier": "D", "content": "$$-$$576"}] | ["D"] | null | <p>Given vertex is (5, 4) and directrix 3x + y $$-$$ 29 = 0</p>
<p>Let foot of perpendicular of (5, 4) on directrix is (x<sub>1</sub>, y<sub>1</sub>)</p>
<p>$${{{x_1} - 5} \over 3} = {{{y_1} - 4} \over 1} = {{ - ( - 10)} \over {10}}$$</p>
<p>$$\therefore$$ $$({x_1},\,{y_1}) \equiv (8,\,5)$$</p>
<p>So, focus of parabola will be $$S = (2,3)$$</p>
<p>Let P(x, y) be any point on parabola, then</p>
<p>$${(x - 2)^2} + {(y - 3)^2} = {{{{(3x + y - 29)}^2}} \over {10}}$$</p>
<p>$$ \Rightarrow 10({x^2} + {y^2} - 4x - 6y + 13) = 9{x^2} + {y^2} + 841 + 6xy - 58y - 174x$$</p>
<p>$$ \Rightarrow {x^2} + 9{y^2} - 6xy + 134x - 2y - 711 = 0$$</p>
<p>and given parabola</p>
<p>$${x^2} + a{y^2} + bxy + cx + dy + k = 0$$</p>
<p>$$\therefore$$ a = 9, b = $$-$$6, c = 134, d = $$-$$2, k = $$-$$711</p>
<p>$$\therefore$$ $$a + b + c + d + k = - 576$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l5aj58pl | maths | parabola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$x = 2t$$, $$y = {{{t^2}} \over 3}$$ be a conic. Let S be the focus and B be the point on the axis of the conic such that $$SA \bot BA$$, where A is any point on the conic. If k is the ordinate of the centroid of the $$\Delta$$SAB, then $$\mathop {\lim }\limits_{t \to 1} k$$ is equal to :</p> | [{"identifier": "A", "content": "$${{17} \\over {18}}$$"}, {"identifier": "B", "content": "$${{19} \\over {18}}$$"}, {"identifier": "C", "content": "$${{11} \\over {18}}$$"}, {"identifier": "D", "content": "$${{13} \\over {18}}$$"}] | ["D"] | null | <p>$$x = 2t,\,y = {2 \over 3}$$</p>
<p>$$t \to 1\,\,\,A \equiv \left( {2,{1 \over 3}} \right)$$</p>
<p>Given conic is $${x^2} = 12y \Rightarrow S \equiv (0,3)$$</p>
<p>Let $$B \equiv (0,\beta )$$</p>
<p>Given $$SA\, \bot \,BA$$</p>
<p>$$\left( {{{{1 \over 3}} \over {2 - 3}}} \right)\left( {{{\beta - {1 \over 3}} \over { - 2}}} \right) = - 1$$</p>
<p>$$ \Rightarrow \left( {\beta - {1 \over 3}} \right){1 \over 3} = - 2$$</p>
<p>$$ \Rightarrow \beta = {1 \over 3}\left( {{{ - 17} \over 3}} \right)$$</p>
<p>$$\mathop {Ordinate\,of\,centroid}\limits_{(as\,t \to 1)} = k = {{\beta + {1 \over 3} + 3} \over 3}$$</p>
<p>$$ = {{{{ - 17} \over 9} + {{10} \over 3}} \over 3} = {{13} \over {18}}$$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5b86e0n | maths | parabola | question-based-on-basic-definition-and-parametric-representation | <p>A particle is moving in the xy-plane along a curve C passing through the point (3, 3). The tangent to the curve C at the point P meets the x-axis at Q. If the y-axis bisects the segment PQ, then C is a parabola with :</p> | [{"identifier": "A", "content": "length of latus rectum 3"}, {"identifier": "B", "content": "length of latus rectum 6"}, {"identifier": "C", "content": "focus $$\\left( {{4 \\over 3},0} \\right)$$"}, {"identifier": "D", "content": "focus $$\\left( {0,{3 \\over 4}} \\right)$$"}] | ["A"] | null | <p>According to the question (Let P(x, y))</p>
<p>$$2x - y{{dx} \over {dy}} = 0$$</p>
<p>($$\because$$ equation of tangent at $$P:y - y = {{dy} \over {dx}}(y - x)$$)</p>
<p>$$\therefore$$ $$2{{dy} \over {y}} = {{dx} \over x}$$</p>
<p>$$ \Rightarrow 2\ln y = \ln x + \ln c$$</p>
<p>$$ \Rightarrow {y^2} = cx$$</p>
<p>$$\because$$ this curve passes through (3, 3)</p>
<p>$$\therefore$$ c = 3</p>
<p>$$\therefore$$ required parabola $${y^2} = 3x$$ and L.R = 3</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5bbcbco | maths | parabola | question-based-on-basic-definition-and-parametric-representation | <p>Let P<sub>1</sub> be a parabola with vertex (3, 2) and focus (4, 4) and P<sub>2</sub> be its mirror image with respect to the line x + 2y = 6. Then the directrix of P<sub>2</sub> is x + 2y = ____________.</p> | [] | null | 10 | <p>Focus = (4, 4) and vertex = (3, 2)</p>
<p>$$\therefore$$ Point of intersection of directrix with axis of parabola = A = (2, 0)</p>
<p>Image of A(2, 0) with respect to line x + 2y = 6 is B(x<sub>2</sub>, y<sub>2</sub>)</p>
<p>$$\therefore$$ $${{{x_2} - 2} \over 1} = {{{y_2} - 0} \over 2} = {{ - 2(2 + 0 - 6)} \over 5}$$</p>
<p>$$\therefore$$ $$B({x_2},\,{y_2}) = \left( {{{18} \over 5},{{16} \over 5}} \right)$$.</p>
<p>Point B is point of intersection of direction with axes of parabola P<sub>2</sub>.</p>
<p>$$\therefore$$ $$x + 2y = \lambda $$ must have point $$\left( {{{18} \over 5},{{16} \over 5}} \right)$$</p>
<p>$$\therefore$$ $$x + 2y = 10$$</p> | integer | jee-main-2022-online-24th-june-evening-shift |
1l6hxz4yr | maths | parabola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$\mathrm{P}$$ and $$\mathrm{Q}$$ be any points on the curves $$(x-1)^{2}+(y+1)^{2}=1$$ and $$y=x^{2}$$, respectively. The distance between $$P$$ and $$Q$$ is minimum for some value of the abscissa of $$P$$ in the interval :</p> | [{"identifier": "A", "content": "$$\\left(0, \\frac{1}{4}\\right)$$"}, {"identifier": "B", "content": "$$\\left(\\frac{1}{2}, \\frac{3}{4}\\right)$$"}, {"identifier": "C", "content": "$$\\left(\\frac{1}{4}, \\frac{1}{2}\\right)$$"}, {"identifier": "D", "content": "$$\\left(\\frac{3}{4}, 1\\right)$$"}] | ["C"] | null | <p>$$y = mx + 2a + {1 \over {{m^2}}}$$ (Equation of normal to $${x^2} = 4ay$$ in slope form) through $$(1, - 1)$$.</p>
<p>$$4{m^3} + 6{m^2} + 1 = 0$$</p>
<p>$$ \Rightarrow m \simeq - 1.6$$</p>
<p>Slope of normal $$ \simeq {{ - 8} \over 5} = \tan \theta $$</p>
<p>$$ \Rightarrow \cos \theta \simeq {{ - 5} \over {\sqrt {89} }},\,\sin \theta \simeq {8 \over {\sqrt {89} }}$$</p>
<p>$${x_p} = 1 + \cos \theta \simeq 1 - {5 \over {\sqrt {89} }} \in \left( {{1 \over 4},{1 \over 2}} \right)$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
luxwclgg | maths | parabola | question-based-on-basic-definition-and-parametric-representation | <p>Let $$A, B$$ and $$C$$ be three points on the parabola $$y^2=6 x$$ and let the line segment $$A B$$ meet the line $$L$$ through $$C$$ parallel to the $$x$$-axis at the point $$D$$. Let $$M$$ and $$N$$ respectively be the feet of the perpendiculars from $$A$$ and $$B$$ on $$L$$. Then $$\left(\frac{A M \cdot B N}{C D}\right)^2$$ is equal to __________.</p> | [] | null | 36 | <p>Equation of $$A B$$</p>
<p>$$y\left(t_1+t_2\right)=2 x+2 a t_1 t_2$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1p45wf/b8641c5f-23f4-4335-a99c-bb1d21171afc/8ac176f0-0f5b-11ef-8792-3d19bf2e18a4/file-1lw1p45wg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1p45wf/b8641c5f-23f4-4335-a99c-bb1d21171afc/8ac176f0-0f5b-11ef-8792-3d19bf2e18a4/file-1lw1p45wg.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) 9th April Evening Shift Mathematics - Parabola Question 9 English Explanation"></p>
<p>$$\begin{aligned}
& \text { For } D, y=2 a t_3 \\
& \Rightarrow x=a\left(t_1 t_3+t_2 t_3-t_1 t_2\right) \\
& C D=\left|a\left(t_1 t_3+t_2 t_3-t_1 t_3\right)-a t_3^2\right| \\
& A M=\left|2 a t_1-2 a t_3\right| \\
& B N=\left|2 a t_3-2 a t_2\right| \\
& \left(\frac{A M \cdot B N}{C D}\right)^2=\left(\frac{4 a^2\left(t_1-t_3\right)\left(t_3-t_2\right)}{a\left(t_1 t_3+t_2 t_3-t_1 t_3-t_3^2\right)}\right)^2 \\
& =\left(\frac{4 a^2\left(t_1-t_3\right)\left(t_3-t_2\right)}{a\left(t_1-t_3\right)\left(t_2-t_3\right)}\right)^2 \\
& =16 a^2=16 \cdot\left(\frac{3}{2}\right)^2=36
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-evening-shift |
KO1xfIr2aFjLe79Q | maths | parabola | tangent-to-parabola | The equation of a tangent to the parabola $${y^2} = 8x$$ is $$y=x+2$$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is : | [{"identifier": "A", "content": "$$(2,4)$$"}, {"identifier": "B", "content": "$$(-2,0)$$"}, {"identifier": "C", "content": "$$(-1,1)$$"}, {"identifier": "D", "content": "$$(0,2)$$"}] | ["B"] | null | Parabola $${y^2} = 8x$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263639/exam_images/o1prumkf5cogacvzecia.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Parabola Question 115 English Explanation">
<br><br>We know that the locus of point of intersection of two perpendicular tangents to a
<br><br>parabola is its directrix. Point must be on the directrix of parabola
<br><br>as equation of directrix $$x + 2 = 0 \Rightarrow x = - 2$$
<br><br>Hence the point is $$\left( { - 2,0} \right)$$ | mcq | aieee-2007 |
D3oDCkmIjGiT6by7 | maths | parabola | tangent-to-parabola | The slope of the line touching both the parabolas $${y^2} = 4x$$ and $${x^2} = - 32y$$ is | [{"identifier": "A", "content": "$${{1 \\over 8}}$$"}, {"identifier": "B", "content": "$${{2 \\over 3}}$$"}, {"identifier": "C", "content": "$${{1 \\over 2}}$$"}, {"identifier": "D", "content": "$${{3 \\over 2}}$$"}] | ["C"] | null | Let tangent to $${y^2} = 4x$$ be $$y = mx + {1 \over m}$$
<br><br>Since this is also tangent to $${x^2} = - 32y$$
<br><br>$$\therefore$$ $${x^2} = - 32\left( {mx + {1 \over m}} \right)$$
<br><br>$$ \Rightarrow {x^2} + 32mx + {{32} \over m} = 0$$
<br><br>Now, $$D=0$$
<br><br>$${\left( {32} \right)^2} - 4\left( {{{32} \over m}} \right) = 0$$
<br><br>$$ \Rightarrow {m^3} = {4 \over {32}} \Rightarrow m = {1 \over 2}$$ | mcq | jee-main-2014-offline |
xsLUk49wH4UD7e5KxYZQp | maths | parabola | tangent-to-parabola | Let P be a point on the parabola, x<sup>2</sup> = 4y. If the distance of P from the center of the circle, x<sup>2</sup> + y<sup>2</sup> + 6x + 8 = 0 is minimum, then the equation of the tangent to the parabola at P, is : | [{"identifier": "A", "content": "x + 4y $$-$$ 2 = 0"}, {"identifier": "B", "content": "x $$-$$ y + 3 = 0"}, {"identifier": "C", "content": "x + y +1 = 0"}, {"identifier": "D", "content": "x + 2y = 0"}] | ["C"] | null | Let P(2t, t<sup>2</sup>) be any point on the parabola.
<br><br>Center of the given circle C = ($$-$$ g, $$-$$f) = ($$-$$3, 0)
<br><br>For PC to be minimum, it must be the normal to the parabola at P.
<br><br>Slope of line PC = $${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$$ = $${{{t^2} - 0} \over {2t + 3}}$$
<br><br>Also, slope of tangent to parabola at P = $${{dy} \over {dx}}$$ = $${x \over 2}$$ = t
<br><br>$$ \therefore $$ Slope of normal = $${{ - 1} \over t}$$
<br><br>$$ \therefore $$ $${{{t^2} - 0} \over {2t + 3}}$$ = $${{ - 1} \over t}$$
<br><br>$$ \Rightarrow $$ t<sup>3</sup> + 2t + 3 = 0
<br><br>$$ \Rightarrow $$ (t+1) (t<sup>2</sup> $$-$$ t + 3) = 0
<br><br>$$\therefore\,\,\,$$ Real roots of above equation is
<br><br>t = $$-$$ 1
<br><br>Coordinate of P = (2t, t<sup>2</sup>) = ($$-$$2, 1)
<br><br>Slope of tangent to parabola at P = t = $$-$$ 1
<br><br>Therefore, equation of tangent is :
<br><br>(y $$-$$ 1) = ($$-$$ 1) (x + 2)
<br><br>$$ \Rightarrow $$ x + y + 1 = 0 | mcq | jee-main-2018-online-16th-april-morning-slot |
tfGX1dpaJzWJ28zbkVzsK | maths | parabola | tangent-to-parabola | The equation of a tangent to the parabola, x<sup>2</sup>
= 8y, which makes an angle $$\theta $$ with the positive directions of x-axis, is : | [{"identifier": "A", "content": "x = y cot $$\\theta $$ \u2013 2 tan $$\\theta $$"}, {"identifier": "B", "content": "y = x tan $$\\theta $$ + 2 cot $$\\theta $$"}, {"identifier": "C", "content": "x = y cot $$\\theta $$ + 2 tan $$\\theta $$"}, {"identifier": "D", "content": "y = x tan $$\\theta $$ \u2013 2 cot $$\\theta $$\n"}] | ["C"] | null | x<sup>2</sup> = 8y
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}} = {x \over 4} = \tan \theta $$
<br><br>$$ \therefore $$ x<sub>1</sub> = 4tan$$\theta $$
<br><br>y<sub>1</sub> = 2 tan<sup>2</sup> $$\theta $$
<br><br>Equation of tangent :-
<br><br>y $$-$$ 2tan<sup>2</sup>$$\theta $$ = tan$$\theta $$ (x $$-$$ 4tan $$\theta $$)
<br><br>$$ \Rightarrow $$ x = y cot $$\theta $$ + 2 tan $$\theta $$ | mcq | jee-main-2019-online-12th-january-evening-slot |
EqKHlIkAheyQcWlaj6wsQ | maths | parabola | tangent-to-parabola | The shortest distance between the line y = x and
the curve y<sup>2</sup> = x – 2 is : | [{"identifier": "A", "content": "$$7\\over 4 \\sqrt2$$"}, {"identifier": "B", "content": "$$7\\over8$$"}, {"identifier": "C", "content": "$$11\\over 4 \\sqrt2$$"}, {"identifier": "D", "content": "2"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265259/exam_images/zea3n7u2udoo4a4xhtuv.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264552/exam_images/fmvqf3yjti1z1cuq9efe.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263645/exam_images/qbyz275lq6l05oar7ina.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Parabola Question 91 English Explanation"></picture>
Slope of the line y = x is 1.
<br><br>$$ \therefore $$ The shortest distance between line y = x and
parabola y<sup>2</sup> = x – 2 is the distance between line y = x and
tangent of parabola having slope 1.
<br><br>Slope of tangent at point P on the parabola y<sup>2</sup> = x – 2 is,
<br><br>2y$${{dy} \over {dx}}$$ = 1
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${1 \over {2y}}$$ = slope
<br><br>$$ \therefore $$ $${1 \over {2y}}$$ = 1
<br><br>$$ \Rightarrow $$ y = $${1 \over 2}$$
<br><br>$$ \therefore $$ x coordinate of point P is,
<br><br>$${1 \over 4}$$ = x – 2
<br><br>$$ \Rightarrow $$ x = $${9 \over 4}$$
<br><br>Shortest distance = length of perpendicular from P on line x – y = 0
<br><br>= $${{\left| {{9 \over 4} - {1 \over 2}} \right|} \over {\sqrt {{1^2} + {1^2}} }}$$ = $${7 \over {4\sqrt 2 }}$$ | mcq | jee-main-2019-online-8th-april-morning-slot |
94sPnL19Hkkpr7KAvIN4n | maths | parabola | tangent-to-parabola | The tangent to the parabola y<sup>2</sup>
= 4x at the point
where it intersects the circle x<sup>2</sup>
+ y<sup>2</sup>
= 5 in the
first quadrant, passes through the point :
| [{"identifier": "A", "content": "$$\\left( { - {1 \\over 4},{1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 3},{4 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { {3 \\over 4},{7 \\over 4}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { {1 \\over 4},{3 \\over 4}} \\right)$$"}] | ["C"] | null | Parabola y<sup>2</sup>
= 4x and circle x<sup>2</sup>
+ y<sup>2</sup>
= 5 intersect with each other.
<br><br>So, x<sup>2</sup> + 4x = 5
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + 5x – x – 5 = 0
<br><br>$$ \Rightarrow $$ x(x + 5) –1(x + 5) = 0
<br><br> x = 1, –5
<br><br>Intersection point in 1<sup>st</sup> quadrant is = (1, 2)
<br><br>Equation of tangent to y<sup>2</sup> = 4x at (1, 2) is
<br><br>y(2) = 2 (x + 1)
<br><br>$$ \Rightarrow $$ y = x + 1 .....(1)
<br><br>By checking each options, you can see
<br><br>point $$\left( { {3 \over 4},{7 \over 4}} \right)$$ lies on equation (1).
| mcq | jee-main-2019-online-8th-april-evening-slot |
OqPt1vG8Q2JLAFLUzI3rsa0w2w9jxaz5osm | maths | parabola | tangent-to-parabola | The tangents to the curve y = (x – 2)<sup>2</sup> – 1 at its points of intersection with the line x – y = 3, intersect at the point :
| [{"identifier": "A", "content": "$$\\left( {{5 \\over 2}, - 1} \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - {5 \\over 2}, - 1} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{5 \\over 2},1} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {5 \\over 2},1} \\right)$$"}] | ["A"] | null | Let the coordinates of C be (h, k)<br><br>
So the chord of contact of C w.r.t y = (x-2)<sup>2</sup> - 1 <br><br>
$$ \Rightarrow $$ y = x<sup>2</sup> - 4x + 3 is T = 0<br><br>
$$y+k \over 2$$ = xh + 3 - 2(x+h)<br><br>
$$ \Rightarrow $$ 2(h-2)x - y = -(6-4h-k)<br><br>
On comparing it with x - y = 3<br><br>
2 (h -2) = 1 $$ \Rightarrow $$ h = $$5 \over 2$$<br><br>
6 - 4h - k = -3 $$ \Rightarrow $$ k = -1<br><br>
$$ \therefore $$ C = ($$5 \over 2$$, -1) | mcq | jee-main-2019-online-12th-april-evening-slot |
Qzo3zWtSFrtM4esOgF7k9k2k5e2t4o5 | maths | parabola | tangent-to-parabola | If y = mx + 4 is a tangent to both the parabolas, y<sup>2</sup> = 4x and x<sup>2</sup> = 2by, then b is equal to : | [{"identifier": "A", "content": "-128"}, {"identifier": "B", "content": "128"}, {"identifier": "C", "content": "-64"}, {"identifier": "D", "content": "-32"}] | ["A"] | null | Given y = mx + 4 is tangent to both the parabolas.
<br><br>$$ \therefore $$ Applying condition of tangent
for y<sup>2</sup>
= 4x, we get
<br><br>$${1 \over m}$$ = 4
<br><br>$$ \Rightarrow $$ m = $${1 \over 4}$$
<br><br>For x<sup>2</sup>
= 2by line y = $${x \over 4}$$ + 4 is tangent
<br><br>$$ \therefore $$ x<sup>2</sup> = 2b$$\left( {{x \over 4} + 4} \right)$$
<br><br>$$ \Rightarrow $$ x<sup>2</sup> - $${{bx} \over 2}$$ - 8b = 0
<br><br>For tangent to the parabola Discriminant = 0
<br><br>$$ \Rightarrow $$ $${{{b^2}} \over 4}$$ + 32b = 0
<br><br>$$ \Rightarrow $$ b = -128 | mcq | jee-main-2020-online-7th-january-morning-slot |
rfhigTYGjSTU3SfhyM7k9k2k5hkcpya | maths | parabola | tangent-to-parabola | Let a line y = mx (m > 0) intersect the parabola,
y<sup>2</sup> = x at a point P, other than the origin. Let
the tangent to it at P meet the x-axis at the point
Q. If area ($$\Delta $$OPQ) = 4 sq. units, then m is equal
to __________. | [] | null | 0.5 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263897/exam_images/dzexdmaraqpgczbig0ch.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Evening Slot Mathematics - Parabola Question 81 English Explanation">
<br><br>let P(t<sup>2</sup>
, t)
<br><br>Tangent at P(t<sup>2</sup>
, t)
<br><br>ty =
$${{x + {t^2}} \over 2}$$
<br><br>$$ \Rightarrow $$2ty = x + t<sup>2</sup>
<br><br>$$ \therefore $$ Q = (-t<sup>2</sup>, 0)
<br><br>Given ($$\Delta $$OPQ) = 4
<br><br>$${1 \over 2}\left| {\matrix{
0 & 0 & 1 \cr
{{t^2}} & t & 1 \cr
{ - {t^2}} & 0 & 1 \cr
} } \right|$$ = 4
<br><br>$$ \Rightarrow $$ $$\left| {{t^3}} \right|$$ = 8
<br><br>$$ \Rightarrow $$ t = 2
<br><br>and P = (4, 2)
<br><br>As y = mx
<br><br>$$ \Rightarrow $$ 2 = 4m
<br><br>$$ \Rightarrow $$ m = 0.5 | integer | jee-main-2020-online-8th-january-evening-slot |
n22hr3tMIhzdm87QTs7k9k2k5khbjd9 | maths | parabola | tangent-to-parabola | If one end of a focal chord AB of the parabola
y<sup>2</sup> = 8x is at $$A\left( {{1 \over 2}, - 2} \right)$$, then the equation of
the tangent to it at B is : | [{"identifier": "A", "content": "2x \u2013 y \u2013 24 = 0"}, {"identifier": "B", "content": "x \u2013 2y + 8 = 0"}, {"identifier": "C", "content": "x + 2y + 8 = 0"}, {"identifier": "D", "content": "2x + y \u2013 24 = 0"}] | ["B"] | null | Given parabola y<sup>2</sup>
= 8x
<br><br> $$ \therefore $$ a = 2
<br><br>Let one end of focal chord is A(at<sup>2</sup>
, 2at) = $$\left( {{1 \over 2}, - 2} \right)$$
<br><br>$$ \therefore $$ 2at = -2
<br><br>$$ \Rightarrow $$ t = $$ - {1 \over 2}$$
<br><br>Other end of focal chord will be B$$\left( {{a \over {{t^2}}}, - {{2a} \over t}} \right)$$ $$ \equiv $$ (8, 8)
<br><br>$$ \therefore $$ Equation of tangent at B is
<br><br>8y = 4(x + 8)
<br><br>$$ \Rightarrow $$ x – 2y + 8 = 0 | mcq | jee-main-2020-online-9th-january-evening-slot |
E3WPivLLU9Lq5xJKxgjgy2xukfuuryjo | maths | parabola | tangent-to-parabola | Let L<sub>1</sub>
be a tangent to the parabola y<sup>2</sup> = 4(x + 1) <br/>and L<sub>2</sub>
be a tangent to the parabola
y<sup>2</sup> = 8(x + 2) <br/>such that L<sub>1</sub>
and L<sub>2</sub>
intersect at right angles. Then L<sub>1</sub>
and L<sub>2</sub>
meet on the straight
line : | [{"identifier": "A", "content": "x + 3 = 0"}, {"identifier": "B", "content": "x + 2y = 0"}, {"identifier": "C", "content": "x + 2 = 0"}, {"identifier": "D", "content": "2x + 1 = 0"}] | ["A"] | null | L<sub>1</sub> : y<sup>2</sup> = 4(x + 1)
<br><br>Equation of tangent y = m(x + 1) + $${1 \over m}$$ ...(1)
<br><br>L<sub>2</sub> : y<sup>2</sup> = 8(x + 2)
<br><br>Equation of tangent y = m'(x + 2) + $${2 \over {m'}}$$
<br><br>$$ \Rightarrow $$ y = m'x + 2$$\left( {m' + {1 \over {m'}}} \right)$$ ....(2)
<br><br>Since lines intersect at right angles
<br><br>$$ \therefore $$ mm' = -1
<br><br>$$ \Rightarrow $$ m' = $${ - {1 \over {m}}}$$
<br><br>Putting it in equation (2)
<br><br>y = $$ - {1 \over m}x + 2\left( { - {1 \over m} - m} \right)$$
<br><br>$$ \Rightarrow $$ y = $$ - {1 \over m}x - 2\left( {m + {1 \over m}} \right)$$ ....(3)
<br><br>From equation (1) and (3)
<br><br>m(x + 1) + $${1 \over m}$$ = $$ - {1 \over m}x - 2\left( {m + {1 \over m}} \right)$$
<br><br>$$ \Rightarrow $$ $$\left( {m + {1 \over m}} \right)x + 3\left( {m + {1 \over m}} \right)$$ = 0
<br><br>$$ \therefore $$ x + 3 = 0 | mcq | jee-main-2020-online-6th-september-morning-slot |
GPFzk7LGXaUuo4pZZRjgy2xukg0ca10s | maths | parabola | tangent-to-parabola | The centre of the circle passing through the
point (0, 1) and touching the parabola <br/>y = x<sup>2</sup> at the point (2, 4) is : | [{"identifier": "A", "content": "$$\\left( {{6 \\over 5},{{53} \\over {10}}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{3 \\over {10}},{{16} \\over 5}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{{ - 53} \\over {10}},{{16} \\over 5}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{{ - 16} \\over 5},{{53} \\over {10}}} \\right)$$"}] | ["D"] | null | Circle passes through A(0, 1) and B(2, 4).
<br><br>y = x<sup>2</sup>
<br><br>$$ \Rightarrow $$ $${\left. {{{dy} \over {dx}}} \right|_B}$$ = 4
<br><br>tangent at (2,4) is
<br><br>(y – 4) = 4(x – 2)
<br><br>4x – y – 4 = 0
<br><br>Equation of circle
<br><br>(x - 2)<sup>2</sup>
+ (y–4)<sup>2</sup>
+ $$\lambda $$(4x–y - 4) = 0
<br><br>Passing through (0,1)
<br><br>$$ \therefore $$ 4 + 9 +
$$\lambda $$(–5) = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${{13} \over 5}$$
<br><br>$$ \therefore $$ Circle is
<br><br>x<sup>2</sup>– 4x + 4 + y<sup>2</sup>
– 8y + 16 +
$${{13} \over 5}$$[4x - y - 4] = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> + $$\left( {{{52} \over 5} - 4} \right)$$x - $$\left( {8 + {{13} \over 5}} \right)$$y + 20 - $${{{52} \over 5}}$$ = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + y<sup>2</sup> + $${{{32} \over 5}x - {{53} \over 5}y}$$ + $${{48} \over 5}$$ = 0
<br><br>$$ \therefore $$ Centre is $$\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)$$ | mcq | jee-main-2020-online-6th-september-evening-slot |
WDNtK3CUUqbHxvqP2a1kls4ew0l | maths | parabola | tangent-to-parabola | A tangent is drawn to the parabola y<sup>2</sup> = 6x which is perpendicular to the line 2x + y = 1. Which of the following points does NOT lie on it? | [{"identifier": "A", "content": "(0, 3)"}, {"identifier": "B", "content": "($$-$$6, 0)"}, {"identifier": "C", "content": "(4, 5)"}, {"identifier": "D", "content": "(5, 4)"}] | ["D"] | null | Equation of tangent : $$y = mx + {3 \over {2m}}$$<br><br>$${m_T} = {1 \over 2}$$ ($$\because$$ perpendicular to line $$2x + y = 1$$)<br><br>$$\therefore$$ tangent is : $$y = {x \over 2} + 3$$<br><br>$$ \Rightarrow x - 2y + 6 = 0$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
jruZYQyLd3qh7eYuel1kmix5bse | maths | parabola | tangent-to-parabola | Let C be the locus of the mirror image of a point on the parabola y<sup>2</sup> = 4x with respect to the line y = x. Then the equation of tangent to C at P(2, 1) is : | [{"identifier": "A", "content": "x $$-$$ y = 1"}, {"identifier": "B", "content": "2x + y = 5"}, {"identifier": "C", "content": "x + 3y = 5"}, {"identifier": "D", "content": "x + 2y = 4"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264947/exam_images/axoyq3cd2jeffcuzpst3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Parabola Question 66 English Explanation">
<br>Image of y<sup>2</sup> = 4x w.r.t. y = x is x<sup>2</sup> = 4y<br><br>tangent from (2, 1)<br><br>xx<sub>1</sub> = 2(y + y<sub>1</sub>)<br><br>2x = 2(y + 1)<br><br>x = y + 1 | mcq | jee-main-2021-online-16th-march-evening-shift |
1krq1l4yn | maths | parabola | tangent-to-parabola | Let y = mx + c, m > 0 be the focal chord of y<sup>2</sup> = $$-$$ 64x, which is tangent to (x + 10)<sup>2</sup> + y<sup>2</sup> = 4. Then, the value of 4$$\sqrt 2 $$ (m + c) is equal to _____________. | [] | null | 34 | y<sup>2</sup> = $$-$$64x<br><br>focus : ($$-$$16, 0)<br><br>y = mx + c is focal chord<br><br>$$\Rightarrow$$ c = 16 m ...........(1)<br><br>y = mx + c is tangent to (x + 10)<sup>2</sup> + y<sup>2</sup> = 4<br><br>$$\Rightarrow$$ y = m(x + 10) $$\pm$$ 2$$\sqrt {1 + {m^2}} $$<br><br>$$\Rightarrow$$ c = 10m $$\pm$$ 2$$\sqrt {1 + {m^2}} $$<br><br>$$\Rightarrow$$ 16m = 10m $$\pm$$ 2$$\sqrt {1 + {m^2}} $$<br><br>$$\Rightarrow$$ 6m = 2$$\sqrt {1 + {m^2}} $$ (m > 0)<br><br>$$\Rightarrow$$ 9m<sup>2</sup> = 1 + m<sup>2</sup><br><br>$$\Rightarrow$$ m = $${1 \over {2\sqrt 2 }}$$ & c = $${8 \over {\sqrt 2 }}$$<br><br>$$4\sqrt 2 (m + c) = 4\sqrt 2 \left( {{{17} \over {2\sqrt 2 }}} \right)$$ = 34 | integer | jee-main-2021-online-20th-july-morning-shift |
1krw132tm | maths | parabola | tangent-to-parabola | Let a parabola b be such that its vertex and focus lie on the positive x-axis at a distance 2 and 4 units from the origin, respectively. If tangents are drawn from O(0, 0) to the parabola P which meet P at S and R, then the area (in sq. units) of $$\Delta$$SOR is equal to : | [{"identifier": "A", "content": "$$16\\sqrt 2 $$"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "$$8\\sqrt 2 $$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264154/exam_images/jj7xl6x9z7rdt8utfb0x.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Parabola Question 60 English Explanation"><br>Clearly RS is latus-rectum<br><br>$$\because$$ VF = 2 = a<br><br>$$\therefore$$ RS = 4a = 8<br><br>Now OF = 2a = 4<br><br>$$\Rightarrow$$ Area of triangle ORS = 16 | mcq | jee-main-2021-online-25th-july-morning-shift |
1l54b84j9 | maths | parabola | tangent-to-parabola | <p>Let P : y<sup>2</sup> = 4ax, a > 0 be a parabola with focus S. Let the tangents to the parabola P make an angle of $${\pi \over 4}$$ with the line y = 3x + 5 touch the parabola P at A and B. Then the value of a for which A, B and S are collinear is :</p> | [{"identifier": "A", "content": "8 only"}, {"identifier": "B", "content": "2 only"}, {"identifier": "C", "content": "$${1 \\over 4}$$ only"}, {"identifier": "D", "content": "any a > 0"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5scup2c/50969c63-d4f1-466d-8464-4fc5bb572b2d/e838e240-077a-11ed-94f6-83604aa63acb/file-1l5scup2d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5scup2c/50969c63-d4f1-466d-8464-4fc5bb572b2d/e838e240-077a-11ed-94f6-83604aa63acb/file-1l5scup2d.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 - Parabola Question 53 English Explanation"></p>
<p>$$\tan {\pi \over 4} = \left| {{{m - 3} \over {1 + 3m}}} \right|$$</p>
<p>$$ \Rightarrow {{m - 3} \over {1 + 3m}} = \pm \,1$$</p>
<p>$$ \Rightarrow {{m - 3} \over {1 + 3m}} = + 1$$ and $${{m - 3} \over {1 + 3m}} = - 1$$</p>
<p>$$ \Rightarrow m - 3 = 1 + 3m$$ and $$m - 3 = - 1 - 3m$$</p>
<p>$$ \Rightarrow 2m = - 4$$ and $$4m = 2$$</p>
<p>$$m = - 2$$ and $$m = {1 \over 2}$$</p>
<p>We know, Equation of tangent to the parabola $${y^2} = 4m$$ is $$y = mx + {a \over m}$$ and point of contact is $$\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)$$</p>
<p>$$\therefore$$ Equation of tangent</p>
<p>$$y = - 2x - {a \over 2}$$</p>
<p>and $$y = {x \over 2} + 2a$$</p>
<p>$$\therefore$$ Point of contact A and B are</p>
<p>$$A\left( {{a \over {{{( - 2)}^2}}},{{2a} \over { - 2}}} \right) = A\left( {{a \over 4}, - a} \right)$$</p>
<p>$$B\left( {{a \over {{{\left( {{1 \over 2}} \right)}^2}}},{{2a} \over {\left( {{1 \over 2}} \right)}}} \right) = B\left( {4a,4a} \right)$$</p>
<p>As points A, B and S are colinear so area of triangle formed by those 3 points are zero.</p>
<p>Area of $$\Delta$$ABS = $${1 \over 2}\left| {\matrix{
{{a \over 4}} & { - a} & 1 \cr
{4a} & {4a} & 1 \cr
a & 0 & 1 \cr
} } \right|$$</p>
<p>$$ = {a \over 4}(4a - 0) + a(4a - a) + 1(0 - 4{a^2})$$</p>
<p>$$ = {a^2} + 3{a^2} - 4{a^2} = 0$$</p>
<p>$$\therefore$$ Area of triangle is independent of value of a.</p>
<p>So, for all value of a > 0 (already given a must be greater than 0) point A, B and S will be collinear.</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l57p81t3 | maths | parabola | tangent-to-parabola | <p>A circle of radius 2 unit passes through the vertex and the focus of the parabola y<sup>2</sup> = 2x and touches the parabola $$y = {\left( {x - {1 \over 4}} \right)^2} + \alpha $$, where $$\alpha$$ > 0. Then (4$$\alpha$$ $$-$$ 8)<sup>2</sup> is equal to ______________.</p> | [] | null | 63 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5qa7f14/20481fc2-2f60-4c10-b833-b651bfbff5ae/fd4dc580-0656-11ed-903e-c9687588b3f3/file-1l5qa7f15.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5qa7f14/20481fc2-2f60-4c10-b833-b651bfbff5ae/fd4dc580-0656-11ed-903e-c9687588b3f3/file-1l5qa7f15.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Parabola Question 50 English Explanation"></p>
<p>Let the equation of circle be</p>
<p>$$x\left( {x - {1 \over 2}} \right) + {y^2} + \lambda y = 0$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} - {1 \over 2}x + \lambda y = 0$$</p>
<p>Radius $$ = \sqrt {{1 \over {16}} + {{{\lambda ^2}} \over 4}} = 2$$</p>
<p>$$ \Rightarrow {\lambda ^2} = {{63} \over 4}$$</p>
<p>$$ \Rightarrow {\left( {x - {1 \over 4}} \right)^2} + {\left( {y + {\lambda \over 2}} \right)^2} = 4$$</p>
<p>$$\because$$ This circle and parabola $$y - \alpha = {\left( {x - {1 \over 4}} \right)^2}$$ touch each other, so</p>
<p>$$\alpha = - {\lambda \over 2} + 2$$</p>
<p>$$ \Rightarrow \alpha - 2 = - {\lambda \over 2}$$</p>
<p>$$ \Rightarrow {(\alpha - 2)^2} = {{{\lambda ^2}} \over 4} = {{63} \over {16}}$$</p>
<p>$$ \Rightarrow {(4\alpha - 8)^2} = 63$$</p> | integer | jee-main-2022-online-27th-june-morning-shift |
1l59kykzv | maths | parabola | tangent-to-parabola | <p>If the line $$y = 4 + kx,\,k > 0$$, is the tangent to the parabola $$y = x - {x^2}$$ at the point P and V is the vertex of the parabola, then the slope of the line through P and V is :</p> | [{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${26 \\over 9}$$"}, {"identifier": "C", "content": "$${5 \\over 2}$$"}, {"identifier": "D", "content": "$${23 \\over 6}$$"}] | ["C"] | null | <p>$$\because$$ Line $$y = kx + 4$$ touches the parabola $$y = x - {x^2}$$.</p>
<p>So, $$kx + 4 = x - {x^2} \Rightarrow {x^2} + (k - 1)x + 4 = 0$$ has only one root</p>
<p>$${(k - 1)^2} = 16 \Rightarrow k = 5$$ or $$-$$3 but $$k > 0$$</p>
<p>So, $$k = 5$$.</p>
<p>And hence $${x^2} + 4x + 4 = 0 \Rightarrow x = - 2$$</p>
<p>So, P($$-$$2, $$-$$6) and V is $$\left( {{1 \over 2},{2 \over 4}} \right)$$</p>
<p>Slope of $$PV = {{{1 \over 4} + 6} \over {{1 \over 2} + 2}} = {5 \over 2}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l6dxgu1l | maths | parabola | tangent-to-parabola | <p>The sum of diameters of the circles that touch (i) the parabola $$75 x^{2}=64(5 y-3)$$ at the point $$\left(\frac{8}{5}, \frac{6}{5}\right)$$ and (ii) the $$y$$-axis, is equal to ______________.</p> | [] | null | 10 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97t0fy1/82aaf9b5-d36f-45fe-b435-ffafbaba2be0/81fa0a90-4b5e-11ed-bfde-e1cb3fafe700/file-1l97t0fy2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97t0fy1/82aaf9b5-d36f-45fe-b435-ffafbaba2be0/81fa0a90-4b5e-11ed-bfde-e1cb3fafe700/file-1l97t0fy2.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 - Parabola Question 40 English Explanation"><br>
Equation of tangent to the parabola at $P\left(\frac{8}{5}, \frac{6}{5}\right)$
<br><br>
$$
\begin{aligned}
&75 x \cdot \frac{8}{5}=160\left(y+\frac{6}{5}\right)-192 \\\\
&\Rightarrow 120 x=160 y \\\\
&\Rightarrow 3 x=4 y
\end{aligned}
$$<br><br>
Equation of circle touching the given parabola at $\mathrm{P}$ can be taken as <br><br>$\left(x-\frac{8}{5}\right)^{2}+\left(y-\frac{6}{5}\right)^{2}+\lambda(3 x-4 y)=0$
<br><br>
If this circle touches $y$-axis then
<br><br>
$$
\begin{aligned}
&\frac{64}{25}+\left(y-\frac{6}{5}\right)^{2}+\lambda(-4 y)=0 \\\\
&\Rightarrow y^{2}-2 y\left(2 \lambda+\frac{6}{5}\right)+4=0 \\\\
&\Rightarrow D=0 \\\\
&\Rightarrow\left(2 \lambda+\frac{6}{6}\right)^{2}=4 \\\\
&\Rightarrow \lambda=\frac{2}{5} \text { or }-\frac{8}{5}
\end{aligned}
$$
<br><br>
Radius $=1$ or 4
<br><br>
Sum of diameter $=10$
| integer | jee-main-2022-online-25th-july-morning-shift |
1l6f1mr40 | maths | parabola | tangent-to-parabola | <p>The tangents at the points $$A(1,3)$$ and $$B(1,-1)$$ on the parabola $$y^{2}-2 x-2 y=1$$ meet at the point $$P$$. Then the area (in unit $${ }^{2}$$ ) of the triangle $$P A B$$ is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <p>Given curve : $${y^2} - 2x - 2y = 1$$.</p>
<p>Can be written as</p>
<p>$${(y - 1)^2} = 2(x + 1)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7bummn7/86c95a3b-4fff-4748-b770-5b447b8c82d7/87ebf730-25ff-11ed-9c74-c5a04899a045/file-1l7bummn8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7bummn7/86c95a3b-4fff-4748-b770-5b447b8c82d7/87ebf730-25ff-11ed-9c74-c5a04899a045/file-1l7bummn8.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Evening Shift Mathematics - Parabola Question 39 English Explanation"></p>
<p>And, the given information can be plotted as shown in figure</p>
<p>Tangent at $$A:2y - x - 5 = 0$$ {using $$T = 0$$}</p>
<p>Intersection with $$y = 1$$ is $$x = - 3$$</p>
<p>Hence, point P is $$( - 3,1)$$</p>
<p>Taking advantage of symmetry</p>
<p>Area of $$\Delta PAB = 2 \times {1 \over 2} \times (1 - ( - 3)) \times (3 - 1)$$</p>
<p>= 8 sq. units</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6jbzo79 | maths | parabola | tangent-to-parabola | <p>Let $$P(a, b)$$ be a point on the parabola $$y^{2}=8 x$$ such that the tangent at $$P$$ passes through the centre of the circle $$x^{2}+y^{2}-10 x-14 y+65=0$$. Let $$A$$ be the product of all possible values of $$a$$ and $$B$$ be the product of all possible values of $$b$$. Then the value of $$A+B$$ is equal to :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "40"}, {"identifier": "D", "content": "65"}] | ["D"] | null | <p>Centre of circle $${x^2} + {y^2} - 10x - 14y + 65 = 0$$ is at (5, 7).</p>
<p>Let the equation of tangent to $${y^2} = 8x$$ is</p>
<p>$$yt = x + 2{t^2}$$</p>
<p>which passes through (5, 7)</p>
<p>$$7t = 5 + 2{t^2}$$</p>
<p>$$ \Rightarrow 2{t^2} - 7t + 5 = 0$$</p>
<p>$$t = 1,{5 \over 2}$$</p>
<p>$$A = 2 \times {1^2} \times 2 \times {\left( {{5 \over 2}} \right)^2} = 25$$</p>
<p>$$B = 2 \times 2 \times 1 \times 2 \times 2 \times {5 \over 2} = 40$$</p>
<p>$$A + B = 65$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6kkcjlr | maths | parabola | tangent-to-parabola | <p>If the length of the latus rectum of a parabola, whose focus is $$(a, a)$$ and the tangent at its vertex is $$x+y=a$$, is 16, then $$|a|$$ is equal to :</p> | [{"identifier": "A", "content": "$$2 \\sqrt{2}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{3}$$"}, {"identifier": "C", "content": "$$4 \\sqrt{2}$$"}, {"identifier": "D", "content": "4"}] | ["C"] | null | <p>Equation of tangent at vertex : $$L \equiv x + y - a = 0$$</p>
<p>Focus : $$F \equiv (a,a)$$</p>
<p>Perpendicular distance of L from F</p>
<p>$$ = \left| {{{a + a - a} \over {\sqrt 2 }}} \right| = \left| {{a \over {\sqrt 2 }}} \right|$$</p>
<p>Length of latus rectum $$ = 4\left| {{a \over {\sqrt 2 }}} \right|$$</p>
<p>Given $$4\,.\,\left| {{a \over {\sqrt 2 }}} \right| = 16$$</p>
<p>$$ \Rightarrow |a| = 4\sqrt 2 $$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6m5skm4 | maths | parabola | tangent-to-parabola | <p>If the tangents drawn at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ on the parabola $$y^{2}=2 x-3$$ intersect at the point $$R(0,1)$$, then the orthocentre of the triangle $$P Q R$$ is :</p> | [{"identifier": "A", "content": "(0, 1)"}, {"identifier": "B", "content": "(2, $$-$$1)"}, {"identifier": "C", "content": "(6, 3)"}, {"identifier": "D", "content": "(2, 1)"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7raxquq/8f209afe-272a-4b1d-9894-cf78a8a0ee0c/10e1a120-2e7f-11ed-8702-156c00ced081/file-1l7raxqur.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7raxquq/8f209afe-272a-4b1d-9894-cf78a8a0ee0c/10e1a120-2e7f-11ed-8702-156c00ced081/file-1l7raxqur.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Parabola Question 33 English Explanation"></p>
<p>Equation of chord $$PQ$$</p>
<p>$$ \Rightarrow y \times 1 = x - 3$$</p>
<p>$$ \Rightarrow x - y = 3$$</p>
<p>For point P & Q</p>
<p>Intersection of PQ with parabola $$P:(6,3)\,Q:(2, - 1)$$</p>
<p>Slope of $$RQ = - 1$$ & Slope of $$PQ = 1$$</p>
<p>Therefore $$\angle PQR = 90^\circ \Rightarrow $$ Orthocentre is at $$Q:(2, - 1)$$</p> | mcq | jee-main-2022-online-28th-july-morning-shift |
ldoadhnb | maths | parabola | tangent-to-parabola | Let $\mathrm{S}$ be the set of all $\mathrm{a} \in \mathrm{N}$ such that the area of the triangle formed by the tangent at the point $\mathrm{P}(\mathrm{b}$, c), b, c $\in \mathbb{N}$, on the parabola $y^{2}=2 \mathrm{a} x$ and the lines $x=\mathrm{b}, y=0$ is $16 $ unit<sup>2</sup>, then $\sum\limits_{\mathrm{a} \in \mathrm{S}} \mathrm{a}$ is equal to : | [] | null | 146 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lefxfag8/8b4831e9-648d-41a1-9ba1-f0f0ff15128a/1d79d280-b2d3-11ed-8169-e1635469e777/file-1lefxfag9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lefxfag8/8b4831e9-648d-41a1-9ba1-f0f0ff15128a/1d79d280-b2d3-11ed-8169-e1635469e777/file-1lefxfag9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Evening Shift Mathematics - Parabola Question 30 English Explanation">
<br>As $\mathrm{P}(\mathrm{b}, \mathrm{c})$ lies on parabola so $\mathrm{c}^{2}=2 \mathrm{ab}---(1)$
<br><br>Now equation of tangent to parabola $\mathrm{y}^{2}=2 \mathrm{ax}$ in point form is <br><br>$\mathrm{yy}_{1}=2 \mathrm{a} \frac{\left(\mathrm{x}+\mathrm{x}_{1}\right)}{2},$
<br><br>Here, $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(\mathrm{b}, \mathrm{c})$
<br><br>$\Rightarrow \mathrm{yc}=\mathrm{a}(\mathrm{x}+\mathrm{b})$
<br><br>For point $\mathrm{B}$, put $\mathrm{y}=0$, now $\mathrm{x}=-\mathrm{b}$
<br><br>So, area of $\triangle \mathrm{PBA}, \frac{1}{2} \times \mathrm{AB} \times \mathrm{AP}=16$
<br><br>$$
\begin{aligned}
& \Rightarrow \frac{1}{2} \times 2 b \times c=16 \\\\
& \Rightarrow b c=16
\end{aligned}
$$
<br><br>As $\mathrm{b}$ and $\mathrm{c}$ are natural number so possible values of $(b, c)$ are $(1,16),(2,8),(4,4),(8,2)$ and $(16,1)$
<br><br>Now from equation (1) $\mathrm{a}=\frac{\mathrm{c}^2}{2 \mathrm{~b}}$ and $\mathrm{a} \in \mathrm{N}$, so values of $(b, c)$ are $(1,16),(2,8)$ and $(4,4)$ now values of are 128,16 and 2 .
<br><br>Hence sum of values of $a$ is 146 . | integer | jee-main-2023-online-31st-january-evening-shift |
1ldptdxi9 | maths | parabola | tangent-to-parabola | <p>Let $$\mathrm{y}=f(x)$$ represent a parabola with focus $$\left(-\frac{1}{2}, 0\right)$$ and directrix $$y=-\frac{1}{2}$$. Then
<br/><br/>$$S=\left\{x \in \mathbb{R}: \tan ^{-1}(\sqrt{f(x)})+\sin ^{-1}(\sqrt{f(x)+1})=\frac{\pi}{2}\right\}$$ :</p> | [{"identifier": "A", "content": "is an empty set"}, {"identifier": "B", "content": "contains exactly one element"}, {"identifier": "C", "content": "contains exactly two elements"}, {"identifier": "D", "content": "is an infinite set"}] | ["C"] | null | $\left(x+\frac{1}{2}\right)^{2}=\left(y+\frac{1}{4}\right)$
<br/><br/>$y=\left(x^{2}+x\right)$
<br/><br/>$\tan ^{-1} \sqrt{\mathrm{x}(\mathrm{x}+1)}+\sin ^{-1} \sqrt{\mathrm{x}^{2}+\mathrm{x}+1}=\pi / 2$
<br/><br/>$0 \leq \mathrm{x}^{2}+\mathrm{x}+1 \leq 1$
<br/><br/>$x^{2}+x \leq 0$
<br/><br/>Also $x^{2}+x \geq 0$
<br/><br/>$\therefore \mathrm{x}^{2}+\mathrm{x}=0 \Rightarrow \mathrm{x}=0,-1$
<br/><br/>$\mathrm{S}$ contains 2 element. | mcq | jee-main-2023-online-31st-january-morning-shift |
1ldsf5fpj | maths | parabola | tangent-to-parabola | <p>If the tangent at a point P on the parabola $$y^2=3x$$ is parallel to the line $$x+2y=1$$ and the tangents at the points Q and R on the ellipse $$\frac{x^2}{4}+\frac{y^2}{1}=1$$ are perpendicular to the line $$x-y=2$$, then the area of the triangle PQR is :</p> | [{"identifier": "A", "content": "$$\\frac{9}{\\sqrt5}$$"}, {"identifier": "B", "content": "$$3\\sqrt5$$"}, {"identifier": "C", "content": "$$5\\sqrt3$$"}, {"identifier": "D", "content": "$$\\frac{3}{2}\\sqrt5$$"}] | ["B"] | null | <p>$$P \equiv \left( {{A \over {{m^2}}},{{2A} \over m}} \right)$$ where $$\left( {A = {3 \over 4},m = {{ - 1} \over 2}} \right)$$</p>
<p>& $$Q,R = \left( { \mp \,{{{a^2}{m_1}} \over {{a^2}m_1^2 + {b^2}}},{{ \mp \,.\,{b^2}} \over {\sqrt {{a^2}m_1^2 + {b^2}} }}} \right)$$</p>
<p>Where $${a^2} = 4,{b^2} = 1$$ and $${m_1} = 1$$</p>
<p>$$\therefore$$ $$P \equiv (3, - 3)$$</p>
<p>$$Q \equiv \left( {{{ - 4} \over {\sqrt 5 }},{{ - 1} \over {\sqrt 5 }}} \right)$$ & $$R\left( {{4 \over {\sqrt 5 }},{1 \over {\sqrt 5 }}} \right)$$</p>
<p>Area $$ = {1 \over 2}\left| {\matrix{
3 & { - 3} & 1 \cr
{{{ - 4} \over {\sqrt 5 }}} & {{{ - 1} \over {\sqrt 5 }}} & 1 \cr
{{4 \over {\sqrt 5 }}} & {{1 \over {\sqrt 5 }}} & 1 \cr
} } \right| = {1 \over {10}}\left| {\matrix{
3 & { - 3} & 1 \cr
{ - 4} & { - 1} & {\sqrt 5 } \cr
0 & 0 & {2\sqrt 5 } \cr
} } \right|$$</p>
<p>$$ = {{2\sqrt 5 } \over {10}}( - 15) = 3\sqrt 5 $$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsfx6la | maths | parabola | tangent-to-parabola | <p>A triangle is formed by the tangents at the point (2, 2) on the curves $$y^2=2x$$ and $$x^2+y^2=4x$$, and the line $$x+y+2=0$$. If $$r$$ is the radius of its circumcircle, then $$r^2$$ is equal to ___________.</p> | [] | null | 10 | <p>Tangent for $${y^2} = 2x$$ at (2, 2) is</p>
<p>$${L_1}:2y = x + 2$$</p>
<p>Tangent for $${x^2} + {y^2} = 4x$$ at (2, 2) is</p>
<p>$${L_2}:y = 2$$</p>
<p>$${L_3}:x + y = 2 = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1let847b1/44924ea8-65a6-4741-9b68-fa3cc8917eba/1e026ad0-ba23-11ed-b1c7-2f0d4a78b053/file-1let847b2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1let847b1/44924ea8-65a6-4741-9b68-fa3cc8917eba/1e026ad0-ba23-11ed-b1c7-2f0d4a78b053/file-1let847b2.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Evening Shift Mathematics - Parabola Question 24 English Explanation"></p>
<p>Radius of circumcircle $$ = {{abc} \over {4\Delta }}$$</p>
<p>$$ = {{(\sqrt {20} )(6)(\sqrt 8 )} \over {4 \times {1 \over 2} \times 6 \times 2}}$$</p>
<p>$$R = \sqrt {10} $$</p>
<p>$$R^2=10$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldwwuxyg | maths | parabola | tangent-to-parabola | <p>The equations of the sides AB and AC of a triangle ABC are $$(\lambda+1)x+\lambda y=4$$ and $$\lambda x+(1-\lambda)y+\lambda=0$$ respectively. Its vertex A is on the y-axis and its orthocentre is (1, 2). The length of the tangent from the point C to the part of the parabola $$y^2=6x$$ in the first quadrant is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2$$\\sqrt2$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\sqrt6$$"}] | ["B"] | null | $$
\begin{aligned}
& \mathrm{AB}:(\lambda+1) x+\lambda y=4 \\\\
& \mathrm{AC}: \lambda x+(1-\lambda) y+\lambda=0 \\\\
& \text { Vertex } A \text { is on } y \text {-axis } \\\\
& \Rightarrow x=0
\end{aligned}
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5hgrn5/44884087-ba3d-4667-bbb2-3f2a5b14abd3/de014610-ad14-11ed-8a8c-4d67f5492755/file-1le5hgrn6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5hgrn5/44884087-ba3d-4667-bbb2-3f2a5b14abd3/de014610-ad14-11ed-8a8c-4d67f5492755/file-1le5hgrn6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Parabola Question 21 English Explanation 1"><br>
So $\mathrm{y}=\frac{4}{\lambda}, \mathrm{y}=\frac{\lambda}{\lambda-1}$<br><br>
$$
\begin{aligned}
& \Rightarrow \frac{4}{\lambda}=\frac{\lambda}{\lambda-1} \\\\
& \Rightarrow \lambda=2 \\\\
& \mathrm{AB}: 3 x+2 y=4 \\\\
& \mathrm{AC}: 2 \mathrm{x}-\mathrm{y}+2=0 \\\\
& \Rightarrow \mathrm{A}(0,2) \text { Let } \mathrm{C}(\alpha, 2 \alpha+2)
\end{aligned}
$$<br><br>
Now (Slope of Altitude through C) $\left(-\frac{3}{2}\right)=-1$<br><br>
$$
\left(\frac{2 \alpha}{\alpha-1}\right)\left(-\frac{3}{2}\right)=-1 \Rightarrow \alpha=-\frac{1}{2}
$$<br><br>
So $\mathrm{C}\left(-\frac{1}{2}, 1\right)$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5hhow2/b6a28f6c-8c4b-40a8-8d7e-2bb4e657d95c/f7b02220-ad14-11ed-8a8c-4d67f5492755/file-1le5hhow3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5hhow2/b6a28f6c-8c4b-40a8-8d7e-2bb4e657d95c/f7b02220-ad14-11ed-8a8c-4d67f5492755/file-1le5hhow3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Parabola Question 21 English Explanation 2"><br>
Let Equation of tangent be $y=m x+\frac{3}{2 m}$<br><br>
$$
\begin{aligned}
& \mathrm{m}^2+2 \mathrm{~m}-3=0 \\\\
& \Rightarrow \mathrm{m}=1,-3
\end{aligned}
$$<br><br>
So tangent which touches in first quadrant at $\mathrm{T}$ is<br><br>
$$
\begin{aligned}
& \mathrm{T} \equiv\left(\frac{\mathrm{a}}{\mathrm{m}^2}, \frac{2 \mathrm{a}}{\mathrm{m}}\right) \\\\
& \equiv\left(\frac{3}{2}, 3\right) \\\\
& \Rightarrow \mathrm{CT}=\sqrt{4+4}=2 \sqrt{2}
\end{aligned}
$$ | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldyamg53 | maths | parabola | tangent-to-parabola | <p>Let a tangent to the curve $$\mathrm{y^2=24x}$$ meet the curve $$xy = 2$$ at the points A and B. Then the mid points of such line segments AB lie on a parabola with the :</p> | [{"identifier": "A", "content": "length of latus rectum 2"}, {"identifier": "B", "content": "directrix 4x = $$-$$3"}, {"identifier": "C", "content": "directrix 4x = 3"}, {"identifier": "D", "content": "length of latus rectum $$\\frac{3}{2}$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2n7sz9/699005b2-a3c2-4567-bfdf-626597fbe0f4/0194ea50-ab85-11ed-bcb9-87e2bc2e0c49/file-1le2n7sza.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2n7sz9/699005b2-a3c2-4567-bfdf-626597fbe0f4/0194ea50-ab85-11ed-bcb9-87e2bc2e0c49/file-1le2n7sza.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Morning Shift Mathematics - Parabola Question 20 English Explanation"></p>
<p>Given parabola $${y^2} = 24x = 4.6x = 4ax$$</p>
<p>$$\therefore$$ a = 6</p>
<p>If $$y = mx + c$$ is a tangent to the parabola then,</p>
<p>$$c = {a \over m}$$</p>
<p>$$ \Rightarrow c = {6 \over m}$$</p>
<p>Let midpoint of chord AB is $$({x_1},{y_1})$$</p>
<p>We know, locus of midpoint of chord for hyperbola,</p>
<p>$$T = {S_1}$$</p>
<p>$$ \Rightarrow {{x{y_1} + {x_1}y} \over 2} - 2 = {x_1}{y_1} - 2$$</p>
<p>$$ \Rightarrow x{y_1} + {x_1}y = 2{x_1}{y_1}$$</p>
<p>$$ \Rightarrow y = \left( { - {{{y_1}} \over {{x_1}}}} \right)x + 2{y_1}$$</p>
<p>This equation is the same tangent to the parabola.</p>
<p>Then comparing with $$y = mx + c$$ we get,</p>
<p>$$c = 2{y_1},\,m = - {{{y_1}} \over {{x_1}}}$$</p>
<p>$$\therefore$$ $$2{y_1} = {6 \over { - {{{y_1}} \over {{x_1}}}}}$$</p>
<p>$$ \Rightarrow - y_1^2 = 3{x_1}$$</p>
<p>$$\therefore$$ Locus of midpoint $$({x_1},{y_1})$$ is,</p>
<p>$$ - {y^2} = 3x$$</p>
<p>$$ \therefore $$ Length of latus rectum = 3</p>
<p>Equation of directrix is $x=\frac{3}{4}$</p> | mcq | jee-main-2023-online-24th-january-morning-shift |
1lgym0ipz | maths | parabola | tangent-to-parabola | <p>Let $$\mathrm{A}(0,1), \mathrm{B}(1,1)$$ and $$\mathrm{C}(1,0)$$ be the mid-points of the sides of a triangle with incentre at the point $$\mathrm{D}$$. If the focus of the parabola $$y^{2}=4 \mathrm{ax}$$ passing through $$\mathrm{D}$$ is $$(\alpha+\beta \sqrt{2}, 0)$$, where $$\alpha$$ and $$\beta$$ are rational numbers, then $$\frac{\alpha}{\beta^{2}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{9}{2}$$"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyvdpob/0009489a-90f2-422f-89e7-7c971c77552a/bebae0b0-5b9f-11ee-b31c-37f6bf9b942e/file-6y3zli1lmyvdpoc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lmyvdpob/0009489a-90f2-422f-89e7-7c971c77552a/bebae0b0-5b9f-11ee-b31c-37f6bf9b942e/file-6y3zli1lmyvdpoc.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) 8th April Evening Shift Mathematics - Parabola Question 15 English Explanation">
<br><br>$$
\begin{aligned}
& \text { So, } \mathrm{D} \equiv\left(\frac{4}{2+2+2 \sqrt{2}}, \frac{4}{2+2+2 \sqrt{2}}\right) \\\\
& \equiv\left(\frac{2}{2+\sqrt{2}}, \frac{2}{2+\sqrt{2}}\right) \\\\
& =\left(\frac{2}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}, \frac{2}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}\right) \\\\
& \equiv(2-\sqrt{2}, 2-\sqrt{2})
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& \because y^2=4 a x \\\\
& (2-\sqrt{2})^2=4 a(2-\sqrt{2}) \\\\
& \Rightarrow 4 a=2-\sqrt{2} \Rightarrow a=\frac{2-\sqrt{2}}{4} \\\\
& \Rightarrow \frac{1}{2}-\frac{\sqrt{2}}{4}=\alpha+\beta \sqrt{2} \\\\
& \Rightarrow \alpha=\frac{1}{2}, \beta=\frac{-1}{4} \\\\
& \text { So, } \frac{\alpha}{\beta^2}=\frac{\frac{1}{2}}{\frac{1}{16}}=8
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh23vqbm | maths | parabola | tangent-to-parabola | <p>Let the tangent to the curve $$x^{2}+2 x-4 y+9=0$$ at the point $$\mathrm{P}(1,3)$$ on it meet the $$y$$-axis at $$\mathrm{A}$$. Let the line passing through $$\mathrm{P}$$ and parallel to the line $$x-3 y=6$$ meet the parabola $$y^{2}=4 x$$ at $$\mathrm{B}$$. If $$\mathrm{B}$$ lies on the line $$2 x-3 y=8$$, then $$(\mathrm{AB})^{2}$$ is equal to ___________.</p> | [] | null | 292 | Given, equation of curve is
<br/><br/>$$x^2+2 x-4 y+9=0$$ ..........(i)
<br/><br/>Equation of tangent at $P(1,3)$ to the given curve (i)
<br/><br/>$$
\begin{array}{rlrl}
& x(1)+2\left(\frac{x+1}{2}\right)-4\left(\frac{y+3}{2}\right)+9 =0 \\\\
& \Rightarrow 2 x+2 x+2-4 y-12+18 =0 \\\\
&\Rightarrow 4 x-4 y+8 =0 \\\\
&\Rightarrow x-y+2 =0
\end{array}
$$
<br/><br/>which is meet the $Y$-axis at $A$
<br/><br/>$$\therefore A \equiv(0,2)$$
<br/><br/>Equation of line passing through $P$ and parallel to $x-3 y=6$ is $x-3 y+8=0$
<br/><br/>Since, line (ii) meet the parabola $y^2=4 x$ at $B$
<br/><br/>$$
\begin{array}{lc}
&\therefore y^2=4(3 y-8) \\\\
&\Rightarrow y^2-12 y+32=0 \\\\
&\Rightarrow (y-4)(y-8)=0
\end{array}
$$
<br/><br/>$\therefore$ Possible co-ordinates of $B$ are $(4,4)$ and $(16,8)$.
<br/><br/>Since, $(4,4)$ does not satisfies line $2 x-3 y=8$
<br/><br/>Thus, $B$ is $(16,8)$
<br/><br/>$$
\therefore (A B)^2=(16-0)^2+(8-2)^2=256+36=292
$$ | integer | jee-main-2023-online-6th-april-morning-shift |
lv9s20p5 | maths | parabola | tangent-to-parabola | <p>Let a line perpendicular to the line $$2 x-y=10$$ touch the parabola $$y^2=4(x-9)$$ at the point P. The distance of the point P from the centre of the circle $$x^2+y^2-14 x-8 y+56=0$$ is __________.</p> | [] | null | 10 | <p>Line perpendicular to $$2 x-y=10$$ have slope $$=\frac{-1}{2}$$</p>
<p>$$\Rightarrow$$ Line tangent to parabola $$y^2=4(x-9)$$ with slope $$m$$ is</p>
<p>$$\begin{aligned}
& y=m(x-9)+\frac{1}{m}, m=\frac{-1}{2} \\
& \Rightarrow y=\frac{-(x-9)}{2}-2 \Rightarrow 2 y=-x+9-4 \\
& \Rightarrow 2 y+x=5
\end{aligned}$$</p>
<p>Solving the tangent and parabola we get point $$P$$</p>
<p>$$\begin{aligned}
& \left(\frac{5-x}{2}\right)^2=4(x-9) \Rightarrow x^2-10 x+25=16 x-144 \\
& \Rightarrow x^2-26 x+169=0 \Rightarrow(x-13)^2=0 \\
& \Rightarrow P \equiv(13,-4)
\end{aligned}$$</p>
<p>Distance of $$P$$ from the centre of circle $$(7,4)$$ is $$\sqrt{(13-7)^2+(-4-4)^2}=\sqrt{36+64}=10$$ units.</p> | integer | jee-main-2024-online-5th-april-evening-shift |
lvc57noi | maths | parabola | tangent-to-parabola | <p>Let a conic $$C$$ pass through the point $$(4,-2)$$ and $$P(x, y), x \geq 3$$, be any point on $$C$$. Let the slope of the line touching the conic $$C$$ only at a single point $$P$$ be half the slope of the line joining the points $$P$$ and $$(3,-5)$$. If the focal distance of the point $$(7,1)$$ on $$C$$ is $$d$$, then $$12 d$$ equals ________.</p> | [] | null | 75 | <p>As per given condition</p>
<p>$$\begin{gathered}
\frac{d y}{d x}=\frac{y+5}{2(x-3)} \\
\Rightarrow \ln (y+5)=\frac{1}{2} \ln (x-3)+c \\
\text { Passes through }(4,-2) \Rightarrow \ln 3=\frac{1}{2} \ln 1+c \\
\Rightarrow c=\ln 3
\end{gathered}$$</p>
<p>$$\Rightarrow$$ Curve is $$(y+5)^2=9(x-3)$$</p>
<p>Focal distance of $$(7,1)=\frac{9}{4}+4=\frac{25}{4}=d$$</p>
<p>$$12 d=75$$</p> | integer | jee-main-2024-online-6th-april-morning-shift |
lvc583gl | maths | parabola | tangent-to-parabola | <p>Let $$L_1, L_2$$ be the lines passing through the point $$P(0,1)$$ and touching the parabola $$9 x^2+12 x+18 y-14=0$$. Let $$Q$$ and $$R$$ be the points on the lines $$L_1$$ and $$L_2$$ such that the $$\triangle P Q R$$ is an isosceles triangle with base $$Q R$$. If the slopes of the lines $$Q R$$ are $$m_1$$ and $$m_2$$, then $$16\left(m_1^2+m_2^2\right)$$ is equal to __________.</p> | [] | null | 68 | <p>$$\begin{aligned}
& 9 x^2+12 x+18 y-14=0 \\
& \left(x+\frac{2}{3}\right)^2=-2(y-1) \ldots(1)
\end{aligned}$$</p>
<p>Equation of tangent to (1)</p>
<p>$$\begin{aligned}
& t\left(x+\frac{2}{3}\right)=-(y-1)+\frac{1}{2} t^2 \text { passes through }(0,1) \\
& \Rightarrow \frac{2}{3} t=\frac{1}{2} t^2 \Rightarrow t=0, \frac{4}{3} \Rightarrow m=0, m=-\frac{4}{3}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd36y63/122dc7d3-2b8c-4a18-982f-a395f4f07780/372994b0-159f-11ef-aabb-5de744d2ad81/file-1lwd36y64.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd36y63/122dc7d3-2b8c-4a18-982f-a395f4f07780/372994b0-159f-11ef-aabb-5de744d2ad81/file-1lwd36y64.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) 6th April Morning Shift Mathematics - Parabola Question 1 English Explanation"></p>
<p>$$\begin{aligned}
& \Rightarrow\left|\frac{m+\frac{4}{3}}{1-\frac{4}{3} m}\right|=|m| \Rightarrow\left|\frac{3 m+4}{3-4 m}\right|=|m| \\
& \Rightarrow 3 m+4=m(3-4 m) \text { or } 3 m+4=-m(3-4 m) \\
& 3 m+4=3 m-4 m^2 \text { or } 3 m+4=-3 m+4 m^2 \\
& 4 m^2+4=0 \text { (not possible) or } 4 m^2-6 m-4=0 \\
& m_1+m_2=\frac{3}{2}, m_1 m_2=-1 \\
& \Rightarrow m_1^2+m_2^2=\frac{17}{4} \\
& 16\left(m_1^2+m_2^2\right)=68
\end{aligned}$$</p> | integer | jee-main-2024-online-6th-april-morning-shift |
UbyEKAsad53b322QnPdNK | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "32"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266952/exam_images/idxhu7dj3nmxsx1ackba.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Permutations and Combinations Question 143 English Explanation">
<br><br>Area = $${1 \over 2}$$ h. k = 50
<br><br>h. k = 100
<br><br>h. k = 2<sup>2</sup> . 5<sup>2</sup>
<br><br>Total divisors
<br><br>= (2 + 1) (2 + 1) = 9
<br><br>if h > 0, k > 0
<br><br>But $${\matrix{
{h > 0,} & {k < 0} \cr
{h < 0,} & {k > 0} \cr
{h < 0,} & {k < 0} \cr
} }$$
<br><br>all are possible so that total no. of positive case
<br><br>9 + 9 + 9 + 9 = 36 | mcq | jee-main-2019-online-9th-january-evening-slot |
eOF6fmsz4KXubATcSH3rsa0w2w9jx2ay4se | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the
top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total
number of beams is : | [{"identifier": "A", "content": "180"}, {"identifier": "B", "content": "210"}, {"identifier": "C", "content": "170"}, {"identifier": "D", "content": "190"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264617/exam_images/eimc4dj1il40k9qseeao.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Evening Slot Mathematics - Permutations and Combinations Question 134 English Explanation">
Any two non-adjacent pillers are joined by beams<br><br>
$$ \therefore $$ number of beams = number of diagonals = <sup>20</sup>C<sub>2</sub> - 20 = 170 | mcq | jee-main-2019-online-10th-april-evening-slot |
I5Ortna4ugLS6ar8gn1kmiwfm5s | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | Consider a rectangle ABCD having 5, 7, 6, 9 points in the interior of the line segments AB, CD, BC, DA respectively. Let $$\alpha$$ be the number of triangles having these points from different sides as vertices and $$\beta$$ be the number of quadrilaterals having these points from different sides as vertices. Then ($$\beta$$ $$-$$ $$\alpha$$) is equal to : | [{"identifier": "A", "content": "717"}, {"identifier": "B", "content": "795"}, {"identifier": "C", "content": "1890"}, {"identifier": "D", "content": "1173"}] | ["A"] | null | $$\alpha = {}^6{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^7{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^9{C_1} + {}^5{C_1}{}^6{C_1}{}^7{C_1} $$
<br><br>$$= 378 + 315 + 270 + 210 = 1173$$<br><br>$$\beta = {}^5{C_1}{}^6{C_1}{}^7{C_1}{}^9{C_1} = 1890$$<br><br>$$ \therefore $$ $$ \beta - \alpha = 1890 - 1173 = 717$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
uTYPsLQsZvZrwoOMNL1kmkl4mb8 | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | If the sides AB, BC and CA of a triangle ABC have 3, 5 and 6 interior points respectively, then the total number of triangles that can be constructed using these points as vertices, is equal to : | [{"identifier": "A", "content": "240"}, {"identifier": "B", "content": "360"}, {"identifier": "C", "content": "333"}, {"identifier": "D", "content": "364"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266804/exam_images/l90mpotax9t4vio9u3us.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Mathematics - Permutations and Combinations Question 106 English Explanation">
<br>Total number of triangles<br><br>= $${}^{14}{C_3} - {}^3{C_3} - {}^5{C_3} - {}^6{C_3}$$
<br><br>= 364 – 31 = 333 | mcq | jee-main-2021-online-17th-march-evening-shift |
1kto82i38 | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | Let P<sub>1</sub>, P<sub>2</sub>, ......, P<sub>15</sub> be 15 points on a circle. The number of distinct triangles formed by points P<sub>i</sub>, P<sub>j</sub>, P<sub>k</sub> such that i +j + k $$\ne$$ 15, is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "419"}, {"identifier": "C", "content": "443"}, {"identifier": "D", "content": "455"}] | ["C"] | null | Total number of triangles = $${}^{15}{C_3}$$<br><br>i + j + k = 15 (Given)<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwoq1bj3/a1cd0f3a-86a8-4d72-98ac-6d132a8be6f4/621a22f0-534d-11ec-9cbb-695a838b20fb/file-1kwoq1bj4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwoq1bj3/a1cd0f3a-86a8-4d72-98ac-6d132a8be6f4/621a22f0-534d-11ec-9cbb-695a838b20fb/file-1kwoq1bj4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Permutations and Combinations Question 90 English Explanation"><br>Number of possible triangles using the vertices P<sub>i</sub>, P<sub>j</sub>, P<sub>k</sub> such that i + j + k $$\ne$$ 15 is equal to $${}^{15}{C_3}$$ $$-$$ 12 = 443<br><br>Option (c) | mcq | jee-main-2021-online-1st-september-evening-shift |
1ldu601xz | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | <p>A triangle is formed by X-axis, Y-axis and the line $$3x+4y=60$$. Then the number of points P(a, b) which lie strictly inside the triangle, where a is an integer and b is a multiple of a, is ____________.</p> | [] | null | 31 | If x = 1, y = $57 \over 4 $ = 14.25<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef6kjms/0eda98c6-e556-46f8-a286-86afedde79c5/1920a240-b26a-11ed-a7d3-67cb923c1f9d/file-1lef6kjmt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lef6kjms/0eda98c6-e556-46f8-a286-86afedde79c5/1920a240-b26a-11ed-a7d3-67cb923c1f9d/file-1lef6kjmt.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 - Permutations and Combinations Question 51 English Explanation"><br>$(1,1)(1,2)-(1,14) \Rightarrow 14$ pts.<br><br>
If $x=2, y=\frac{27}{2}=13.5$<br><br>
$(2,2)(2,4) \ldots(2,12) \quad \Rightarrow 6$ pts.<br><br>
If $\mathrm{x}=3, \mathrm{y}=\frac{51}{4}=12.75$<br><br>
$(3,3)(3,6)-(3,12) \Rightarrow 4$ pts.<br><br>
If $x=4, y=12$<br><br>
$(4,4)(4,8) \quad \Rightarrow 2$ pts.<br><br>
If $x=5 . y=\frac{45}{4}=11.25$<br><br>
$(5,5),(5,10) \Rightarrow 2$ pts.<br><br>
If $\mathrm{x}=6, \mathrm{y}=\frac{21}{2}=10.5$<br><br>
$(6,6) \quad \Rightarrow 1 \mathrm{pt}$.<br><br>
If $x=7, y=\frac{39}{4}=9.75$<br><br>
$(7,7) \Rightarrow 1 \mathrm{pt}$.<br><br>
If $x=8, y=9$<br><br>
$(8,8) \quad \Rightarrow 1$ pt.<br><br>
If $\mathrm{x}=9 \mathrm{y}=\frac{33}{4}=8.25 \Rightarrow$ no $\mathrm{pt}$.<br><br>
Total $=31$ pts. | integer | jee-main-2023-online-25th-january-evening-shift |
1lgxwhg04 | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | <p>Some couples participated in a mixed doubles badminton tournament. If the number of matches played, so that no couple played in a match, is 840, then the total number of persons, who participated in the tournament, is ___________.</p> | [] | null | 16 | Let, $n$ be the total number of couples who participated in the tournament.
<br/><br/>According to the question, $2 \times{ }^n C_2 \times{ }^{n-2} C_2=840$
<br/><br/>$$
\begin{aligned}
& \Rightarrow{ }^n C_2 \times{ }^{n-2} C_2=420 \\\\
& \Rightarrow \frac{n !}{2 !(n-2) !} \times \frac{(n-2) !}{(n-4) ! 2 !}=420 \\\\
& \Rightarrow \frac{n(n-1)(n-2)(n-3)}{4}=420
\end{aligned}
$$
<br/><br/>Put $n=8$ satisfied the equation.
<br/><br/>So, $n=8$
<br/><br/>Hence, total number of players who participated<br/>in the tournament $=2 \times 8=16$ | integer | jee-main-2023-online-10th-april-morning-shift |
lvc57pix | maths | permutations-and-combinations | application-of-permutations-and-combination-in-geometry | <p>The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is</p> | [{"identifier": "A", "content": "56"}, {"identifier": "B", "content": "16"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "48"}] | ["B"] | null | <p>To solve this problem, we need to determine the number of triangles formed by the vertices of a regular octagon such that none of the sides of the triangle is also a side of the octagon.</p>
<p>Let's start by counting the total number of triangles that can be formed using the 8 vertices of the octagon. The number of ways to choose 3 vertices out of 8 is given by the combination formula:</p>
<p>
<p>$$ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 $$</p>
</p>
<p>Next, we need to exclude those triangles that have at least one side coinciding with a side of the octagon. Let's analyze how many such invalid triangles there can be.</p>
<p>Consider each side of the octagon. For any given side, there are exactly 5 other vertices remaining (since we must exclude the two vertices that form the current side). Out of these 5 vertices, we can choose any 1 to form a triangle that has one side common with the octagon. Hence, for each side of the octagon, there are 5 such triangles.</p>
<p>Since the octagon has 8 sides, the total number of triangles that have at least one side as a side of the octagon is:</p>
<p>
<p>$$ 8 \times 5 = 40 $$</p>
</p>
<p>Therefore, the number of triangles whose sides do not coincide with any sides of the octagon is:</p>
<p>
<p>$$ 56 - 40 = 16 $$</p>
</p>
<p>So, the correct answer is:</p>
<p><b>Option B: 16</b></p> | mcq | jee-main-2024-online-6th-april-morning-shift |
EkC57Q2rx3Nbk6Jx | maths | permutations-and-combinations | circular-permutations | The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by | [{"identifier": "A", "content": "$$7!\\, \\times 5!\\,\\,$$ "}, {"identifier": "B", "content": "$$6!\\, \\times 5!$$ "}, {"identifier": "C", "content": "$$30!$$ "}, {"identifier": "D", "content": "$$5!\\, \\times 4!$$ "}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266135/exam_images/ihwshbrvypvoyv0j1fui.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Permutations and Combinations Question 172 English Explanation 1">
<br><br>6 men can sit at the round table = $$\left( {6 - 1} \right)! = 5!$$ ways
<br><br>Now at the round table among 6 men there are 6 empty places and 5 women can sit at those 6 empty positions.<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266027/exam_images/pu5xcxrx6roxmo36hk63.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Permutations and Combinations Question 172 English Explanation 2">
<br><br>So total no of ways 6 men and 5 women can dine at the round table
<br><br> = $$5!\, \times {}^6{C_5} \times 5!$$
<br><br>= $$5!\, \times 6 \times 5!$$
<br><br>= $$5!\, \times 6!$$ | mcq | aieee-2003 |
J7rxWs0HNWj60oLGyuAiC | maths | permutations-and-combinations | circular-permutations | The number of ways in which 5 boys and 3 girls can be seated on a round table if a
particular boy B<sub>1</sub> and a particular girl G<sub>1</sub> never sit adjacent to each other, is : | [{"identifier": "A", "content": "5 $$ \\times $$ 6!"}, {"identifier": "B", "content": "6 $$ \\times $$ 6!"}, {"identifier": "C", "content": "7!"}, {"identifier": "D", "content": "5 $$ \\times $$ 7!"}] | ["A"] | null | Number of ways = Total - when B<sub>1</sub> and G<sub>1</sub> sit together
<br><br>Total ways to seat 8 people on round table = (8 - 1)! = 7!
<br><br>When B<sub>1</sub> and G<sub>1</sub> sit together then assume B<sub>1</sub> and G<sub>1</sub> are one people, so total 7 people are there and among B<sub>1</sub> and G<sub>1</sub> they can sit 2! ways.
<br><br>So total no of ways when B<sub>1</sub> and G<sub>1</sub> sit together
<br>= (7 - 1)! $$ \times $$ 2! = 6! $$ \times $$ 2!
<br><br>Number of ways = 7! - 6! $$ \times $$ 2! = 6!$$ \times $$(7 - 2) = 5 $$ \times $$ 6! | mcq | jee-main-2017-online-9th-april-morning-slot |
5PYzxgZ7ICeBa0uH9Ijgy2xukez5kjog | maths | permutations-and-combinations | circular-permutations | Let n > 2 be an integer. Suppose that there are
n Metro stations in a city located along a
circular path. Each pair of stations is connected
by a straight track only. Further, each pair of
nearest stations is connected by blue line,
whereas all remaining pairs of stations are
connected by red line. If the number of red lines
is 99 times the number of blue lines, then the
value of n is : | [{"identifier": "A", "content": "201"}, {"identifier": "B", "content": "199"}, {"identifier": "C", "content": "101"}, {"identifier": "D", "content": "200"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266330/exam_images/ui7cjccg9xqw9a5eyvuj.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267793/exam_images/ejhknnuxiul6gerikcuo.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267655/exam_images/sqrjvd7hnsr1xwffmqw7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Evening Slot Mathematics - Permutations and Combinations Question 123 English Explanation"></picture><br>
Number of blue lines = Number of sides = n<br><br>
Number of red lines = number of diagonals = <sup>n</sup>C<sub>2</sub> – n<br><br>
According to question, <br><br>
$${}^n{C_2} - n = 99n$$<br><br>
$$ \Rightarrow {{n\left( {n - 1} \right)} \over 2} - n = 99n$$<br><br>
$$ \Rightarrow {{n - 1} \over 2} = 99$$<br><br>
$$ \Rightarrow n = 201$$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
1lgzycfb1 | maths | permutations-and-combinations | circular-permutations | <p>The number of ways, in which 5 girls and 7 boys can be seated at a round table so that no two girls sit together, is :</p> | [{"identifier": "A", "content": "720"}, {"identifier": "B", "content": "$$7(360)^{2}$$"}, {"identifier": "C", "content": "$$7(720)^{2}$$"}, {"identifier": "D", "content": "$$126(5 !)^{2}$$"}] | ["D"] | null | We have,
<br><br>Number of girls $=5$
<br><br>Number of boys $=7$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmltts5a/3a0ee79a-7b48-44a1-98c7-740fb4585474/7ae256d0-5473-11ee-9283-c929f40dddd4/file-6y3zli1lmltts5b.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lmltts5a/3a0ee79a-7b48-44a1-98c7-740fb4585474/7ae256d0-5473-11ee-9283-c929f40dddd4/file-6y3zli1lmltts5b.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) 8th April Morning Shift Mathematics - Permutations and Combinations Question 26 English Explanation">
<br><br>So, number of ways of arranging boys <br><br>around the table $=6$ ! and 5 girls can be arranged in 7 gaps in ${ }^7 \mathrm{P}_5$ ways.
<br><br>$\therefore$ Required no. of ways $=6 ! \times{ }^7 \mathrm{P}_5$ $=126 \times(5 !)^2$ | mcq | jee-main-2023-online-8th-april-morning-shift |
Ya9LDYYS0XiNoh28 | maths | permutations-and-combinations | conditional-combinations | A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is | [{"identifier": "A", "content": "346"}, {"identifier": "B", "content": "140"}, {"identifier": "C", "content": "196"}, {"identifier": "D", "content": "280"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264044/exam_images/uld4peekswqjropomsze.webp" loading="lazy" alt="AIEEE 2003 Mathematics - Permutations and Combinations Question 173 English Explanation">
<br><br><b>Case 1 :</b>
<br><br>No of ways student can answer 10 questions = $${}^5{C_4} \times {}^8{C_6}$$ = 140
<br><br><b>Case 2 :</b>
<br><br>No of ways student can answer 10 questions = $${}^5{C_5} \times {}^8{C_5}$$ = 56
<br><br>$$\therefore$$ Total ways = 140 + 56 = 196 | mcq | aieee-2003 |
CcpStCuUJqkQIn23GBxDt | maths | permutations-and-combinations | conditional-combinations | Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can
be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same
team, is :
| [{"identifier": "A", "content": "500"}, {"identifier": "B", "content": "350"}, {"identifier": "C", "content": "200"}, {"identifier": "D", "content": "300"}] | ["D"] | null | From 5 girls 2 girls can be selected
<br><br>= <sup>5</sup>C<sub>2</sub> ways
<br><br>From 7 boys 3 boys can be selected
<br><br>= <sup>7</sup>C<sub>3</sub> way
<br><br>$$ \therefore $$ Total number of ways we can select 2 girls and 3 boys
<br><br>= <sup>5</sup>C<sub>2</sub> $$ \times $$ <sup>7</sup>C<sub>3</sub> ways
<br><br>When two boys A and B are chosen in a team then one more boy will be chosen from remaining 5 boys.
<br><br>So, no of ways 3 boys can be chosen when A and B should must be chosen = <sup>5</sup>C<sub>1</sub> ways
<br><br>$$ \therefore $$ Total number of ways a team of 2 girl and 3 boys can be made where boy A and B must be in the team = <sup>5</sup><sup></sup>C<sub>1</sub> $$ \times $$ <sup>5</sup>C<sub>2</sub> ways
<br><br>$$ \therefore $$ Required number of ways
<br>= Total number of ways $$-$$ when A and B are always included.
<br><br>= <sup>5</sup>C<sub>2</sub> $$ \times $$ <sup>7</sup>C<sub>3</sub> $$-$$ <sup>5</sup>C<sub>1</sub> $$ \times $$ <sup>5</sup>C<sub>2</sub>
<br><br>= 300 | mcq | jee-main-2019-online-9th-january-morning-slot |
iH5MBhbvLwnvCtEId43rsa0w2w9jxb03xdv | maths | permutations-and-combinations | conditional-combinations | A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can
randomly be selected from this group such that there is at least one boy and at least one girl in each team, is
1750, then n is equal to : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "28"}] | ["B"] | null | Given that 5 Boy, n girls.<br><br>
(1B, 2G) + (2B, 1G)<br><br>
$${}^5{C_1}.{}^n{C_2} + {}^5{C_2}.{}^n{C_1} = 1750$$<br><br>
$$ \Rightarrow 5.{{n\left( {n - 1} \right)} \over 2} + 10.n = 1750$$<br><br>
$$ \Rightarrow {{n\left( {n - 1} \right)} \over 2} + 2n = 350$$<br><br>
$$ \Rightarrow {n^2} - n + 4n = 700$$<br><br>
$$ \Rightarrow {n^2} + 3n - 700 = 0$$<br><br>
$$ \Rightarrow (n + 28)(n - 25) = 0$$<br><br>
$$ \Rightarrow n = 25, -28$$
| mcq | jee-main-2019-online-12th-april-evening-slot |
ynFU9TI4bBJDhzTbt7TZv | maths | permutations-and-combinations | conditional-combinations | A committee of 11 members is to be formed from
8 males and 5 females. If m is the number of ways
the committee is formed with at least 6 males and
n is the number of ways the committee is formed
with at least 3 females, then : | [{"identifier": "A", "content": "n = m \u2013 8"}, {"identifier": "B", "content": "m = n = 78"}, {"identifier": "C", "content": "m + n = 68"}, {"identifier": "D", "content": "m = n = 68"}] | ["B"] | null | At least 6 males means in the committee there can be 6 males or 7 males or 8 males.
<br><br>$$ \therefore $$ m = $${}^8{C_6} \times {}^5{C_5} + {}^8{C_7} \times {}^5{C_4} + {}^8{C_8} \times {}^5{C_3}$$ = 78
<br><br>At least 3 females means in the committee there can be 3 females or 4 females or 5 females.
<br><br>$$ \therefore $$ n = $${}^5{C_3} \times {}^8{C_8} + {}^5{C_4} \times {}^8{C_7} + {}^5{C_5} \times {}^8{C_6}$$ = 78
<br><br>So, m = n = 78 | mcq | jee-main-2019-online-9th-april-morning-slot |
dGIKlgBoeVJkXafbxljgy2xukfjjsiul | maths | permutations-and-combinations | conditional-combinations | Four fair dice are thrown independently 27 times. Then the expected number of times, at
least two dice show up a three or a five, is _________. | [] | null | 11 | 4 dice are independently thrown. Each die has probability to show 3 or 5 is <br><br>$$P = {2 \over 6} = {1 \over 3}$$<br><br>$$ \therefore $$ $$q = 1 - {1 \over 3} = {2 \over 3}$$ (not showing 3 or 5)<br><br>Experiment is performed with 4 dices independently<br><br>$$ \therefore $$ Their binomial distribution is <br><br>$${(q + p)^4} = {(q)^4} + {}^4{C_1}{q^3}p + {}^4{C_2}{q^2}{p^2} + {}^4{C_3}q{p^3} + {}^4{C_4}{P^4}$$<br><br>$$ \therefore $$ In one throw of each dice probability of showing 3 or 5 at least twice is<br><br>= $${p^4} + {}^4{C_3}q{p^3} + {}^4{C_2}{q^2}{p^2}$$<br><br>$$ = {{33} \over {81}}$$<br><br>Given such experiment performed 27 times<br><br>$$ \therefore $$ So expected outcomes = np<br><br>= $${{33} \over {81}} \times 27$$<br><br>= 11 | integer | jee-main-2020-online-5th-september-morning-slot |
ogAhHOb39XAuEA27w1jgy2xukfqbzbjj | maths | permutations-and-combinations | conditional-combinations | There are 3 sections in a question paper and
each section contains 5 questions. A candidate
has to answer a total of 5 questions, choosing
at least one question from each section. Then
the number of ways, in which the candidate
can choose the questions, is : | [{"identifier": "A", "content": "2250"}, {"identifier": "B", "content": "2255"}, {"identifier": "C", "content": "3000"}, {"identifier": "D", "content": "1500"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265719/exam_images/cbjd8rqt5qx4eis3hsmv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Mathematics - Permutations and Combinations Question 117 English Explanation">
<br><br>$$ \therefore $$ Total number of selection of 5 questions
<br><br>= $$\left( {{}^5{C_1}.{}^5{C_2}.{}^5{C_2}} \right)$$$$ \times $$3 + $$\left( {{}^5{C_1}.{}^5{C_1}.{}^5{C_3}} \right)$$ $$ \times $$ 3
<br><br>= 5 $$ \times $$10$$ \times $$10$$ \times $$3 + 5$$ \times $$5$$ \times $$10$$ \times $$3
<br><br>= 2250 | mcq | jee-main-2020-online-5th-september-evening-slot |
kWZNAunq6lVQD0QwtP1klrg0wg0 | maths | permutations-and-combinations | conditional-combinations | A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at
least 2 Indians and double the number of foreigners as Indians. Then the number of ways,
the committee can be formed, is : | [{"identifier": "A", "content": "1050"}, {"identifier": "B", "content": "575"}, {"identifier": "C", "content": "560"}, {"identifier": "D", "content": "1625"}] | ["D"] | null | Given,<br/><br/>Number of Indians = 6<br/><br/>Number of foreigners = 8<br/><br/>Committee of at least 2 Indians and double number of foreigners is to be formed. Hence, the required cases are<br/><br/>(2I, 4F) + (3I, 6F) + (4I, 8F)<br/><br/>= $${}^6{C_2} \times {}^8{C_4} + {}^6{C_3} \times {}^8{C_6} + {}^6{C_4} \times {}^8{C_8}$$<br/><br/>= (15 $$\times$$ 70) + (20 $$\times$$ 28) + (15 $$\times$$ 1)<br/><br/>= 1050 + 560 + 15 = 1625 | mcq | jee-main-2021-online-24th-february-morning-slot |
xFuZjzUptaNFdoUgdM1klrmy2in | maths | permutations-and-combinations | conditional-combinations | The students S<sub>1</sub>, S<sub>2</sub>, ....., S<sub>10</sub> are to be divided into 3 groups A, B and C such that each group has at least one student and the group C has at most 3 students. Then the total number of possibilities of forming such groups is ___________. | [] | null | 31650 | If group C has one student then number of
groups
<br><br>= <sup>10</sup>C<sub>1</sub>
[2<sup>9</sup>
– 2] = 5100
<br><br>If group C has two students then number of
groups
<br><br>= <sup>10</sup>C<sub>2</sub>
[2<sup>8</sup>
– 2] = 11430
<br><br>If group C has three students then number of
groups
<br><br>= <sup>10</sup>C<sub>3</sub>
× [2<sup>7</sup>
– 2] = 15120
<br><br>So total groups = 31650 | integer | jee-main-2021-online-24th-february-evening-slot |
OtCzg7Sq3gsThouMtK1kls4fv1n | maths | permutations-and-combinations | conditional-combinations | The total number of positive integral solutions (x, y, z) such that xyz = 24 is : | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "45"}, {"identifier": "D", "content": "30"}] | ["D"] | null | $$x.y.z = 24$$<br><br>$$x.y.z = {2^3}.\,{3^1}$$<br><br>Three 2 has to be distributed among x, y and z<br><br>Each may receive none, one or two<br><br>$$\therefore$$ Number of ways = $${}^{3 + 3 - 1}{C_{3 - 1}}$$ = $$^5{C_2}$$ ways<br><br>Similarly one 3 has to be distributed among x, y and z<br><br>$$ \therefore $$ Number of ways = $${}^{1 + 3 - 1}{C_{3 - 1}}$$ = $$^3{C_2}$$ ways<br><br>Total ways = $$^5{C_2}\,.{\,^3}{C_2}$$ = 30 | mcq | jee-main-2021-online-25th-february-morning-slot |
eNLKKMO5cYrCM8Jz6S1kluge0sz | maths | permutations-and-combinations | conditional-combinations | The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only is : | [{"identifier": "A", "content": "35"}, {"identifier": "B", "content": "42"}, {"identifier": "C", "content": "82"}, {"identifier": "D", "content": "77"}] | ["D"] | null | (I) First possibility is 1, 1, 1, 1, 1, 2, 3<br><br>required number = $${{7!} \over {5!}}$$ = 7 $$\times$$ 6 = 42<br><br>(II) Second possibility is 1, 1, 1, 1, 2, 2, 2<br><br>required number = $${{7!} \over {4!3!}} = {{7 \times 6 \times 5} \over 6} = 35$$<br><br>Total = 42 + 35 = 77 | mcq | jee-main-2021-online-26th-february-morning-slot |
1krq1dvgf | maths | permutations-and-combinations | conditional-combinations | There are 15 players in a cricket team, out of which 6 are bowlers, 7 are batsman and 2 are wicketkeepers. The number of ways, a team of 11 players be selected from them so as to include at least 4 bowlers, 5 batsman and 1 wicketkeeper, is ______________. | [] | null | 777 | 15 : Players<br><br>6 : Bowlers<br><br>7 : Batsman<br><br>2 : Wicket keepers<br><br>Total number of ways for :<br><br>at least 4 bowler, 5 batsman & 1 wicket keeper<br><br>= $${}^6{C_4}({}^7{C_6} \times {}^2{C_1} + {}^7{C_5} \times {}^2{C_2}) + {}^6{C_5} \times {}^7{C_5} \times {}^2{C_1}$$<br><br>$$ = 777$$ | integer | jee-main-2021-online-20th-july-morning-shift |
1krw30swu | maths | permutations-and-combinations | conditional-combinations | There are 5 students in class 10, 6 students in class 11 and 8 students in class 12. If the number of ways, in which 10 students can be selected from them so as to include at least 2 students from each class and at most 5 students from the total 11 students of class 10 and 11 is 100 k, then k is equal to _____________. | [] | null | 238 | Class $$\matrix{
{{{10}^{th}}} & {{{11}^{th}}} & {{{12}^{th}}} \cr
} $$<br><br>Total student $$\matrix{
5 & 6 & 8 \cr
} $$<br><br>$$\matrix{
2 & 3 & 5 \cr
} \Rightarrow $$ $${}^5{C_2} \times {}^6{C_3} \times {}^8{C_5}$$<br><br>Number of selection $$\matrix{
2 & 2 & 6 \cr
} \Rightarrow {}^5{C_2} \times {}^6{C_2} \times {}^8{C_6}$$<br><br>$$\matrix{
3 & 2 & 5 \cr
} \Rightarrow {}^5{C_3} \times {}^6{C_2} \times {}^8{C_5}$$<br><br>$$\Rightarrow$$ Total number of ways = 23800<br><br>According to question 100 K = 23800<br><br>$$\Rightarrow$$ K = 238 | integer | jee-main-2021-online-25th-july-morning-shift |
1ktgp4gdp | maths | permutations-and-combinations | conditional-combinations | Let S = {1, 2, 3, 4, 5, 6, 9}. Then the number of elements in the set T = {A $$ \subseteq $$ S : A $$\ne$$ $$\phi$$ and the sum of all the elements of A is not a multiple of 3} is _______________. | [] | null | 80 | 3n type $$\to$$ 3, 6, 9 = P<br><br>3n $$-$$ 1 type $$\to$$ 2, 5 = Q<br><br>3n $$-$$ 2 type $$\to$$ 1, 4 = R<br><br>number of subset of S containing one element which are not divisible by 3 = $${}^2$$C<sub>1</sub> + $${}^2$$C<sub>1</sub> = 4<br><br>number of subset of S containing two numbers whose some is not divisible by 3<br><br>= $${}^3$$C<sub>1</sub> $$\times$$ $${}^2$$C<sub>1</sub> + $${}^3$$C<sub>1</sub> $$\times$$ $${}^2$$C<sub>1</sub> + $${}^2$$C<sub>2</sub> + $${}^2$$C<sub>2</sub> = 14<br><br>number of subsets containing 3 elements whose sum is not divisible by 3<br><br>= $${}^3$$C<sub>2</sub> $$\times$$ $${}^4$$C<sub>1</sub> + ($${}^2$$C<sub>2</sub> $$\times$$ $${}^2$$C<sub>1</sub>)2 + $${}^3$$C<sub>1</sub>($${}^2$$C<sub>2</sub> + $${}^2$$C<sub>2</sub>) = 22<br><br>number of subsets containing 4 elements whose sum is not divisible by 3<br><br>= $${}^3$$C<sub>3</sub> $$\times$$ $${}^4$$C<sub>1</sub> + $${}^3$$C<sub>2</sub>($${}^2$$C<sub>2</sub> + $${}^2$$C<sub>2</sub>) + ($${}^3$$C<sub>1</sub>$${}^2$$C<sub>1</sub> $$\times$$ $${}^2$$C<sub>2</sub>)2<br><br>= 4 + 6 + 12 = 22<br><br>number of subsets of S containing 5 elements whose sum is not divisible by 3.<br><br>= $${}^3$$C<sub>3</sub>($${}^2$$C<sub>2</sub> + $${}^2$$C<sub>2</sub>) + ($${}^3$$C<sub>2</sub>$${}^2$$C<sub>1</sub> $$\times$$ $${}^2$$C<sub>2</sub>) $$\times$$ 2 = 2 + 12 = 14<br><br>number of subsets of S containing 6 elements whose sum is not divisible by 3 = 4<br><br>$$\Rightarrow$$ Total subsets of Set A whose sum of digits is not divisible by 3 = 4 + 14 + 22 + 22 + 14 + 4 = 80. | integer | jee-main-2021-online-27th-august-evening-shift |
1l55h4ytt | maths | permutations-and-combinations | conditional-combinations | <p>The number of ways to distribute 30 identical candies among four children C<sub>1</sub>, C<sub>2</sub>, C<sub>3</sub> and C<sub>4</sub> so that C<sub>2</sub> receives at least 4 and at most 7 candies, C<sub>3</sub> receives at least 2 and at most 6 candies, is equal to :</p> | [{"identifier": "A", "content": "205"}, {"identifier": "B", "content": "615"}, {"identifier": "C", "content": "510"}, {"identifier": "D", "content": "430"}] | ["D"] | null | <p>By multinomial theorem, no. of ways to distribute 30 identical candies among four children C<sub>1</sub>, C<sub>2</sub> and C<sub>3</sub>, C<sub>4</sub></p>
<p>= Coefficient of x<sup>30</sup> in (x<sup>4</sup> + x<sup>5</sup> + .... + x<sup>7</sup>) (x<sup>2</sup> + x<sup>3</sup> + .... + x<sup>6</sup>) (1 + x + x<sup>2</sup> ....)<sup>2</sup></p>
<p>= Coefficient of x<sup>24</sup> in $${{(1 - {x^4})} \over {1 - x}}{{(1 - {x^5})} \over {1 - x}}{{{{(1 - {x^{31}})}^2}} \over {{{(1 - x)}^2}}}$$</p>
<p>= Coefficient of x<sup>24</sup> in $$(1 - {x^4} - {x^5} + {x^9}){(1 - x)^{ - 4}}$$</p>
<p>$$ = {}^{27}{C_{24}} - {}^{23}{C_{20}} - {}^{22}{C_{19}} + {}^{18}{C_{15}} = 430$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l56rx9ym | maths | permutations-and-combinations | conditional-combinations | <p>Let A be a matrix of order 2 $$\times$$ 2, whose entries are from the set {0, 1, 2, 3, 4, 5}. If the sum of all the entries of A is a prime number p, 2 < p < 8, then the number of such matrices A is ___________.</p> | [] | null | 180 | <p>$$\because$$ Sum of all entries of matrix A must be prime p such that 2 < p < 8 then sum of entries may be 3, 5 or 7.</p>
<p>If sum is 3 then possible entries are (0, 0, 0, 3), (0, 0, 1, 2) or (0, 1, 1, 1).</p>
<p>$$\therefore$$ Total number of matrices = 4 + 4 + 12 = 20</p>
<p>If sum of 5 then possible entries are</p>
<p>(0, 0, 0, 5), (0, 0, 1, 4), (0, 0, 2, 3), (0, 1, 1, 3), (0, 1, 2, 2) and (1, 1, 1, 2).</p>
<p>$$\therefore$$ Total number of matrices = 4 + 12 + 12 + 12 + 12 + 4 = 56</p>
<p>If sum is 7 then possible entries are</p>
<p>(0, 0, 2, 5), (0, 0, 3, 4), (0, 1, 1, 5), (0, 3, 3, 1), (0, 2, 2, 3), (1, 1, 1, 4), (1, 2, 2, 2), (1, 1, 2, 3) and (0, 1, 2, 4).</p>
<p>Total number of matrices with sum 7 = 104</p>
<p>$$\therefore$$ Total number of required matrices</p>
<p>= 20 + 56 + 104</p>
<p>= 180</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l57p1hn1 | maths | permutations-and-combinations | conditional-combinations | <p>The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is _____________.</p> | [] | null | 56 | First we arrange 5 red cubes in a row and assume
x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub> and x<sub>6</sub> number of blue cubes
between them<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc3zecxi/95ec386f-1c04-46f9-b1e8-eca213646d74/ff4a6070-84a8-11ed-9e2c-bf48676ab5eb/file-1lc3zecxj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc3zecxi/95ec386f-1c04-46f9-b1e8-eca213646d74/ff4a6070-84a8-11ed-9e2c-bf48676ab5eb/file-1lc3zecxj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Permutations and Combinations Question 84 English Explanation"><br><br>
Here, $x_1+x_2+x_3+x_4+x_5+x_6=11$ and $x_2, x_3, x_4, x_5 \geq 2$<br><br>
So $x_1+x_2+x_3+x_4+x_5+x_6=3$<br><br>
No. of solutions $={ }^8 C_5=56$ | integer | jee-main-2022-online-27th-june-morning-shift |
1l58afhi6 | maths | permutations-and-combinations | conditional-combinations | <p>There are ten boys B<sub>1</sub>, B<sub>2</sub>, ......., B<sub>10</sub> and five girls G<sub>1</sub>, G<sub>2</sub>, ........, G<sub>5</sub> in a class. Then the number of ways of forming a group consisting of three boys and three girls, if both B<sub>1</sub> and B<sub>2</sub> together should not be the members of a group, is ___________.</p> | [] | null | 1120 | <p>Number of ways when B<sub>1</sub> and B<sub>2</sub> are not together</p>
<p>= Total number of ways of selecting 3 boys $$-$$ B<sub>1</sub> and B<sub>2</sub> are together</p>
<p>= <sup>10</sup>C<sub>3</sub> $$-$$ <sup>8</sup>C<sub>1</sub></p>
<p>= $${{10\,.\,9\,.\,8} \over {1\,.\,2\,.\,3}} - 8$$</p>
<p>= 112</p>
<p>Number of ways to select 3 girls = <sup>5</sup>C<sub>3</sub> = 10</p>
<p>$$\therefore$$ Total number of ways = 112 $$\times$$ 10 = 1120</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l5ajwu4h | maths | permutations-and-combinations | conditional-combinations | <p>Let A be a 3 $$\times$$ 3 matrix having entries from the set {$$-$$1, 0, 1}. The number of all such matrices A having sum of all the entries equal to 5, is ___________.</p> | [] | null | 414 | <b>Case-I</b>: <br/><br/>$\begin{aligned} 1 & \rightarrow 7 \text { times } \\\\ \text { and }-1 & \rightarrow 2 \text { times } \end{aligned}$<br/><br/>
number of possible marrix $=\frac{9 !}{7 ! 2 !}=36$<br/><br/>
<b>Case-II</b>: <br/><br/>$1 \rightarrow 6$ times,<br/><br/>
$-1 \rightarrow 1$ times<br/><br/>
and $0 \rightarrow 2$ times<br/><br/>
number of possible marrix $=\frac{9 !}{6 ! 2 !}=252$<br/><br/>
<b>Case-III</b>: <br/><br/>$ 1 \rightarrow 5$ times,<br/><br/>
and $0 \rightarrow 4$ times<br/><br/>
number of possible marrix $=\frac{9 !}{5 ! 4 !}=126$<br/><br/>
Hence total number of all such matrix $A$ $=414$ | integer | jee-main-2022-online-25th-june-morning-shift |
1l5c20i6d | maths | permutations-and-combinations | conditional-combinations | <p>In an examination, there are 5 multiple choice questions with 3 choices, out of which exactly one is correct. There are 3 marks for each correct answer, $$-$$2 marks for each wrong answer and 0 mark if the question is not attempted. Then, the number of ways a student appearing in the examination gets 5 marks is ____________.</p> | [] | null | 40 | Let student marks $x$ correct answers and $y$ incorrect. So
<br/><br/>
$3 x-2 y=5$ and $x+y \leq 5$ where $x, y \in \mathrm{W}$
<br/><br/>
Only possible solution is $(x, y)=(3,2)$
<br/><br/>
Students can mark correct answers by only one choice but for an incorrect answer, there are two choices. So total number of ways of scoring 5 marks $={ }^{5} C_{3}(1)^{3} \cdot(2)^{2}=40$ | integer | jee-main-2022-online-24th-june-morning-shift |
1ldo7e9ya | maths | permutations-and-combinations | conditional-combinations | <p>Number of integral solutions to the equation $$x+y+z=21$$, where $$x \ge 1,y\ge3,z\ge4$$, is equal to ____________.</p> | [] | null | 105 | $\begin{aligned} & x+y+z=21 \\\\ & \because \quad x \geq 1, y \geq 3, y \geq 4 \\\\ & \therefore \quad x_1+y_1+z_1=13 \\\\ & \text { Number of solutions }={ }^{13+3-1} C_{3-1} \\\\ & ={ }^{15} C_2=\frac{15 \times 14}{2}=7 \times 15 \\\\ & =105\end{aligned}$ | integer | jee-main-2023-online-1st-february-evening-shift |
ldqvb3cg | maths | permutations-and-combinations | conditional-combinations | The number of ways of selecting two numbers $a$ and $b, a \in\{2,4,6, \ldots ., 100\}$ and $b \in\{1,3,5, \ldots . ., 99\}$ such that 2 is the remainder when $a+b$ is divided by 23 is : | [{"identifier": "A", "content": "186"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "108"}, {"identifier": "D", "content": "268"}] | ["C"] | null | <p>$$a+b=23\lambda+2$$</p>
<p>$$\lambda=0,1,2,$$ ...., but $$\lambda$$ cannot be even as $$a+b$$ is odd</p>
<p>$$\lambda=1$$ $$(a, b)\to12$$ pairs</p>
<p>$$\lambda=3$$ $$(a,b)\to35$$ pairs</p>
<p>$$\lambda=5$$ $$(a,b)\to42$$ pairs</p>
<p>$$\lambda=7$$ $$(a,b)\to19$$ pairs</p>
<p>$$\lambda=9$$ $$(a,b)\to0$$ pairs</p>
<p>$$ \vdots $$</p>
<p>Total $$=12+35+42+19=108$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldsetdwf | maths | permutations-and-combinations | conditional-combinations | <p>The number of 3 digit numbers, that are divisible by either 3 or 4 but not divisible by 48, is :</p> | [{"identifier": "A", "content": "400"}, {"identifier": "B", "content": "472"}, {"identifier": "C", "content": "507"}, {"identifier": "D", "content": "432"}] | ["D"] | null | <p>Number divisible by 3 = 300</p>
<p>Number divisible by 4 = 225</p>
<p>Number divisible by 12 = 75</p>
<p>Number divisible by 48 = 18</p>
<p>Total required number = 300 + 225$$-$$ 75 $$-$$ 18 = 432</p>
<p>$$\therefore$$ Option (1) is correct.</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldu63yht | maths | permutations-and-combinations | conditional-combinations | <p>Suppose Anil's mother wants to give 5 whole fruits to Anil from a basket of 7 red apples, 5 white apples and 8 oranges. If in the selected 5 fruits, at least 2 oranges, at least one red apple and at least one white apple must be given, then the number of ways, Anil's mother can offer 5 fruits to Anil is ____________</p> | [] | null | 6860 OR 3 | Total 8 oranges, 5 white apple and 7 red apple. 5 fruits needs to be selected.
<br/><br/>
<b>Case I</b>: 3 orange $+1$ red apple $+1$ white apple
<br/><br/>
$$
={ }^{8} C_{3} \times{ }^{7} C_{1} \times{ }^{5} C_{1}=1960
$$
<br/><br/>
<b>Case II</b> : 2 oranges $+2$ red apples $+1$ white apple.
<br/><br/>
$$
={ }^{8} C_{2} \times{ }^{7} C_{2} \times{ }^{5} C_{1}=2940
$$
<br/><br/>
<b>Case III</b> : 2 oranges $+1$ red apples $+2$ white apple.
<br/><br/>
$$
\begin{aligned}
& ={ }^{8} C_{2} \times{ }^{7} C_{1} \times{ }^{5} C_{2} \\\\
& =1960 \\\\
\text { Total } & =1960+2940+1960 \\\\
& =6860
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-evening-shift |
1ldwwxq9r | maths | permutations-and-combinations | conditional-combinations | <p>The number of square matrices of order 5 with entries from the set {0, 1}, such that the sum of all the elements in each row is 1 and the sum of all the elements in each column is also 1, is :</p> | [{"identifier": "A", "content": "125"}, {"identifier": "B", "content": "150"}, {"identifier": "C", "content": "225"}, {"identifier": "D", "content": "120"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5h0vkb/6ab5456d-64a1-4237-89f3-e400627057ab/241652a0-ad13-11ed-8a8c-4d67f5492755/file-1le5h0vkc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5h0vkb/6ab5456d-64a1-4237-89f3-e400627057ab/241652a0-ad13-11ed-8a8c-4d67f5492755/file-1le5h0vkc.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Permutations and Combinations Question 45 English Explanation"></p>
<p>$\because$ In every row and every column there would be exactly one 1 and four zeroes.
<br><br>
Number of matrices $={ }^{5} C_{1} \cdot{ }^{4} C_{1} \cdot{ }^{3} C_{1} \cdot{ }^{2} C_{1} \cdot{ }^{1} C_{1}=$ 120</p> | mcq | jee-main-2023-online-24th-january-evening-shift |
1ldyc2w6e | maths | permutations-and-combinations | conditional-combinations | <p>A boy needs to select five courses from 12 available courses, out of which 5 courses are language courses. If he can choose at most two language courses, then the number of ways he can choose five courses is __________</p> | [] | null | 546 | <p>Among 12 courses, 5 courses are of language.</p>
<p>$$\therefore$$ Remaining 7 are different courses.</p>
<p>Now, number of ways to select 5 courses where at most 2 language courses present.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top}
.tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 448px">
<colgroup>
<col style="width: 106px">
<col style="width: 87px">
<col style="width: 86px">
<col style="width: 169px">
</colgroup>
<thead>
<tr>
<th class="tg-7btt"></th>
<th class="tg-7btt">Language</th>
<th class="tg-7btt">Different</th>
<th class="tg-7btt">Number of ways</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">Case 1</td>
<td class="tg-c3ow">0</td>
<td class="tg-c3ow">5</td>
<td class="tg-c3ow">$${}^5{C_0} \times {}^7{C_5}$$</td>
</tr>
<tr>
<td class="tg-c3ow">Case 2</td>
<td class="tg-c3ow">1</td>
<td class="tg-c3ow">4</td>
<td class="tg-c3ow">$${}^5{C_1} \times {}^7{C_4}$$</td>
</tr>
<tr>
<td class="tg-c3ow">Case 3</td>
<td class="tg-c3ow">2</td>
<td class="tg-c3ow">3</td>
<td class="tg-c3ow">$${}^5{C_2} \times {}^7{C_3}$$</td>
</tr>
</tbody>
</table></p>
<p>$$\therefore$$ Total number of ways</p>
<p>$$ = {}^5{C_0} \times {}^7{C_5} + {}^5{C_1} \times {}^7{C_4} + {}^5{C_2} \times {}^7{C_3}$$</p>
<p>$$ = 546$$</p> | integer | jee-main-2023-online-24th-january-morning-shift |
1ldyc8xzj | maths | permutations-and-combinations | conditional-combinations | <p>The number of 9 digit numbers, that can be formed using all the digits of the number 123412341 so that the even digits occupy only even places, is ______________.</p> | [] | null | 60 | <p>Here, even digits are 2 and 4.</p>
<p>Number of digit "2" presents = 2</p>
<p>Number of digit "4" presents = 2</p>
<p>$$\therefore$$ Total even digits = 4</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2pxjq6/e260bc6f-c1c6-485e-abbc-aa68e85a10dd/9fbb03e0-ab8f-11ed-a599-53c07234da0d/file-1le2pxjq7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2pxjq6/e260bc6f-c1c6-485e-abbc-aa68e85a10dd/9fbb03e0-ab8f-11ed-a599-53c07234da0d/file-1le2pxjq7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Morning Shift Mathematics - Permutations and Combinations Question 44 English Explanation"></p>
<p>$$\therefore$$ Total 4 even places presents.</p>
<p>Number of ways to put those 4 digits in those 4 places $$ = {{{}^4{C_4} \times 4!} \over {2!2!}} = {{4!} \over {2!2!}}$$</p>
<p>Now, remaining 5 digits (three 1 and two 3) can be put in those 5 places in $$ = {{{}^5{C_5} \times 5!} \over {3!2!}} = {{5!} \over {3!2!}}$$ ways.</p>
<p>$$\therefore$$ Total possible 9 digit numbers</p>
<p>$$ = {{4!} \over {2!2!}} \times {{5!} \over {3!2!}} = 60$$</p> | integer | jee-main-2023-online-24th-january-morning-shift |
1lgq0urew | maths | permutations-and-combinations | conditional-combinations | <p>The number of seven digit positive integers formed using the digits $$1,2,3$$ and $$4$$ only and sum of the digits equal to $$12$$ is ___________.</p> | [] | null | 413 | $$
x_1+x_2+x_3+\ldots x_7=12
$$. This equation represents the number of ways to distribute 12 identical items (the sum of the digits) into 7 distinct boxes (the seven digits of the number), where each box can contain one of the numbers 1, 2, 3, or 4.
<br/><br/>Number of solutions
<br/><br/>$$
\begin{aligned}
& =\text { Coefficient of } x^{12} \text { in }\left(x^1+x^2+x^3+x^4\right)^7 \\\\
& =\text { Coefficient of } x^5 \text { in }\left(1+x+x^2+x^3\right)^7 \\\\
& =\text { Coefficient of } x^5 \text { in }\left(1-x^4\right)^7(1-x)^{-7} \\\\
& =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right)(1-x)^{-7} \\\\
& =\text { Coefficient of } x^5 \text { in }\left(1-7 x^4\right) \sum_{r=0}^{\infty}{ }^{7+r-1} C_r \cdot x^r \\\\
& ={ }^{11} C_5-7 \times{ }^7 C_1 \\\\
& =462-49=413
\end{aligned}
$$ | integer | jee-main-2023-online-13th-april-morning-shift |
1lguw0e0f | maths | permutations-and-combinations | conditional-combinations | <p>The number of triplets $$(x, \mathrm{y}, \mathrm{z})$$, where $$x, \mathrm{y}, \mathrm{z}$$ are distinct non negative integers satisfying $$x+y+z=15$$, is :</p> | [{"identifier": "A", "content": "136"}, {"identifier": "B", "content": "80"}, {"identifier": "C", "content": "92"}, {"identifier": "D", "content": "114"}] | ["D"] | null | We have, $x+y+z=15$
<br/><br/>$$
\begin{aligned}
\text { Total number of solution } & ={ }^{15+3-1} C_{3-1} \\\\
& ={ }^{17} C_2=\frac{17 \times 16}{1 \times 2}=136
\end{aligned}
$$
<br/><br/>Now, we need to exclude the solutions where two of $(x, y, z)$ are the same.
<br/><br/>1) For the case $x = y \neq z$ :
<br/><br/>$ 2x + z = 15 $
<br/><br/>The solutions are :
<br/><br/>$ x = 0, z = 15 $
<br/><br/>$ x = 1, z = 13 $
<br/><br/>$ x = 2, z = 11 $
<br/><br/>$ x = 3, z = 9 $
<br/><br/>$ x = 4, z = 7 $
<br/><br/>$ x = 5, z = 5 $ (Not valid as all are the same)
<br/><br/>$ x = 6, z = 3 $
<br/><br/>$ x = 7, z = 1 $
<br/><br/>Out of these, 7 are valid.
<br/><br/>Similarly, for the cases $y = z \neq x$ and $z = x \neq y$, there will be 7 valid solutions for each, so a total of $ 7 \times 3 = 21 $ solutions where two of the variables are equal.
<br/><br/>Thus, the number of triplets where all are distinct is :
<br/><br/>$ 136 - 21 = 115 $
<br/><br/>There is one solution in which $\mathrm{x}=\mathrm{y}=\mathrm{z}$
<br/><br/>Required answer $=136-21-1=114$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lh245n1s | maths | permutations-and-combinations | conditional-combinations | <p>The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is ___________.</p> | [] | null | 3483638676 | <li><p><strong>Total ways without any restrictions :</strong>
<br/><br/>There are $3^{20}$ ways to distribute the oranges to the 3 children.</p>
</li>
<li><p><strong>Number of ways one child receives no orange :</strong>
<br/><br/>Choose 1 child out of the 3 to not receive any orange in ${ }^3 C_1 = 3$ ways. Distribute 20 oranges to the remaining 2 children in $2^{20}$ ways. However, we've included the scenarios where the 2 children each get all the oranges. So, we subtract the 2 ways where one of the two remaining children gets all the oranges.
$ { }^3 C_1(2^{20} - 2) $</p>
</li>
<li><p><strong>Number of ways two children receive no orange :</strong>
<br/><br/>Choose 2 children out of the 3 to not receive any oranges in ${ }^3 C_2 = 3$ ways. The third child will receive all 20 oranges in $1^{20} = 1$ way.
$ { }^3 C_2 \times 1^{20} = 3 $</p>
</li>
</ol>
<br/>Number of ways <br/><br/>$=$ Total $-($ One child receive no orange + two child receive no orange)
<br/><br/>$$
\begin{aligned}
& =3^{20}-\left({ }^3 C_1\left(2^{20}-2\right)+{ }^3 C_2 1^{20}\right) \\\\
& =3483638676
\end{aligned}
$$ | integer | jee-main-2023-online-6th-april-morning-shift |
jaoe38c1lsd39r8j | maths | permutations-and-combinations | conditional-combinations | <p>The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is</p> | [{"identifier": "A", "content": "130"}, {"identifier": "B", "content": "136"}, {"identifier": "C", "content": "142"}, {"identifier": "D", "content": "406"}] | ["B"] | null | <p>To solve this problem, we can use a classic combinatorics method known as "stars and bars" (or "balls and bins"), which is a way to solve problems involving distributing identical items into distinct groups with certain restrictions.</p>
<p>First, since each child must get at least 2 apples, let's give 2 apples to each child right away. That accounts for 6 apples (2 apples for each of the 3 children). Now, we have 21 - 6 = 15 apples left to distribute freely among the three children.</p>
<p>The "stars and bars" technique involves representing the apples as stars (*) and the divisions between children as bars (|). For example, if we had 5 apples to distribute among three children, one possible distribution could be represented as **|*|**. This means the first child gets 2 apples, the second child gets 1 apple, and the third child gets 2 apples.</p>
<p>In our case, we need to distribute 15 apples (stars) among the three children with 2 bars to create the partitions. We arrange 15 stars and 2 bars in a row, where the arrangement of stars and bars corresponds to a distribution of the apples.</p>
<p>The total number of objects we're arranging is 15 apples + 2 bars = 17 objects. We need to choose 2 positions out of these 17 to place the bars. The remaining positions will be occupied by the stars (apples).</p>
<p>The number of ways to choose 2 positions out of 17 for the bars is given by the binomial coefficient:</p>
$$
\text{Number of ways} = \binom{17}{2} = \frac{17!}{2!(17-2)!} = \frac{17 \times 16}{2 \times 1} = 136
$$
<p>Thus, there are 136 ways to distribute the 21 identical apples among three children such that each child gets at least 2 apples. The correct answer is Option B: 136.</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
luxweoza | maths | permutations-and-combinations | conditional-combinations | <p>The number of integers, between 100 and 1000 having the sum of their digits equals to 14 , is __________.</p> | [] | null | 70 | <p>Number in this range will be 3-digit number.</p>
<p>$$N=\overline{a b c}$$ such that $$a+b+c=14$$</p>
<p>Also, $$a \geq 1, \quad a, b, c \in\{0,1,2, \ldots 9\}$$</p>
<p>Case I</p>
<p>All 3-digit same</p>
<p>$$\Rightarrow 3 a=14$$ not possible</p>
<p>Case II</p>
<p>Exactly 2 digit same:</p>
<p>$$\Rightarrow 2 a+c=14$$</p>
<p>$$\begin{aligned}
& (a, c) \in\{(3,8),(4,6),(5,4),(6,2),(7,0)\} \\
& \Rightarrow\left(\frac{3!}{2!}\right) \text { ways } \Rightarrow 5 \times 3-1 \\
& =15-1=14
\end{aligned}$$</p>
<p>Case III</p>
<p>All digits are distinct</p>
<p>$$a+b+c=14$$</p>
<p>without losing generality $$a > b > c$$</p>
<p>$$\begin{aligned}
& (a, b, c) \in\left\{\begin{array}{l}
(9,5,0),(9,4,1),(9,3,2) \\
(8,6,0),(8,5,1),(8,4,2) \\
(7,6,1),(7,5,2),(7,4,3) \\
(6,5,3)
\end{array}\right. \\
& \Rightarrow 8 \times 3!+2(3!-2!)=48+8=56 \\
& =0+14+56=70
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-evening-shift |
lv3ve4dd | maths | permutations-and-combinations | conditional-combinations | <p>The number of ways five alphabets can be chosen from the alphabets of the word MATHEMATICS, where the chosen alphabets are not necessarily distinct, is equal to:</p> | [{"identifier": "A", "content": "179"}, {"identifier": "B", "content": "177"}, {"identifier": "C", "content": "175"}, {"identifier": "D", "content": "181"}] | ["A"] | null | <p>$$\begin{aligned}
& 2 M \\
& 2 A \\
& 2 T \\
& H, E, I, C, S
\end{aligned}$$</p>
<p>Case-I</p>
<p>2 Alike 2 Alike 1 Diff</p>
<p>$${ }^3 C_2 \times{ }^6 C_1=18$$</p>
<p>Case-II</p>
<p>2 Alike + 3 Diff</p>
<p>$${ }^3 C_1 \times{ }^7 C_3=105$$</p>
<p>Case-III</p>
<p>All different</p>
<p>$${ }^8 C_5=56$$</p>
<p>Total ways $$=179$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
TIV9ByukNGgS1zIC | maths | permutations-and-combinations | conditional-permutations | Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is : | [{"identifier": "A", "content": "125"}, {"identifier": "B", "content": "105"}, {"identifier": "C", "content": "374"}, {"identifier": "D", "content": "625"}] | ["C"] | null | There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed
<br><br><b>Case 1 :</b> First digit is 1 = 1 _ _ _
<br><br>Possible numbers starting with 1 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125
<br><br>But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.
<br><br><b>Case 2 :</b> First digit is 2 = 2 _ _ _
<br><br>Possible numbers starting with 2 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125
<br><br><b>Case 3 :</b> First digit is 3 = 3 _ _ _
<br><br>Possible numbers starting with 3 = 1$$ \times $$5$$ \times $$5$$ \times $$5 = 125
<br><br>Total possible numbers = 124 + 125 + 125 = 374 | mcq | aieee-2002 |
e1EyFwNJvi8vSFaE | maths | permutations-and-combinations | conditional-permutations | Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are : | [{"identifier": "A", "content": "216"}, {"identifier": "B", "content": "375"}, {"identifier": "C", "content": "400"}, {"identifier": "D", "content": "720"}] | ["D"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263746/exam_images/x99fiwum87l1tzmfgxkv.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Permutations and Combinations Question 177 English Explanation">
<br>$$\therefore$$ Total no of ways = 5$$ \times $$6$$ \times $$6$$ \times $$$${}^4{C_1}$$ = 720 | mcq | aieee-2002 |
shaQfGQgffjfRATa | maths | permutations-and-combinations | conditional-permutations | Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are : | [{"identifier": "A", "content": "312"}, {"identifier": "B", "content": "3125"}, {"identifier": "C", "content": "120"}, {"identifier": "D", "content": "216"}] | ["D"] | null | <b>Note :</b> For a number to be divisible by 3, the sum of digits should be divisible by 3.
<br><br>Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.
<br><br><b>Set 1 :</b> Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267485/exam_images/cosyxpmaifzqeuqzagre.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Permutations and Combinations Question 175 English Explanation 1">
<br><br>So total no of arrangement = 1$$ \times $$2$$ \times $$3$$ \times $$4$$ \times $$5 = 5!
<br><br><b>Set 2 :</b> Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)
<img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264157/exam_images/nmuqaapkyzrj5ufunbg2.webp" loading="lazy" alt="AIEEE 2002 Mathematics - Permutations and Combinations Question 175 English Explanation 2">
<br><br>So total no of arrangement = 4$$ \times $$4$$ \times $$3$$ \times $$2$$ \times $$1 = 4.4!
<br><br>$$\therefore$$ Total arrangement = 5! + 4.4! = 216 | mcq | aieee-2002 |
n1oWTwJqCyxG22uT | maths | permutations-and-combinations | conditional-permutations | How many ways are there to arrange the letters in the word GARDEN with vowels in alphabetical order | [{"identifier": "A", "content": "480"}, {"identifier": "B", "content": "240"}, {"identifier": "C", "content": "360"}, {"identifier": "D", "content": "120"}] | ["C"] | null | In the word ''GARDEN'', there are two vowels A and E present, and A should come always before E.
<br><br>$$\therefore\,\,\,$$ Total no of ways = $${{6!} \over {2!}}$$ = 360
<br><br>Here A and E has fixed order that is why we divide by 2!. | mcq | aieee-2004 |
CtCfGCORzMr3qyUf | maths | permutations-and-combinations | conditional-permutations | How many different words can be formed by jumbling the letters in the word MISSISSIPPI in which no two S are adjacent? | [{"identifier": "A", "content": "$$8.{}^6{C_4}.{}^7{C_4}$$ "}, {"identifier": "B", "content": "$$6.7.{}^8{C_4}$$ "}, {"identifier": "C", "content": "$$6.8.{}^7{C_4}$$. "}, {"identifier": "D", "content": "$$7.{}^6{C_4}.{}^8{C_4}$$ "}] | ["D"] | null | <p>This problem is solved using gap method. As here no 'S' is adjacent to each other so we have to put them in the gap. So first write all the letters other than 'S' such a way that there is a gap between two letters.</p>
<p>Given word is MISSISSIPPI.</p>
<p>Here, I = 4 times, S = 4 times, P = 2 times, M = 1 time</p>
<p>_M_I_I_I_I_P_P_</p>
<p>Those seven letters M, I, I, I, I, P, P can be arranged in $${{7!} \over {4!2!}}$$ ways</p>
<p>Those seven letters creates 8 gaps and we have to choose 4 gaps from those 8 gaps to put those four 'S' letters.</p>
<p>This can be done $${}^8{C_4}$$ ways.</p>
<p>After placing those four 'S' letters we can arrange them in $${{4!} \over {4!}}$$ ways.</p>
<p>Therefore, required number of words</p>
<p>$$ = {{7!} \over {4!2!}} \times {}^8{C_4} \times {{4!} \over {4!}}$$</p>
<p>$$ = {{7\,.\,6!} \over {4!4!}} \times {}^8{C_4}$$</p>
<p>$$ = 7\,.\,{}^6{C_4}\,.\,{}^8{C_4}$$</p> | mcq | aieee-2008 |
pOakikkdQhU84tLNpvMvZ | maths | permutations-and-combinations | conditional-permutations | If the four letter words (need not be meaningful ) are to be formed using the
letters from the word “MEDITERRANEAN” such that the first letter is R and the fourth letter is E, then the total number of all such words is :
| [{"identifier": "A", "content": "$${{11!} \\over {{{\\left( {2!} \\right)}^3}}}$$"}, {"identifier": "B", "content": "110"}, {"identifier": "C", "content": "56"}, {"identifier": "D", "content": "59"}] | ["D"] | null | Here total no of different letters present are,
<br><br>(1) One M
<br><br>(2) Three E (E E E)
<br><br>(3) One D
<br><br>(4) One I
<br><br>(5) One T
<br><br>(6) Two R (R R)
<br><br>(7) Two A (A A)
<br><br>(8) Two N (N N)
<br><br>In the four letter word first letter is R and last letter is E.
<br><br>$$ \therefore $$ Word is = R _ _ E
<br><br>Now remaining letters are,
<br><br>M, EE, D, I, T, R, AA, NN
<br><br>Those 2 empty places can be filled with identical letters [EE, AA, NN] in 3 ways.
<br><br>Or two empty places can be filled with distinct letters [M, E, D, I, T, R, A, N] in $${}^8{C_2} \times 2!$$ ways.
<br><br>$$ \therefore $$ Total no of words = 3 + $${}^8{C_2} \times 2!$$ = 59 | mcq | jee-main-2016-online-9th-april-morning-slot |
va6hXgEJsJa8x7wVuocWG | maths | permutations-and-combinations | conditional-permutations | The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to : | [{"identifier": "A", "content": "374"}, {"identifier": "B", "content": "372"}, {"identifier": "C", "content": "375"}, {"identifier": "D", "content": "250"}] | ["A"] | null | Total no 1 digit numbers possible = 4 (allowed digits 1, 3, 7, 9)
<br><br>Total no 2 digit numbers possible = 4$$ \times $$5 = 20
<br><br>Total no 3 digit numbers possible = 4$$ \times $$5$$ \times $$5 = 100
<br><br>Total no 4 digit numbers possible = 2$$ \times $$5$$ \times $$5$$ \times $$5 = 250
<br><br>So the number of natural numbers less than 7,000 possible are
<br>= 4 + 20 + 100 + 250 = 374 | mcq | jee-main-2019-online-9th-january-evening-slot |
VyKFCxCKFLUFzxl6WeXIp | maths | permutations-and-combinations | conditional-permutations | All possible numbers are formed using the digits
1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number
of such numbers in which the odd digits occupy
even places is : | [{"identifier": "A", "content": "175"}, {"identifier": "B", "content": "162"}, {"identifier": "C", "content": "160"}, {"identifier": "D", "content": "180"}] | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266512/exam_images/lapq8as1rpgqmd4kignb.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267401/exam_images/ckx4aga8bpukg4jpltqh.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263663/exam_images/glftgnpjo21ixfswlqfw.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265431/exam_images/gsv9svdr1bkpr8acenak.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264452/exam_images/qpbjyyxzzuan68yf9kxb.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267399/exam_images/xgnodv9vbpjc1qwfcggz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Permutations and Combinations Question 138 English Explanation"></picture>
<br><br>For those three odd digit numbers 1, 1, 3 we can choose any three positions out of the four even positions.
<br><br>$$ \therefore $$ No of ways we can choose 3 positions out of the 4 positions = <sup>4</sup>C<sub>3</sub>
<br><br>After choosing those three positions, number of ways we can arrange three odd digit numbers = <sup>4</sup>C<sub>3</sub> $$ \times $$ $${{3!} \over {2!}}$$
<br><br>Then the remaining 6 digits can be arrange = $${{6!} \over {2!4!}}$$ ways
<br><br>$$ \therefore $$ Total number of 9 digit numbers = <sup>4</sup>C<sub>3</sub> $$ \times $$ $${{3!} \over {2!}}$$ $$ \times $$ $${{6!} \over {2!4!}}$$ = 180
| mcq | jee-main-2019-online-8th-april-morning-slot |
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