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1l5aifh9z | maths | matrices-and-determinants | symmetric-and-skew-symmetric-matrices | <p>Let $$A = \left[ {\matrix{
0 & { - 2} \cr
2 & 0 \cr
} } \right]$$. If M and N are two matrices given by $$M = \sum\limits_{k = 1}^{10} {{A^{2k}}} $$ and $$N = \sum\limits_{k = 1}^{10} {{A^{2k - 1}}} $$ then MN<sup>2</sup> is :</p> | [{"identifier": "A", "content": "a non-identity symmetric matrix"}, {"identifier": "B", "content": "a skew-symmetric matrix"}, {"identifier": "C", "content": "neither symmetric nor skew-symmetric matrix"}, {"identifier": "D", "content": "an identity matrix"}] | ["A"] | null | <p>$$A = \left[ {\matrix{
0 & { - 2} \cr
2 & 0 \cr
} } \right]$$</p>
<p>$${A^2} = \left[ {\matrix{
0 & { - 2} \cr
2 & 0 \cr
} } \right]\left[ {\matrix{
0 & { - 2} \cr
2 & 0 \cr
} } \right] = \left[ {\matrix{
{ - 4} & 0 \cr
0 & { - 4} \cr
} } \right] = - 4I$$</p>
<p>$$M = {A^2} + {A^4} + {A^6} + \,\,.....\,\, + \,\,{A^{20}}$$</p>
<p>$$ = - 4I + 16I - 64I\,\, + $$ ..... upto 10 terms</p>
<p>$$ = - I$$ [$$4 - 16 + 64$$ .... + upto 10 terms]</p>
<p>$$ = - I\,.\,4\left[ {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right] = {4 \over 5}({2^{20}} - 1)I$$</p>
<p>$$N = {A^1} + {A^3} + {A^5} + \,\,....\,\, + \,\,{A^{19}}$$</p>
<p>$$ = A - 4A + 16A\,\, + $$ ..... upto 10 terms</p>
<p>$$ = A\left( {{{{{( - 4)}^{10}} - 1} \over { - 4 - 1}}} \right) = - \left( {{{{2^{20}} - 1} \over 5}} \right)A$$</p>
<p>$${N^2} = {{{{({2^{20}} - 1)}^2}} \over {{2^5}}}{A^2} = {{ - 4} \over {24}}{({2^{20}} - 1)^2}I$$</p>
<p>$$M{N^2} = {{ - 16} \over {125}}{({2^{20}} - 1)^3}I = KI\,\,\,\,\,(K \ne \pm \,1)$$</p>
<p>$${(M{N^2})^T} = {(KI)^T} = KI$$</p>
<p>$$\therefore$$ A is correct</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l6nl9px1 | maths | matrices-and-determinants | symmetric-and-skew-symmetric-matrices | <p>Let $$\mathrm{A}$$ and $$\mathrm{B}$$ be any two $$3 \times 3$$ symmetric and skew symmetric matrices respectively. Then which of the following is NOT true?</p> | [{"identifier": "A", "content": "$$\\mathrm{A}^{4}-\\mathrm{B}^{4}$$ is a smmetric matrix"}, {"identifier": "B", "content": "$$\\mathrm{AB}-\\mathrm{BA}$$ is a symmetric matrix"}, {"identifier": "C", "content": "$$\\mathrm{B}^{5}-\\mathrm{A}^{5}$$ is a skew-symmetric matrix"}, {"identifier": "D", "content": "$$\\mathrm{AB}+\\mathrm{BA}$$ is a skew-symmetric matrix"}] | ["C"] | null | <p>(A) $$M = {A^4} - {B^4}$$</p>
<p>$${M^T} = {({A^4} - {B^4})^T} = {({A^T})^4} - {({B^T})^4}$$</p>
<p>$$ = {A^4} - {( - B)^4} = {A^4} - {B^4} = M$$</p>
<p>(B) $$M = AB - BA$$</p>
<p>$${M^T} = {(AB - BA)^T} = {(AB)^T} - {(BA)^T}$$</p>
<p>$$ = {B^T}{A^T} - {A^T}{B^T}$$</p>
<p>$$ = - BA - A( - B)$$</p>
<p>$$ = AB - BA = M$$</p>
<p>(C) $$M = {B^5} - {A^5}$$</p>
<p>$${M^T} = {({B^T})^5} - {({A^T})^5} = - ({B^5} + {A^5}) \ne - M$$</p>
<p>(D) $$M = AB + BA$$</p>
<p>$${M^T} = {(AB)^T} + {(BA)^T}$$</p>
<p>$$ = {B^T}{A^T} + {A^T}{B^T} = - BA - AB = - M$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldsgfp13 | maths | matrices-and-determinants | symmetric-and-skew-symmetric-matrices | <p>Let A be a symmetric matrix such that $$\mathrm{|A|=2}$$ and $$\left[ {\matrix{
2 & 1 \cr
3 & {{3 \over 2}} \cr
} } \right]A = \left[ {\matrix{
1 & 2 \cr
\alpha & \beta \cr
} } \right]$$. If the sum of the diagonal elements of A is $$s$$, then $$\frac{\beta s}{\alpha^2}$$ is equal to __________.</p> | [] | null | 5 | <p>$$A = \left( {\matrix{
a & c \cr
c & b \cr
} } \right)$$</p>
<p>$$|A| = ab - {c^2} = 2$$ ...... (1)</p>
<p>$$\left( {\matrix{
2 & 1 \cr
3 & {{3 \over 2}} \cr
} } \right)\left( {\matrix{
a & c \cr
c & b \cr
} } \right) = \left( {\matrix{
1 & 2 \cr
\alpha & \beta \cr
} } \right)$$</p>
<p>$$2a + c = 1$$ ..... (2)</p>
<p>$$2c + b = 2$$ ..... (3)</p>
<p>$$3a + {3 \over 2}c = \alpha $$ .... (4)</p>
<p>$$3c + {3 \over 2}b = \beta $$ ..... (5)</p>
<p>From (1), (2) and (3)</p>
<p>$$a = {3 \over 4},b = 3,c = - {1 \over 2}$$</p>
<p>$$\Rightarrow$$ Now $$\alpha = {6 \over 4}$$</p>
<p>$$\beta = 3$$</p>
<p>$$s = {{15} \over 4}$$</p>
<p>$${{\beta s} \over {{\alpha ^2}}} = {{3 \times {{15} \over 4}} \over {{{\left( {{6 \over 4}} \right)}^2}}} = {{{{45} \over 4}} \over {{9 \over 4}}} = 5$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldu4mkm0 | maths | matrices-and-determinants | symmetric-and-skew-symmetric-matrices | <p>Let A, B, C be 3 $$\times$$ 3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements</p>
<p>(S1) A$$^{13}$$ B$$^{26}$$ $$-$$ B$$^{26}$$ A$$^{13}$$ is symmetric</p>
<p>(S2) A$$^{26}$$ C$$^{13}$$ $$-$$ C$$^{13}$$ A$$^{26}$$ is symmetric</p>
<p>Then,</p> | [{"identifier": "A", "content": "Only S2 is true"}, {"identifier": "B", "content": "Only S1 is true"}, {"identifier": "C", "content": "Both S1 and S2 are false"}, {"identifier": "D", "content": "Both S1 and S2 are true"}] | ["A"] | null | $A^{T}=A, B^{T}=-B, C^{T}=-C$
<br/><br/>
$$
\begin{aligned}
P & =A^{13} B^{26}-B^{26} A^{13} \\\\
P^{T} & =\left(A^{13} B^{26}-B^{26} A^{13}\right)^{T}=\left(A^{13} B^{26}\right)^{T}-\left(B^{26} A^{B}\right)^{T} \\\\
& =\left(B^{26}\right)^{T}\left(A^{13}\right)^{T}-\left(A^{13}\right)^{T}\left(B^{26}\right)^{T} \\\\
& =\left(B^{T}\right)^{26}\left(A^{T}\right)^{13}-\left(A^{T}\right)^{13}\left(A^{T}\right)^{26} \\\\
& =B^{26} A^{13}-A^{13} B^{26}=-\left(A^{13} B^{26}-B^{26} A^{13}\right)=-P
\end{aligned}
$$
<br/><br/>
$P$ is skew-symmetric matrix $\Rightarrow S_{1}$ is false
<br/><br/>
$Q=A^{26} C^{13}-C^{13} A^{26}=Q^{T}=\left(A^{26} C^{13}-C^{13} A^{26}\right)^{T}$
<br/><br/>
$Q=\left(A^{26} C^{13}\right)^{T}-\left(C^{13} A^{26}\right)^{T}=\left(C^{13}\right)^{T}\left(A^{26}\right)^{T}-\left(A^{26}\right)^{T}\left(C^{13}\right)^{T}$
<br/><br/>
$=\left(C^{T}\right)^{13}\left(A^{T}\right)^{26}-\left(A^{T}\right)^{26}\left(C^{T}\right)^{13}=-C^{13} A^{26}+A^{26} C^{13}$
<br/><br/>
$=A^{26} C^{13}+C^{13} A^{26}$
<br/><br/>
$\Rightarrow Q^{T}=Q \Rightarrow Q$ is symmetric matrix $\Rightarrow S_{2}$ is true. | mcq | jee-main-2023-online-25th-january-evening-shift |
1lgq0m4hk | maths | matrices-and-determinants | symmetric-and-skew-symmetric-matrices | <p>The number of symmetric matrices of order 3, with all the entries from the set $$\{0,1,2,3,4,5,6,7,8,9\}$$ is :</p> | [{"identifier": "A", "content": "$$10^{9}$$"}, {"identifier": "B", "content": "$$9^{10}$$"}, {"identifier": "C", "content": "$$10^{6}$$"}, {"identifier": "D", "content": "$$6^{10}$$"}] | ["C"] | null | <p>Sure! A symmetric matrix is a square matrix that is equal to its transpose. For a matrix to be symmetric, the element at row i and column j must be equal to the element at row j and column i. In other words, $$A_{ij} = A_{ji}$$. </p>
<p>For a 3 $$ \times $$ 3 symmetric matrix, it looks like this:</p>
<p>$$
\begin{pmatrix}
a & b & c \\
b & d & e \\
c & e & f \\
\end{pmatrix}
$$
</p>
<p>Notice that there are only 6 unique elements we need to fill because of the symmetry:</p>
<ol>
<li>$$a$$ in the (1,1) position</li>
<li>$$b$$ in the (1,2) and (2,1) positions</li>
<li>$$c$$ in the (1,3) and (3,1) positions</li>
<li>$$d$$ in the (2,2) position</li>
<li>$$e$$ in the (2,3) and (3,2) positions</li>
<li>$$f$$ in the (3,3) position</li>
</ol>
<p>Each of these unique elements can take a value from the set $${0,1,2,3,4,5,6,7,8,9}$$, which has 10 elements. </p>
<p>We have 10 choices for each of the 6 unique elements, so the total number of symmetric matrices can be calculated as:</p>
<p>$$10 \times 10 \times 10 \times 10 \times 10 \times 10 = 10^{6}$$</p>
<p>Thus, the total number of symmetric matrices of order 3 with entries from this set is $$10^{6}$$.</p>
| mcq | jee-main-2023-online-13th-april-morning-shift |
adqLo6ekAzwOBx7J | maths | matrices-and-determinants | trace-of-a-matrix | Let $$A$$ be $$a\,2 \times 2$$ matrix with real entries. Let $$I$$ be the $$2 \times 2$$ identity matrix. Denote by tr$$(A)$$, the sum of diagonal entries of $$a$$. Assume that $${a^2} = I.$$
<br/><b>Statement-1 :</b> If $$A \ne I$$ and $$A \ne - I$$, then det$$(A)=-1$$
<br/><b>Statement- 2 :</b> If $$A \ne I$$ and $$A \ne - I$$, then tr $$(A)$$ $$ \ne 0$$. | [{"identifier": "A", "content": "statement - 1 is false, statement -2 is true "}, {"identifier": "B", "content": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1."}, {"identifier": "C", "content": "statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1."}, {"identifier": "D", "content": "statement - 1 is true, statement - 2 is false."}] | ["D"] | null | Let $$A = \left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$ $$\,\,\,$$ then $${A^2} = 1$$
<br><br>$$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$$
<br><br>$$ac + cd = 0\,\,\,\,bc + {d^2} = 1$$
<br><br>From these four relations,
<br><br>$${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$$
<br><br>and $$\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d$$
<br><br>We can take $$a = 1,b = 0,c = 0,d = - 1$$
<br><br>as one possible set of values, then
<br><br>$$A = \left[ {\matrix{
1 & 0 \cr
0 & { - 1} \cr
} } \right]$$
<br><br>Clearly $$A \ne I\,\,\,$$ and $$\,\,\,\,A \ne - I\,\,$$ and $$\,\,\,A = - 1$$
<br><br>$$\therefore$$ $$\,\,\,\,\,$$ Statement $$1$$ is true.
<br><br>Also if $$A \ne I\,\,\,\,\,tr\left( A \right) = 0$$
<br><br>$$\therefore$$ $$\,\,\,\,\,$$ Statement $$2$$ is false. | mcq | aieee-2008 |
hHBIhG3bWMTtRnpl | maths | matrices-and-determinants | trace-of-a-matrix | Let $$A$$ be a $$\,2 \times 2$$ matrix with non-zero entries and let $${A^2} = I,$$
<br/>where $$I$$ is $$2 \times 2$$ identity matrix. Define
<br/>$$Tr$$$$(A)=$$ sum of diagonal elements of $$A$$ and $$\left| A \right| = $$ determinant of matrix $$A$$.
<br/><b>Statement- 1:</b> $$Tr$$$$(A)=0$$.
<br/><b>Statement- 2:</b> $$\left| A \right| = 1$$ . | [{"identifier": "A", "content": "statement - 1 is true, statement - 2 is true; statement - 2 is <b>not</b> a correct explanation for statement - 1. "}, {"identifier": "B", "content": "statement - 1 is true, statement - 2 is false. "}, {"identifier": "C", "content": "statement - 1 is false, statement -2 is true "}, {"identifier": "D", "content": "statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1."}] | ["B"] | null | Let $$A = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)$$ where $$a,b,c,d$$ $$ \ne 0$$
<br><br>$${A^2} = \left( {\matrix{
a & b \cr
c & d \cr
} } \right)\left( {\matrix{
a & b \cr
c & d \cr
} } \right)$$
<br><br>$$ \Rightarrow {A^2} = \left( {\matrix{
{{a^2} + bc} & {ab + bd} \cr
{ac + cd} & {bc + {d^2}} \cr
} } \right)$$
<br><br>$$ \Rightarrow {a^2} + bc = 1,\,bc + {d^2} = 1$$
<br><br>$$ab + bd = ac + cd = 0$$
<br><br>$$c \ne 0\,\,\,\,\,b \ne 0$$
<br><br>$$ \Rightarrow a + d = 0 \Rightarrow Tr\left( A \right) = 0$$
<br><br>$$\left| A \right| = ad - bc = - {a^2} - bc = - 1$$ | mcq | aieee-2010 |
0siKIhXARnH2944i8v7k9k2k5gzyxdt | maths | matrices-and-determinants | trace-of-a-matrix | The number of all 3 × 3 matrices A, with
enteries from the set {–1, 0, 1} such that the sum
of the diagonal elements of AA<sup>T</sup> is 3, is | [] | null | 672 | Let A = $$\left[ {\matrix{
{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \cr
{{a_{21}}} & {{a_{22}}} & {{a_{23}}} \cr
{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \cr
} } \right]$$
<br><br>$$ \therefore $$ A<sup>T</sup> = $$\left[ {\matrix{
{{a_{11}}} & {{a_{21}}} & {{a_{31}}} \cr
{{a_{12}}} & {{a_{22}}} & {{a_{32}}} \cr
{{a_{13}}} & {{a_{23}}} & {{a_{33}}} \cr
} } \right]$$
<br><br>diagonal elements of AA<sup>T</sup>
are $$a_{11}^2 + a_{12}^2 + a_{13}^2$$
, <br>$$a_{21}^2 + a_{22}^2 + a_{23}^2$$
, $$a_{31}^2 + a_{32}^2 + a_{33}^2$$
<br><br>Given Sum = ($$a_{11}^2 + a_{12}^2 + a_{13}^2$$) + <br>($$a_{21}^2 + a_{22}^2 + a_{23}^2$$) + ($$a_{31}^2 + a_{32}^2 + a_{33}^2$$) = 3
<br><br>This is only possible when three enteries must be either 1 or – 1 and all other six enteries are 0.
<br><br>$$ \therefore $$ Number of matrices = <sup>9</sup>C<sub>3</sub> $$ \times $$ 2 $$ \times $$ 2 $$ \times $$ 2
<br><br>= 672 | integer | jee-main-2020-online-8th-january-morning-slot |
Awf3Xo25ZkDyIfzHoijgy2xukewq8sg6 | maths | matrices-and-determinants | trace-of-a-matrix | Let A be a 2 $$ \times $$ 2 real matrix with entries from
{0, 1} and |A|
$$ \ne $$ 0. Consider the following two
statements :
<br/><br/>(P) If A $$ \ne $$ I<sub>2</sub>
, then |A| = –1
<br/>(Q) If |A| = 1, then tr(A) = 2,
<br/><br/>where I<sub>2</sub>
denotes 2 $$ \times $$ 2 identity matrix and tr(A)
denotes the sum of the diagonal entries of A. Then :
| [{"identifier": "A", "content": "(P) is true and (Q) is false"}, {"identifier": "B", "content": "Both (P) and (Q) are false"}, {"identifier": "C", "content": "Both (P) and (Q) are true"}, {"identifier": "D", "content": "(P) is false and (Q) is true"}] | ["D"] | null | Let A = $$\left[ {\matrix{
a & b \cr
c & d \cr
} } \right]$$, where a, b, c, d $$ \in $$ {0, 1}
<br><br>$$ \Rightarrow $$ |A| = ad – bc
<br><br>$$ \therefore $$ ad = 0 or 1 and bc = 0 or 1
<br><br>So possible values of |A| are 1, 0 or –1
<br><br>(P) If A $$ \ne $$ I<sub>2</sub>
then |A| is either 0 or –1
<br><br>(Q) If |A| = 1 then ad = 1 and bc = 0
<br><br>$$ \Rightarrow $$ a = d = 1 $$ \Rightarrow $$ Tr(A) = 2 | mcq | jee-main-2020-online-2nd-september-morning-slot |
YI4Yw7Wd7pYuECq7od1kmhzsuuy | maths | matrices-and-determinants | trace-of-a-matrix | The total number of 3 $$\times$$ 3 matrices A having entries from the set {0, 1, 2, 3} such that the sum of all the diagonal entries of AA<sup>T</sup> is 9, is equal to _____________. | [] | null | 766 | $$A{A^T} = \left[ {\matrix{
x & y & z \cr
a & b & c \cr
d & e & f \cr
} } \right]\left[ {\matrix{
x & a & d \cr
y & b & e \cr
z & c & f \cr
} } \right]$$<br><br>$$ = \left[ {\matrix{
{{x^2} + {y^2} + {z^2}} & {ax + by + cz} & {dx + ey + fz} \cr
{ax + by + cz} & {{a^2} + {b^2} + {c^2}} & {ad + be + cf} \cr
{dx + ey + fz} & {ad + be + cf} & {{d^2} + {e^2} + {f^2}} \cr
} } \right]$$<br><br>$$Tr(A{A^T}) = {x^2} + {y^2} + {z^2} + {a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} = 9$$<br><br>Case-I : Nine ones = 1 case<br><br>Case-II : 8 zeroes and one entry is 3 = $${{{9!} \over {8!}} = 9}$$ cases<br><br>Case-III : Two 2’s, one 1’s and 6 zeroes = $${{9!} \over {2!6!}} = 63 \times 4 = 252$$<br><br>Case IV : one 2, five 1, rest 0 $${{9!} \over {5!3!}} = 63 \times 8 = 504$$<br><br>$$ \therefore $$ Total cases = 9 + 252 + 504 + 1 = 766 | integer | jee-main-2021-online-16th-march-morning-shift |
p9s7EngSW3BQIbwlq71kmlj61ex | maths | matrices-and-determinants | trace-of-a-matrix | Let $$A + 2B = \left[ {\matrix{
1 & 2 & 0 \cr
6 & { - 3} & 3 \cr
{ - 5} & 3 & 1 \cr
} } \right]$$ and $$2A - B = \left[ {\matrix{
2 & { - 1} & 5 \cr
2 & { - 1} & 6 \cr
0 & 1 & 2 \cr
} } \right]$$. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) $$-$$ Tr(B) has value equal to | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "3"}] | ["B"] | null | $$A = {1 \over 5}((A + 2B) + 2(2A - B))$$<br><br>$$ = {1 \over 5}\left( {\left[ {\matrix{
1 & 2 & 0 \cr
6 & { - 3} & 3 \cr
{ - 5} & 3 & 1 \cr
} } \right] + \left[ {\matrix{
4 & { - 2} & {10} \cr
4 & { - 2} & {12} \cr
0 & 2 & 4 \cr
} } \right]} \right)$$<br><br>$$ = {1 \over 5}\left[ {\matrix{
5 & 0 & {10} \cr
{10} & { - 5} & {15} \cr
{ - 5} & 5 & 5 \cr
} } \right] \Rightarrow tr(A) = 1$$<br><br>Similarly,<br><br>$$B = {1 \over 5}(2(A + 2B) - (2A - B))$$<br><br>$$ = {1 \over 5}\left( {\left[ {\matrix{
2 & 4 & 0 \cr
{12} & { - 6} & 6 \cr
{ - 10} & 6 & 2 \cr
} } \right] - \left[ {\matrix{
2 & { - 1} & 5 \cr
2 & { - 1} & 6 \cr
0 & 1 & 2 \cr
} } \right]} \right)$$<br><br>$$ = {1 \over 5}\left[ {\matrix{
0 & 6 & { - 5} \cr
{10} & { - 5} & 0 \cr
{ - 10} & 5 & 0 \cr
} } \right] \Rightarrow tr(B) = - 1$$<br><br>$$Tr(A) - Tr(B) = 1 - ( - 1) = 2$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
1ldpsteev | maths | matrices-and-determinants | trace-of-a-matrix | <p>Let $$A = \left( {\matrix{
1 & 0 & 0 \cr
0 & 4 & { - 1} \cr
0 & {12} & { - 3} \cr
} } \right)$$. Then the sum of the diagonal elements of the matrix $${(A + I)^{11}}$$ is equal to :</p> | [{"identifier": "A", "content": "4094"}, {"identifier": "B", "content": "2050"}, {"identifier": "C", "content": "6144"}, {"identifier": "D", "content": "4097"}] | ["D"] | null | $A^{2}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3\end{array}\right]$
<br/><br/>$$
=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & 4 & -1 \\
0 & 12 & -3
\end{array}\right]=A
$$
<br/><br/>$$\Rightarrow \mathrm{A}_{3}=\mathrm{A}_{4}=.......=\mathrm{A}$$
<br/><br/>Now,
<br/><br/>$$
\begin{aligned}
(A+I)^{11} & ={ }^{11} C_{0} A^{11}+{ }^{11} C_{1} A^{10}+\ldots{ }^{11} C_{11} I \\\\
& =A\left({ }^{11} C_{0}+{ }^{11} C_{1} \ldots{ }^{11} C_{10}\right)+I \\\\
& =A\left(2^{11}-1\right)+I
\end{aligned}
$$
<br/><br/>Trace of
<br/><br/>$$
\begin{aligned}
(A+I)^{11} & =2^{11}+4\left(2^{11}-1\right)+1-3\left(2^{11}-1\right)+1 \\\\
& =2 \times 2^{11}-4+3+2 \\\\
& =2^{12}+1 \\\\
& =4097
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-morning-shift |
jaoe38c1lscom6c6 | maths | matrices-and-determinants | trace-of-a-matrix | <p>Let $$A$$ be a $$2 \times 2$$ real matrix and $$I$$ be the identity matrix of order 2. If the roots of the equation $$|\mathrm{A}-x \mathrm{I}|=0$$ be $$-1$$ and 3, then the sum of the diagonal elements of the matrix $$\mathrm{A}^2$$ is</p> | [] | null | 10 | <p>$$|A-x I|=0$$</p>
<p>Roots are $$-$$1 and 3</p>
<p>Sum of roots $$=\operatorname{tr}(A)=2$$</p>
<p>Product of roots $$=|\mathrm{A}|=-3$$</p>
<p>Let $$A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$$</p>
<p>We have $$\mathrm{a}+\mathrm{d}=2$$</p>
<p>$$\mathrm{ad}-\mathrm{bc}=-3$$</p>
<p>$$A^2=\left[\begin{array}{ll}a & b \\
c & d
\end{array}\right] \times\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{ll}
a^2+b c & a b+b d \\
a c+c d & b c+d^2
\end{array}\right]$$</p>
<p>We need $$a^2+b c+b c+d^2$$</p>
<p>$$\begin{aligned}
& =a^2+2 b c+d^2 \\
& =(a+d)^2-2 a d+2 b c \\
& =4-2(a d-b c) \\
& =4-2(-3) \\
& =4+6 \\
& =10
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-evening-shift |
1lsg3zn8d | maths | matrices-and-determinants | trace-of-a-matrix | <p>Let $$R=\left(\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right)$$ be a non-zero $$3 \times 3$$ matrix, where $$x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0, \theta \in(0,2 \pi)$$. For a square matrix $$M$$, let trace $$(M)$$ denote the sum of all the diagonal entries of $$M$$. Then, among the statements:</p>
<p>(I) Trace $$(R)=0$$</p>
<p>(II) If trace $$(\operatorname{adj}(\operatorname{adj}(R))=0$$, then $$R$$ has exactly one non-zero entry.</p> | [{"identifier": "A", "content": "Only (I) is true\n"}, {"identifier": "B", "content": "Only (II) is true\n"}, {"identifier": "C", "content": "Both (I) and (II) are true\n"}, {"identifier": "D", "content": "Neither (I) nor (II) is true"}] | ["D"] | null | <p>$$\begin{aligned}
& x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0 \\
& \Rightarrow x, y, z \neq 0
\end{aligned}$$</p>
<p>Also,</p>
<p>$$\begin{aligned}
& \sin \theta+\sin \left(\theta+\frac{2 \pi}{3}\right)+\sin \left(\theta+\frac{4 \pi}{3}\right)=0 \forall \theta \in \mathrm{R} \\
& \Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=0 \\
& \Rightarrow \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=0
\end{aligned}$$</p>
<p>(i) $$\quad \operatorname{Trace}(\mathrm{R})=\mathrm{x}+\mathrm{y}+\mathrm{z}$$</p>
<p>If $$x+y+z=0$$ and $$x y+y z+z x=0$$</p>
<p>$$\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=0$$</p>
<p>Statement (i) is False</p>
<p>(ii) $$\quad \operatorname{Adj}(\operatorname{Adj}(\mathrm{R}))=|\mathrm{R}| \mathrm{R}$$</p>
<p>Trace $$(\operatorname{Adj}(\operatorname{Adj}(\mathrm{R})))$$</p>
<p>$$=x y z(x+y+z) \neq 0$$</p>
<p>Statement (ii) is also False</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
lv0vxdis | maths | matrices-and-determinants | trace-of-a-matrix | <p>Let $$A$$ be a square matrix of order 2 such that $$|A|=2$$ and the sum of its diagonal elements is $$-$$3 . If the points $$(x, y)$$ satisfying $$\mathrm{A}^2+x \mathrm{~A}+y \mathrm{I}=\mathrm{O}$$ lie on a hyperbola, whose transverse axis is parallel to the $$x$$-axis, eccentricity is $$\mathrm{e}$$ and the length of the latus rectum is $$l$$, then $$\mathrm{e}^4+l^4$$ is equal to ________.</p> | [] | null | 25 | <p>$$|A|=2 \sum \mathrm{dia}=-3$$</p>
<p>$$\therefore \quad$$ character equation : $$A^2+3 A+2 I=0$$</p>
<p>$$\Rightarrow x=3 \quad y=2$$</p>
<p>$$\because$$ We are getting only one point $$(3,2)$$ but its given many points satisfy this equation.</p>
<p>Moreover hyperbola whose transverse axis is $$x$$ axis and passing through $$(3,2)$$ is not unique.</p>
<p>$$\therefore$$ multiple value of '$$e$$' and $$L(L R)$$ is possible.</p>
<p>We'll not get a unique result.</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lv5grwjl | maths | matrices-and-determinants | trace-of-a-matrix | <p>Let $$A=\left[\begin{array}{cc}2 & -1 \\ 1 & 1\end{array}\right]$$. If the sum of the diagonal elements of $$A^{13}$$ is $$3^n$$, then $$n$$ is equal to ________.</p> | [] | null | 7 | <p>$$\begin{aligned}
& A=\left[\begin{array}{cc}
2 & -1 \\
1 & 1
\end{array}\right] \\
& A^2=\left[\begin{array}{cc}
2 & -1 \\
1 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & -1 \\
1 & 1
\end{array}\right] \\
& A^2=\left[\begin{array}{cc}
3 & -3 \\
3 & 0
\end{array}\right]=3\left[\begin{array}{cc}
1 & -1 \\
1 & 0
\end{array}\right] \\
& A^4=9\left[\begin{array}{ll}
0 & -1 \\
1 & -1
\end{array}\right] \\
& A^8=81\left[\begin{array}{ll}
-1 & 1 \\
-1 & 0
\end{array}\right] \\
& A^{12}=729\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& A^{13}=729\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & -1 \\
1 & 1
\end{array}\right] \\
& A^{13}=\left[\begin{array}{cc}
1458 & -729 \\
729 & 729
\end{array}\right]
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Sum }=2187=3^n \\
& 3^7=3^n \\
& n=7
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-morning-shift |
amsDpowk5keRB2xC | maths | matrices-and-determinants | transpose-of-a-matrix | If $$A = \left[ {\matrix{
1 & 2 & 2 \cr
2 & 1 & { - 2} \cr
a & 2 & b \cr
} } \right]$$ is a matrix satisfying the equation
<br/><br/>$$A{A^T} = 9\text{I},$$ where $$I$$ is $$3 \times 3$$ identity matrix, then the ordered
<br><br/>pair $$(a, b)$$ is equal to :</br> | [{"identifier": "A", "content": "$$(2, 1)$$"}, {"identifier": "B", "content": "$$(-2, -1)$$"}, {"identifier": "C", "content": "$$(2, -1)$$"}, {"identifier": "D", "content": "$$(-2, 1)$$"}] | ["B"] | null | $$\left[ {\matrix{
1 & 2 & 2 \cr
2 & 1 & { - 2} \cr
a & 2 & b \cr
} } \right]\left[ {\matrix{
1 & 2 & a \cr
2 & 1 & 2 \cr
2 & { - 2} & b \cr
} } \right] = \left[ {\matrix{
9 & 0 & 0 \cr
0 & 9 & 0 \cr
0 & 0 & 9 \cr
} } \right]$$
<br><br>$$ \Rightarrow \left[ {\matrix{
{1 + 4 + 4} & {2 + 2 - 4} & {a + 4 + 2b} \cr
{2 + 2 - 4} & {4 + 1 + 4} & {2a + 2 - 2b} \cr
{a + 4 + 2b} & {2a + 2 - 2b} & {{a^2} + 4 + {b^2}} \cr
} } \right]$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\matrix{
9 & 0 & 0 \cr
0 & 9 & 0 \cr
0 & 0 & 9 \cr
} } \right]$$
<br><br>$$ \Rightarrow a + 4 + 2b = 0$$ $$ \Rightarrow a + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$2a + 2 - 2b = 0 \Rightarrow 2a - 2b = - 2$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a - b = - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>On solving $$(i)$$ and $$(ii)$$ we get
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 + b + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$b=-1$$ and $$a=-2$$
<br><br>$$\left( {a,b} \right) = \left( { - 2, - 1} \right)$$ | mcq | jee-main-2015-offline |
qS95ij4pasrt0BwY2C0oD | maths | matrices-and-determinants | transpose-of-a-matrix | If P = $$\left[ {\matrix{
{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr
{ - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right],A = \left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]\,\,\,$$
<br/><br/>Q = PAP<sup>T</sup>, then P<sup>T</sup> Q<sup>2015</sup> P is : | [{"identifier": "A", "content": "$$\\left[ {\\matrix{\n 0 & {2015} \\cr \n 0 & 0 \\cr \n\n } } \\right]$$"}, {"identifier": "B", "content": "$$\\left[ {\\matrix{\n {2015} & 1 \\cr \n 0 & {2015} \\cr \n\n } } \\right]$$ "}, {"identifier": "C", "content": "$$\\left[ {\\matrix{\n {2015} & 0 \\cr \n 1 & {2015} \\cr \n\n } } \\right]$$"}, {"identifier": "D", "content": "$$\\left[ {\\matrix{\n 1 & {2015} \\cr \n 0 & 1 \\cr \n\n } } \\right]$$"}] | ["D"] | null | P = $$\left[ {\matrix{
{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr
{ - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right]$$
<br><br>$$ \therefore $$ P<sup>T</sup> = $$\left[ {\matrix{
{{{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr
{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr
} } \right]$$
<br><br>As PP<sup>T</sup> = P<sup>T</sup>P = I
<br><br>given, Q = PAP<sup>T</sup>
<br><br>$$ \therefore $$ P<sup>T</sup>Q = P<sup>T</sup>P AP<sup>T</sup>
<br><br>$$ \Rightarrow $$ P<sup>T</sup>Q = IAP<sup>T</sup> = AP<sup>T</sup> [ as P<sup>T</sup>P = I]
<br><br>Now,
<br><br>P<sup>T</sup> Q<sup>2015</sup> P
<br><br>= P<sup>T</sup>Q . Q<sup>2014</sup> . P
<br><br>= AP<sup>T</sup> Q<sup>2014</sup> P
<br><br>= AP<sup>T</sup> . Q . Q<sup>2013</sup> . P
<br><br>= A<sup>2</sup>P<sup>T</sup> . Q<sup>2013</sup> . P
<br><br>.
<br><br>.
<br><br>.
<br><b>= A<sup>2014</sup> . P<sup>T</sup>QP
<br><br>= A<sup>2014</sup> . AP<sup>T</sup>P
<br><br>= A<sup>2015</sup>
<br><br>As A = $$\left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]$$
<br><br>$$ \therefore $$ A<sup>2</sup> = $$\left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]\left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right]$$
<br><br>A<sup>3</sup> =
$$\left[ {\matrix{
1 & 2 \cr
0 & 1 \cr
} } \right]$$ $$\left[ {\matrix{
1 & 1 \cr
0 & 1 \cr
} } \right]$$ = $$\left[ {\matrix{
1 & 3 \cr
0 & 1 \cr
} } \right]$$
<br><br>A<sup>2015</sup> = $$\left[ {\matrix{
1 & {2015} \cr
0 & 1 \cr
} } \right]$$</b> | mcq | jee-main-2016-online-9th-april-morning-slot |
2fwdocUS7iYG0xGqpGtLt | maths | matrices-and-determinants | transpose-of-a-matrix | For two 3 × 3 matrices A and B, let A + B = 2B<sup>T</sup> and 3A + 2B = I<sub>3</sub>, where B<sup>T</sup> is
the transpose of B and I<sub>3</sub> is 3 × 3 identity matrix. Then : | [{"identifier": "A", "content": "5A + 10B = 2I<sub>3</sub> "}, {"identifier": "B", "content": "10A + 5B = 3I<sub>3</sub>"}, {"identifier": "C", "content": "B + 2A = I<sub>3</sub>"}, {"identifier": "D", "content": "3A + 6B = 2I<sub>3</sub>"}] | ["B"] | null | Given, A + B = 2B<sup>T</sup> .......(1)
<br><br>$$ \Rightarrow $$ (A + B)<sup>T</sup> = (2B<sup>T</sup>)<sup>T</sup>
<br><br>$$ \Rightarrow $$ A<sup>T</sup> + B<sup>T</sup> = 2B
<br><br>$$ \Rightarrow $$ B = $${{{A^T} + {B^T}} \over 2}$$
<br><br>Now put this in equation (1)
<br><br>So, A + $${{{A^T} + {B^T}} \over 2}$$ = 2B<sup>T</sup>
<br><br>$$ \Rightarrow $$2A + A<sup>T</sup> = 3B<sup>T</sup>
<br><br>$$ \Rightarrow $$ A = $${{3{B^T} - {A^T}} \over 2}$$
<br><br>Also, 3A + 2B = I<sub>3</sub> .......(2)
<br><br>$$ \Rightarrow $$ $$3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)$$ = I<sub>3</sub>
<br><br>$$ \Rightarrow $$ 11B<sup>T</sup> - A<sup>T</sup> = 2I<sub>3</sub>
<br><br>$$ \Rightarrow $$ (11B<sup>T</sup> - A<sup>T</sup>)<sup>T</sup> = (2I<sub>3</sub>)<sup>T</sup>
<br><br>$$ \Rightarrow $$ 11B - A = 2I<sub>3</sub> ........(3)
<br><br>Multiply (3) by 3 and then adding (2) and (3) we get,
<br><br>35B = 7I<sub>3</sub>
<br><br>$$ \Rightarrow $$ B = $${{{I_3}} \over 5}$$
<br><br>From (3), 11$${{{I_3}} \over 5}$$ - A = 2I<sub>3</sub>
<br><br>$$ \Rightarrow $$ A = $${{{I_3}} \over 5}$$
<br><br>$$ \therefore $$ 5A = 5B = I<sub>3</sub>
<br><br>$$ \Rightarrow $$ 10A + 5B = 3I<sub>3</sub> | mcq | jee-main-2017-online-9th-april-morning-slot |
UJzvMeYq6IQdNHywI6sjl | maths | matrices-and-determinants | transpose-of-a-matrix | Let A = $$\left( {\matrix{
0 & {2q} & r \cr
p & q & { - r} \cr
p & { - q} & r \cr
} } \right).$$ If AA<sup>T</sup> = I<sub>3</sub>, then $$\left| p \right|$$ is : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 5 }}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 6 }}$$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 3 }}$$"}] | ["A"] | null | A is orthogonal matrix
<br><br>$$ \Rightarrow $$ 0<sup>2</sup> + p<sup>2</sup> + p<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ $$\left| p \right| = {1 \over {\sqrt 2 }}$$ | mcq | jee-main-2019-online-11th-january-morning-slot |
K7MM9Pd6ljHdsVeV1i18hoxe66ijvwvdh35 | maths | matrices-and-determinants | transpose-of-a-matrix | The total number of matrices<br/>
$$A = \left( {\matrix{
0 & {2y} & 1 \cr
{2x} & y & { - 1} \cr
{2x} & { - y} & 1 \cr
} } \right)$$<br/>
(x, y $$ \in $$ R,x $$ \ne $$ y) for which A<sup>T</sup>A = 3I<sub>3</sub> is :- | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "6"}] | ["B"] | null | Given A<sup>T</sup>A = 3I<sub>3</sub>
<br><br>$$ \Rightarrow $$ $$\left[ {\matrix{
0 & {2x} & {2x} \cr
{2y} & y & { - y} \cr
1 & { - 1} & 1 \cr
} } \right]\left[ {\matrix{
0 & {2y} & 1 \cr
{2x} & y & { - 1} \cr
{2x} & { - y} & 1 \cr
} } \right]$$
<br><br> = $$3\left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$
<br><br>$$ \Rightarrow $$ $$\left[ {\matrix{
{8{x^2}} & 0 & 0 \cr
0 & {6{y^2}} & 0 \cr
0 & 0 & 3 \cr
} } \right]$$ = $$\left[ {\matrix{
3 & 0 & 0 \cr
0 & 3 & 0 \cr
0 & 0 & 3 \cr
} } \right]$$
<br><br>$$ \therefore $$ 8x<sup>2</sup> = 3, 6y<sup>2</sup> = 3
<br><br>$$ \Rightarrow $$ x = $$ \pm \sqrt {{3 \over 8}} $$, y = $$ \pm \sqrt {{1 \over 2}} $$
<br><br>Total possible combination of x and y = 2 $$ \times $$ 2 = 4 | mcq | jee-main-2019-online-9th-april-evening-slot |
oCcGJycx3AUOdA82Pnjgy2xukez6fr99 | maths | matrices-and-determinants | transpose-of-a-matrix | Let a, b, c $$ \in $$ R be all non-zero and satisfy
<br/>a<sup>3</sup> + b<sup>3</sup> + c<sup>3</sup> = 2. If the matrix
<br/><br/>A = $$\left( {\matrix{
a & b & c \cr
b & c & a \cr
c & a & b \cr
} } \right)$$
<br/><br/>satisfies A<sup>T</sup>A = I, then a value of <b>abc</b> can be : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$${1 \\over 3}$$"}, {"identifier": "C", "content": "-$${1 \\over 3}$$"}, {"identifier": "D", "content": "$${2 \\over 3}$$"}] | ["B"] | null | Given, <br>
$${a^3} + {b^3} + {c^3} = 2$$<br><br>
$${A^T}A = I$$<br><br>
$$ \Rightarrow \left[ {\matrix{
a & b & c \cr
b & c & a \cr
c & a & b \cr
} } \right]\left[ {\matrix{
a & b & c \cr
b & c & a \cr
c & a & b \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 1 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$<br><br>
$$ \Rightarrow {a^2} + {b^2} + {c^2} = 1$$<br><br>
and ab + bc + ca = 0<br><br>
Now (a + b + c)<sup>2</sup> = 1<br><br>
$$ \Rightarrow (a + b + c) = \pm 1$$<br><br>
So, $${a^3} + {b^3} + {c^3} - 3abc$$<br><br>
$$ \Rightarrow (a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$<br><br>
$$ \Rightarrow \pm 1(1 - 0) = \pm 1$$<br><br>$$ \therefore 2 - 3abc = \pm 1$$<br><br>
$$ \Rightarrow 3abc = 2 \pm 1 = 3,1$$<br><br>
$$ \Rightarrow abc = 1,{1 \over 3}$$
| mcq | jee-main-2020-online-2nd-september-evening-slot |
LQSeGCs3mNm7ThykWi1klrju1ju | maths | matrices-and-determinants | transpose-of-a-matrix | Let M be any 3 $$ \times $$ 3 matrix with entries from the set {0, 1, 2}. The maximum number of such matrices, for which the sum of diagonal elements of M<sup>T</sup>M is seven, is ________. | [] | null | 540 | $$\left[ {\matrix{
a & b & c \cr
d & e & f \cr
g & h & i \cr
} } \right]\left[ {\matrix{
a & d & g \cr
b & e & h \cr
c & f & i \cr
} } \right]$$<br><br>$${a^2} + {b^2} + {c^2} + {d^2} + {e^2} + {f^2} + {g^2} + {h^2} + {i^2} = 7$$<br><br><b>Case I :</b> Seven (1's) and two (0's)<br><br>Number of such matrices = $${}^9{C_2} = 36$$<br><br><b>Case II :</b> One (2) and three (1's) and five (0's)<br><br>Number of such matrices = $${{9!} \over {5!3!}} = 504$$<br><br>$$ \therefore $$ Total = 540 | integer | jee-main-2021-online-24th-february-morning-slot |
Be5SSizFcRv9rQls9P1klt99nb8 | maths | matrices-and-determinants | transpose-of-a-matrix | If for the matrix, $$A = \left[ {\matrix{
1 & { - \alpha } \cr
\alpha & \beta \cr
} } \right]$$, $$A{A^T} = {I_2}$$, then the value of $${\alpha ^4} + {\beta ^4}$$ is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}] | ["C"] | null | $$\left[ {\matrix{
1 & { - \alpha } \cr
\alpha & \beta \cr
} } \right]\left[ {\matrix{
1 & \alpha \cr
{ - \alpha } & \beta \cr
} } \right] = \left[ {\matrix{
{1 + {\alpha ^2}} & {\alpha - \alpha \beta } \cr
{\alpha - \alpha \beta } & {{\alpha ^2} + {\beta ^2}} \cr
} } \right] = \left[ {\matrix{
1 & 0 \cr
0 & 1 \cr
} } \right]$$<br><br>1 + $$\alpha$$<sup>2</sup> = 1<br><br>$$\alpha$$<sup>2</sup> = 0<br><br>$$\alpha$$<sup>2</sup> + $$\beta$$<sup>2</sup> = 1<br><br>$$\beta$$<sup>2</sup> = 1<br><br>$$\alpha$$<sup>4</sup> = 0<br><br>$$\beta$$<sup>4</sup> = 1<br><br>$$\alpha$$<sup>4</sup> + $$\beta$$<sup>4</sup> = 1 | mcq | jee-main-2021-online-25th-february-evening-slot |
1l6jekbdu | maths | matrices-and-determinants | transpose-of-a-matrix | <p>Let $$S$$ be the set containing all $$3 \times 3$$ matrices with entries from $$\{-1,0,1\}$$. The total number of matrices $$A \in S$$ such that the sum of all the diagonal elements of $$A^{\mathrm{T}} A$$ is 6 is ____________.</p> | [] | null | 5376 | <p>Sum of all diagonal elements is equal to sum of square of each element of the matrix.</p>
<p>i.e., $$A = \left[ {\matrix{
{{a_1}} & {{a_2}} & {{a_3}} \cr
{{b_1}} & {{b_2}} & {{b_3}} \cr
{{c_1}} & {{c_2}} & {{c_3}} \cr
} } \right]$$</p>
<p>then $${t_r}\,(A\,.\,{A^T})$$</p>
<p>$$ = a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 + c_1^2 + c_2^2 + c_3^2$$</p>
<p>$$\because$$ $${a_i},{b_i},{c_i} \in \{ - 1,0,1\} $$ for $$i = 1,2,3$$</p>
<p>$$\therefore$$ Exactly three of them are zero and rest are 1 or $$-$$1.</p>
<p>Total number of possible matrices $${}^9{C_3} \times {2^6}$$</p>
<p>$$ = {{9 \times 8 \times 7} \over 6} \times 64$$</p>
<p>$$ = 5376$$</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1lgzzvyiu | maths | matrices-and-determinants | transpose-of-a-matrix | <p>Let $$P=\left[\begin{array}{cc}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right], A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$$ and $$Q=P A P^{T}$$. If $$P^{T} Q^{2007} P=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$$, then $$2 a+b-3 c-4 d$$ equal to :</p> | [{"identifier": "A", "content": "2004"}, {"identifier": "B", "content": "2006"}, {"identifier": "C", "content": "2007"}, {"identifier": "D", "content": "2005"}] | ["D"] | null | $$
\text { Here, } P=\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{-1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right], A=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]
$$
<br/><br/>$$
\text { Here, } \mathrm{PP}^{\mathrm{T}}=\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{-1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right]\left[\begin{array}{cc}
\frac{\sqrt{3}}{2} & \frac{-1}{2} \\
\frac{1}{2} & \frac{\sqrt{3}}{2}
\end{array}\right]
$$
<br/><br/>$$
=\left[\begin{array}{cc}
\frac{3}{4}+\frac{1}{4} & -\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4} \\
\frac{-\sqrt{3}}{4}+\frac{\sqrt{3}}{4} & \frac{1}{4}+\frac{3}{4}
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\mathrm{I}
$$
<br/><br/>Similarly $P^T P=1$
<br/><br/>$$ \because $$ $$Q=P A P^{T}$$
<br/><br/>Now, $Q^{2007}=\left(P A P^T\right)\left(P A P^T\right) \ldots 2007$ times $=P A^{2007} P^T$
<br/><br/>$$
\begin{aligned}
& \text { As, } A=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right] \\\\
& \Rightarrow A^2=\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1+0 & 1+1 \\
0+0 & 0+1
\end{array}\right]=\left[\begin{array}{ll}
1 & 2 \\
0 & 1
\end{array}\right]
\end{aligned}
$$
<br/><br/>$$
A^3=\left[\begin{array}{ll}
1 & 3 \\
0 & 1
\end{array}\right]
$$
<br/>.
<br/>.
<br/>.
<br/>.
<br/>$$
A^{2007}=\left[\begin{array}{cc}
1 & 2007 \\
0 & 1
\end{array}\right]
$$
<br/><br/>$$
\begin{aligned}
& \text { Hence, } \mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2007} \mathrm{P}=\mathrm{A}^{2007}=\left[\begin{array}{cc}
1 & 2007 \\
0 & 1
\end{array}\right] \\\\
& \Rightarrow\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{cc}
1 & 2007 \\
0 & 1
\end{array}\right]
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow a=1, b=2007, c=0, d=1 \\\\
& \therefore 2 a+b-3 c-4 d=2(1)+2007-3(0)-4(1) \\\\
& =2+2007-4=2005
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
lsaob5vy | maths | matrices-and-determinants | transpose-of-a-matrix | If $\mathrm{A}=\left[\begin{array}{cc}\sqrt{2} & 1 \\ -1 & \sqrt{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right], \mathrm{C}=\mathrm{ABA}^{\mathrm{T}}$ and $\mathrm{X}=\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{~A}$, then $\operatorname{det} \mathrm{X}$ is equal to : | [{"identifier": "A", "content": "243"}, {"identifier": "B", "content": "729"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "891"}] | ["B"] | null | <p>The solution involves understanding matrix operations and properties such as multiplication, transpose, and determinant. Given $\mathrm{A}$, $\mathrm{B}$, and that $\mathrm{C} = \mathrm{ABA}^{\mathrm{T}}$, and $\mathrm{X} = \mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}$, we find $\operatorname{det} \mathrm{X}$ as follows:</p>
<p>First, we express $|\mathrm{C}|$, the determinant of $\mathrm{C}$, in terms of the determinants of $\mathrm{A}$ and $\mathrm{B}$, using the property that $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$:</p>
<p>$\begin{aligned} |\mathrm{C}| &= |\mathrm{ABA}^{\mathrm{T}}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{A}^{\mathrm{T}}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{B}| \quad \text{since } |\mathrm{A}^{\mathrm{T}}| = |\mathrm{A}|. \end{aligned}$</p>
<p>Next, we find $|\mathrm{X}|$, using the property $|\mathrm{ABC}| = |\mathrm{A}| \cdot |\mathrm{B}| \cdot |\mathrm{C}|$ and substituting $|\mathrm{C}|$:</p>
<p>$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}^{\mathrm{T}} \mathrm{C}^2 \mathrm{A}| = |\mathrm{A}^{\mathrm{T}}| \cdot |\mathrm{C}|^2 \cdot |\mathrm{A}| \\\\ &= |\mathrm{A}|^2 \cdot |\mathrm{C}|^2. \end{aligned}$</p>
<p>Substituting the expression for $|\mathrm{C}|$ obtained earlier:</p>
<p>$\begin{aligned} |\mathrm{X}| &= |\mathrm{A}|^2 \cdot (|\mathrm{A}|^2 \cdot |\mathrm{B}|)^2 \\\\ &= (|\mathrm{A}|^3 \cdot |\mathrm{B}|)^2. \end{aligned}$</p>
<p>The determinants of $\mathrm{A}$ and $\mathrm{B}$ are calculated as:</p>
$$
|A|=\left|\begin{array}{cc}
\sqrt{2} & 1 \\
-1 & \sqrt{2}
\end{array}\right|=2+1=3
$$
<br/><br/>$$
|B|=\left|\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right|=1
$$
<p>Finally, substituting these values into our expression for $|\mathrm{X}|$:</p>
<p>$\begin{aligned} |\mathrm{X}| &= (3^3 \cdot 1)^2 \\\\ &= 729. \end{aligned}$</p>
<p>Therefore, $\operatorname{det} \mathrm{X} = 729$.</p> | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lsezyqu2 | maths | matrices-and-determinants | transpose-of-a-matrix | <p>Let $$\mathrm{A}$$ be a square matrix such that $$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}$$. Then $$\frac{1}{2} A\left[\left(A+A^T\right)^2+\left(A-A^T\right)^2\right]$$ is equal to</p> | [{"identifier": "A", "content": "$$\\mathrm{A}^2+\\mathrm{A}^{\\mathrm{T}}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{A}^3+\\mathrm{I}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{A}^3+\\mathrm{A}^{\\mathrm{T}}$$\n"}, {"identifier": "D", "content": "$$\\mathrm{A}^2+\\mathrm{I}$$"}] | ["C"] | null | <p>$$\mathrm{AA}^{\mathrm{T}}=\mathrm{I}=\mathrm{A}^{\mathrm{T}} \mathrm{A}$$</p>
<p>On solving given expression, we get</p>
<p>$$\begin{aligned}
& \frac{1}{2} \mathrm{~A}\left[\mathrm{~A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2+2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}+\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2-2 \mathrm{~A} \mathrm{~A}^{\mathrm{T}}\right] \\
& =\mathrm{A}\left[\mathrm{A}^2+\left(\mathrm{A}^{\mathrm{T}}\right)^2\right]=\mathrm{A}^3+\mathrm{A}^{\mathrm{T}}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
hrCdUD5PJzdt0puJ8q5I7 | maths | parabola | chord-of-contact | Tangents drawn from the point ($$-$$8, 0) to the parabola y<sup>2</sup> = 8x touch the parabola at $$P$$ and $$Q.$$ If F is the focus of the parabola, then the area of the triangle PFQ (in sq. units) is equal to : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "32"}, {"identifier": "C", "content": "48"}, {"identifier": "D", "content": "64"}] | ["C"] | null | Equation of the chord of contact PQ is given by : T=0
<br><br>or T $$ \equiv $$ yy<sub>1</sub> $$-$$ 4(x + x<sub>1</sub>), where (x<sub>1</sub>, y<sub>1</sub><sub></sub>) $$ \equiv $$ ($$-$$8, 0)
<br><br>$$\therefore\,\,\,$$Equation becomes : x = 8
<br><br>& chord of contact is x = 8
<br><br>$$\therefore\,\,\,$$ Coordinates of point P and Q are (8, 8) and (8, $$-$$ 8)
<br><br>and focus of the parabola is F (2, 0)
<br><br>$$\therefore\,\,\,$$ Area of triangle PQF = $${1 \over 2}$$ $$ \times $$ (8 $$-$$ 2) $$ \times $$ (8 + 8) = 48 sq. units | mcq | jee-main-2018-online-15th-april-evening-slot |
7YsDSgdMqevO5bt7 | maths | parabola | chord-of-parabola | If $$a \ne 0$$ and the line $$2bx+3cy+4d=0$$ passes through the points of intersection of the parabolas $${y^2} = 4ax$$ and $${x^2} = 4ay$$, then : | [{"identifier": "A", "content": "$${d^2} + {\\left( {3b - 2c} \\right)^2} = 0$$ "}, {"identifier": "B", "content": "$${d^2} + {\\left( {3b + 2c} \\right)^2} = 0$$ "}, {"identifier": "C", "content": "$${d^2} + {\\left( {2b - 3c} \\right)^2} = 0$$ "}, {"identifier": "D", "content": "$${d^2} + {\\left( {2b + 3c} \\right)^2} = 0$$ "}] | ["D"] | null | Solving equations of parabolas
<br><br>$${y^2} = 4ax$$ and $${x^2} = 4ay$$
<br><br>we get $$(0,0)$$ and $$(4a, 4a)$$
<br><br>Substituting in the given equation of line
<br><br>$$2bx+3cy+4d=0,$$
<br><br>we get $$d=0$$
<br><br>and $$2b+3c=0$$ $$ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$$ | mcq | aieee-2004 |
MRfcmYZHlTsDOJCop3OsB | maths | parabola | chord-of-parabola | The length of the chord of the parabola x<sup>2</sup> $$=$$ 4y having equation x – $$\sqrt 2 y + 4\sqrt 2 = 0$$ is - | [{"identifier": "A", "content": "$$8\\sqrt 2 $$"}, {"identifier": "B", "content": "$$6\\sqrt 3 $$"}, {"identifier": "C", "content": "$$3\\sqrt 2 $$"}, {"identifier": "D", "content": "$$2\\sqrt {11} $$"}] | ["B"] | null | x<sup>2</sup> = 4y
<br><br>x $$-$$ $$\sqrt 2 $$y + 4$$\sqrt 2 $$ = 0
<br><br>Solving together we get
<br><br>x<sup>2</sup> = 4$$\left( {{{x + 4\sqrt 2 } \over {\sqrt 2 }}} \right)$$
<br><br>$$\sqrt 2 $$x<sup>2</sup> + 4x + 16$$\sqrt 2 $$
<br><br>$$\sqrt 2 $$x<sup>2</sup> $$-$$ 4x $$-$$ 16$$\sqrt 2 $$ = 0
<br><br>x<sub>1</sub> + x<sub>2</sub> = 2$$\sqrt 2 $$; x<sub>1</sub>x<sub>2</sub> = $${{ - 16\sqrt 2 } \over {\sqrt 2 }}$$ = $$-$$ 16
<br><br>Similarly,
<br><br>($$\sqrt 2 $$y $$-$$ 4$$\sqrt 2 $$)<sup>2</sup> = 4y
<br><br>2y<sup>2</sup> + 32 $$-$$ 16y = 4y
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267738/exam_images/dj4d0ou4immj3t18yq1g.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Parabola Question 96 English Explanation 1">
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266175/exam_images/abwa7naonbnbqkrjrncp.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Parabola Question 96 English Explanation 2">
<br>$$\ell $$<sub>AB</sub> = $$\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $$
<br><br>$$ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)} $$
<br><br>$$ = \sqrt {8 + 64 + 100 - 64} $$
<br><br>$$ = \sqrt {108} = 6\sqrt 3 $$ | mcq | jee-main-2019-online-10th-january-evening-slot |
rtn4tbqmzmPGU6Czgf18hoxe66ijvwpvxez | maths | parabola | chord-of-parabola | If one end of a focal chord of the parabola,
y<sup>2</sup> = 16x is at (1, 4), then the length of this focal
chord is : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "20"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "22"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267371/exam_images/fs6tfqb2vks175baakkh.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267826/exam_images/w5w1xiqap8u9c8okt7io.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267183/exam_images/gu3kbdwmjr2pxmljyejz.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Morning Slot Mathematics - Parabola Question 89 English Explanation"></picture>
<br>For this parabola y<sup>2</sup> = 16x,
<br><br>a = 4
<br><br>Here PQ is focal cord.
<br><br>Let P(at<sub>1</sub><sup>2</sup>, 2at<sub>1</sub>) and Q(at<sub>2</sub><sup>2</sup>, 2at<sub>2</sub>).
<br><br>Given P(1, 4),
<br><br>$$ \therefore $$ at<sub>1</sub><sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ 4t<sub>1</sub><sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ t<sub>1</sub><sup>2</sup> = $${1 \over 4}$$
<br><br>$$ \Rightarrow $$ t<sub>1</sub> = $${1 \over 2}$$
<br><br>In parabola if the parameter of one end point of the focal cord is t<sub>1</sub> then parameter of the other end point t<sub>2</sub> = $$ - {1 \over {{t_1}}}$$
<br><br>Here parameter for point Q t<sub>2</sub> = - 2
<br><br>$$ \therefore $$ Length of focal cord
<br><br>|PQ| = a$${\left( {{t_1} - {t_2}} \right)^2}$$ = 4$${\left( {{1 \over 2} + 2} \right)^2}$$ = 25 | mcq | jee-main-2019-online-9th-april-morning-slot |
YdDyAxUKkRDsCFMrbHjgy2xukf3yz3ju | maths | parabola | chord-of-parabola | Let the latus ractum of the parabola y<sup>2</sup>
= 4x be
the common chord to the circles C<sub>1</sub>
and C<sub>2</sub>
each of them having radius 2$$\sqrt 5 $$. Then, the
distance between the centres of the circles C<sub>1</sub>
and C<sub>2</sub>
is : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "$$8\\sqrt 5 $$"}, {"identifier": "D", "content": "$$4\\sqrt 5 $$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266972/exam_images/bzjcfypz4zfwsyd69i2a.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Evening Slot Mathematics - Parabola Question 77 English Explanation">
<br>For parabola y<sup>2</sup> = 4x,
<br><br>Length of latus rectum = 4a = 4<br><br>C<sub>1</sub>C<sub>2</sub> = 2(C<sub>1</sub>A)<br><br>C<sub>1</sub>A = $$\sqrt {{{(2\sqrt 5 )}^2} - {2^2}} = 4$$<br><br>$$ \therefore $$ C<sub>1</sub>C<sub>2</sub> = 2(4) = 8 | mcq | jee-main-2020-online-3rd-september-evening-slot |
1l545nh5v | maths | parabola | chord-of-parabola | <p>Let PQ be a focal chord of the parabola y<sup>2</sup> = 4x such that it subtends an angle of $${\pi \over 2}$$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse $$E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $${a^2} > {b^2}$$. If e is the eccentricity of the ellipse E, then the value of $${1 \over {{e^2}}}$$ is equal to :</p> | [{"identifier": "A", "content": "$$1 + \\sqrt 2 $$"}, {"identifier": "B", "content": "$$3 + 2\\sqrt 2 $$"}, {"identifier": "C", "content": "$$1 + 2\\sqrt 3 $$"}, {"identifier": "D", "content": "$$4 + 5\\sqrt 3 $$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5niod7a/d422eadb-ccc2-408b-8388-e3a333afd463/c3bb6e70-04d1-11ed-93b8-936002ac8631/file-1l5niod7b.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5niod7a/d422eadb-ccc2-408b-8388-e3a333afd463/c3bb6e70-04d1-11ed-93b8-936002ac8631/file-1l5niod7b.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th June Morning Shift Mathematics - Parabola Question 54 English Explanation"></p>
<p>As $$\angle PRQ = {\pi \over 2}$$</p>
<p>$$\left( {{{{2 \over t}} \over {3 - {1 \over {{t^2}}}}}} \right)\,.\,\left( {{{ - 2t} \over {3 - {t^2}}}} \right) = - 1$$</p>
<p>$$ \Rightarrow t = \pm \,1$$</p>
<p>$$\therefore$$ $$P \equiv (1,2)$$ & $$Q(1, - 2)$$</p>
<p>$$\therefore$$ for ellipse $${1 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$ and $$ae = 1$$</p>
<p>$$ \Rightarrow {1 \over {{a^2}}} + {4 \over {{a^2}(1 - {e^2})}} = 1$$</p>
<p>$$ \Rightarrow 1 + {4 \over {(1 - {e^2})}} = {1 \over {{e^2}}}$$</p>
<p>$$ \Rightarrow (5 - {e^2}){e^2} = 1 - {e^2}$$</p>
<p>$$ \Rightarrow {e^4} - 6{e^2} + 1 = 0$$</p>
<p>$$ \Rightarrow {e^2} = {1 \over {3 - 2\sqrt 2 }} \Rightarrow {1 \over {{e^2}}} = 3 + 2\sqrt 2 $$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l5w1gh8w | maths | parabola | chord-of-parabola | <p>Let PQ be a focal chord of length 6.25 units of the parabola y<sup>2</sup> = 4x. If O is the vertex of the parabola, then 10 times the area (in sq. units) of $$\Delta$$POQ is equal to ___________.</p> | [] | null | 25 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l6dmy0zj/4987e864-b411-4a25-aaa9-467edd1daa02/d5c9d0f0-132e-11ed-941a-4dd6502f33e3/file-1l6dmy0zk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l6dmy0zj/4987e864-b411-4a25-aaa9-467edd1daa02/d5c9d0f0-132e-11ed-941a-4dd6502f33e3/file-1l6dmy0zk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Parabola Question 41 English Explanation"></p>
<p>Given parabola $${y^2} = 4x$$</p>
<p>$$\therefore$$ a = 1</p>
<p>Here, P, S, Q points are collinear.</p>
<p>$$\therefore$$ Slope of PS = Slope of QS</p>
<p>$$ \Rightarrow {{2{t_1} - 0} \over {t_1^2 - 1}} = {{0 - 2{t_2}} \over {1 - t_2^2}}$$</p>
<p>$$ \Rightarrow {{2{t_1}} \over {t_1^2 - 1}} = {{2{t_2}} \over {t_2^2 - 1}}$$</p>
<p>$$ \Rightarrow {t_1}(t_2^2 - 1) = {t_2}(t_1^2 - 1)$$</p>
<p>$$ \Rightarrow t_2^2{t_1} - {t_1} = t_1^2{t_2} - {t_2}$$</p>
<p>$$ \Rightarrow t_2^2{t_1} - t_1^2{t_2} - {t_1} + {t_2} = 0$$</p>
<p>$$ \Rightarrow {t_1}{t_2}({t_2} - {t_1}) + ({t_2} - {t_1}) = 0$$</p>
<p>$$ \Rightarrow ({t_2} - {t_1})({t_1}{t_2} + 1) = 0$$</p>
<p>As $${t_2} - {t_1} \ne 0$$</p>
<p>$$\therefore$$ $${t_1}{t_2} + 1 = 0$$</p>
<p>$${t_1}{t_2} = - 1$$</p>
<p>Now, lenght of PQ</p>
<p>$$ = \sqrt {{{\left( {t_1^2 - t_2^2} \right)}^2} + {{\left( {2{t_1} - 2{t_2}} \right)}^2}} $$</p>
<p>$$ = \sqrt {{{\left( {{t_1} + {t_2}} \right)}^2}{{\left( {{t_1} - {t_2}} \right)}^2} + 4{{\left( {{t_1} - {t_2}} \right)}^2}} $$</p>
<p>$$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} + {t_2}} \right)}^2} + 4} $$</p>
<p>$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2{t_1}{t_2} + 4} $$</p>
<p>$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2( - 1) + 4} $$</p>
<p>$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2} $$</p>
<p>$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2( - 1)} $$</p>
<p>$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2{t_1}{t_2}} $$</p>
<p>$$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} - {t_2}} \right)}^2}} $$</p>
<p>$$ = {\left( {{t_1} - {t_2}} \right)^2}$$</p>
<p>Given, length of $$PQ = {\left( {{t_1} - {t_2}} \right)^2} = 6.25$$</p>
<p>$$ \Rightarrow {t_1} - {t_2} = 2.5$$</p>
<p>Now, Area of $$\Delta OPQ$$</p>
<p>$$ = \left| {{1 \over 2}\left| {\matrix{
{t_1^2} & {2{t_1}} & 1 \cr
{t_2^2} & {2{t_2}} & 1 \cr
0 & 0 & 1 \cr
} } \right|} \right|$$</p>
<p>$$ = \left| {{1 \over 2}\left( {2{t_2}t_1^2 - 2{t_1}t_2^2} \right)} \right|$$</p>
<p>$$ = \left| {{1 \over 2} \times 2{t_1}{t_2}\left( {{t_1} - {t_2}} \right)} \right|$$</p>
<p>$$ = \left| {{t_1}{t_2} \times \left( {{t_1} - {t_2}} \right)} \right|$$</p>
<p>$$ = \left| { - 1 \times 2.5} \right|$$</p>
<p>$$ = 2.5$$</p>
<p>$$\therefore$$ 10$$\Delta$$OPQ =</p>
<p>$$ = 10 \times {{25} \over {10}} = 25$$</p> | integer | jee-main-2022-online-30th-june-morning-shift |
1l6p2o6sm | maths | parabola | chord-of-parabola | <p>Let the focal chord of the parabola $$\mathrm{P}: y^{2}=4 x$$ along the line $$\mathrm{L}: y=\mathrm{m} x+\mathrm{c}, \mathrm{m}>0$$ meet the parabola at the points M and N. Let the line L be a tangent to the hyperbola $$\mathrm{H}: x^{2}-y^{2}=4$$. If O is the vertex of P and F is the focus of H on the positive x-axis, then the area of the quadrilateral OMFN is :</p> | [{"identifier": "A", "content": "$$2 \\sqrt{6}$$"}, {"identifier": "B", "content": "$$2 \\sqrt{14}$$"}, {"identifier": "C", "content": "$$4 \\sqrt{6}$$"}, {"identifier": "D", "content": "$$4 \\sqrt{14}$$"}] | ["B"] | null | <p>$$H:{{{x^2}} \over 4} - {{{y^2}} \over 4} = 1$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssn4pj/b0858943-62f1-4335-ac1e-b62256a92364/16a87a70-2f51-11ed-85dd-19dc023e9ad1/file-1l7ssn4pk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ssn4pj/b0858943-62f1-4335-ac1e-b62256a92364/16a87a70-2f51-11ed-85dd-19dc023e9ad1/file-1l7ssn4pk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Morning Shift Mathematics - Parabola Question 32 English Explanation"></p>
<p>Focus $$(ae,0)$$</p>
<p>$$F\left( {2\sqrt 2 ,\,0} \right)$$</p>
<p>$$y = mx + c$$ passes through (1, 0)</p>
<p>$$0 = m + C$$ ...... (i)</p>
<p>L is tangent to hyperbola</p>
<p>$$C = \, \pm \,\sqrt {4{m^2} - 4} $$</p>
<p>$$ - m = \, \pm \,\sqrt {4{m^2} - 4} $$</p>
<p>$${m^2} = 4{m^2} - 4$$</p>
<p>$$m = {2 \over {\sqrt 3 }}$$</p>
<p>$$C = {{ - 2} \over {\sqrt 3 }}$$</p>
<p>$$T:y = {2 \over {\sqrt 3 }}x - {2 \over {\sqrt 3 }}$$</p>
<p>$$P:{y^2} = 4x$$</p>
<p>$${y^2} = 4\left( {{{\sqrt 3 y + 2} \over 2}} \right)$$</p>
<p>$${y^2} - 2\sqrt 3 y - 4 = 0$$</p>
<p>Area</p>
<p>$${1 \over 2}\left| {\matrix{
0 & 0 \cr
{{x_1}} & {{y_1}} \cr
{2\sqrt 2 } & 0 \cr
{{x_2}} & {{y_2}} \cr
0 & 0 \cr
} } \right|$$</p>
<p>$$ = \left| {{1 \over 2}\left( { - 2\sqrt 2 {y_1} + 2\sqrt 2 {y_2}} \right)} \right|$$</p>
<p>$$ = \sqrt 2 \left| {{y_2} - {y_1}} \right| = \sqrt 2 \sqrt {{{({y_1} + {y_2})}^2} - 4{y_1}{y_2}} $$</p>
<p>$$ = \sqrt {56} $$</p>
<p>$$ = 2\sqrt {14} $$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo71096 | maths | parabola | chord-of-parabola | <p>If the $$x$$-intercept of a focal chord of the parabola $$y^{2}=8x+4y+4$$ is 3, then the length of this chord is equal to ____________.</p> | [] | null | 16 | $$
\begin{aligned}
& y^2=8 x+4 y+4 \\\\
& (y-2)^2=8(x+1) \\\\
& Y^2=4 a X \\\\
& a=2, X=x+1, Y=y-2 \\\\
& \text { focus }(1,2) \\\\
& y-2=m(x-1)
\end{aligned}
$$
<br/><br/>Put $(3,0)$ in the above line $\mathrm{m}=-1$
<br/><br/>Length of focal chord $=16$ | integer | jee-main-2023-online-1st-february-evening-shift |
1lgpxl5r8 | maths | parabola | chord-of-parabola | <p>Let $$\mathrm{PQ}$$ be a focal chord of the parabola $$y^{2}=36 x$$ of length 100 , making an acute angle with the positive $$x$$-axis. Let the ordinate of $$\mathrm{P}$$ be positive and $$\mathrm{M}$$ be the point on the line segment PQ such that PM : MQ = 3 : 1. Then which of the following points does NOT lie on the line passing through M and perpendicular to the line $$\mathrm{PQ}$$?</p> | [{"identifier": "A", "content": "$$(6,29)$$"}, {"identifier": "B", "content": "$$(-3,43)$$"}, {"identifier": "C", "content": "$$(3,33)$$"}, {"identifier": "D", "content": "$$(-6,45)$$"}] | ["B"] | null | The given parabola is of the form $$y^2 = 4ax$$. Here, 4a = 36, which means a = 9.
<br><br>Length of focal chord at $(t)=a\left(t+\frac{1}{t}\right)^2=100$
<br><br>Where $a=9$
<br><br>$$
\begin{gathered}
t+\frac{1}{t}= \pm \frac{10}{3} \\\\
\therefore \quad t=3, \frac{1}{3},-3, \frac{-1}{3}
\end{gathered}
$$
<br><br>Since ordinate of $P$ is $+\mathrm{ve}$
<br><br>$$
\therefore t=3 \text { or } \frac{1}{3}
$$
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh8cf7fa/cc7db376-968a-4280-8ff8-3d1538f92ccf/25044f60-ea0c-11ed-9dfd-37e68ac0819e/file-1lh8cf7fb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh8cf7fa/cc7db376-968a-4280-8ff8-3d1538f92ccf/25044f60-ea0c-11ed-9dfd-37e68ac0819e/file-1lh8cf7fb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 13th April Morning Shift Mathematics - Parabola Question 19 English Explanation">
<p>For each value of t, we can find the coordinates of the points P and Q on the parabola using the parametric form of the parabola $y^2=4ax$, where a is 9. The parametric form is given by $x=at^2$ and $y=2at$.</p>
<p>Let's consider $t=3$ first.</p>
<ol>
<li><p>For $t=3$, the coordinates of P are given by $P(at^2, 2at) = P(81, 54)$.</p>
</li>
<li><p>Since $t_1t_2=-1$ for a focal chord, the parameter value for Q is $t_2=-1/t_1=-1/3$. So the coordinates of Q are given by $Q(at_2^2, 2at_2) = Q(1, -6)$.</p>
</li>
</ol>
<p>Now, let's find the coordinates of the point M that divides the line segment PQ in the ratio 3:1.</p>
<ol>
<li><p>The x-coordinate of M is given by $\frac{3Q_x+P_x}{4} = \frac{3*1+81}{4} = 21$.</p>
</li>
<li><p>The y-coordinate of M is given by $\frac{3Q_y+P_y}{4} = \frac{3*(-6)+54}{4} = 9$.</p>
</li>
</ol>
<p>So, the coordinates of M are M$(21, 9)$.</p>
<p>We can repeat these steps for $t=1/3$ to find the other set of points P, Q, and M. However, since the ordinate of P is positive and PQ makes an acute angle with the positive x-axis, $t=3$ is the appropriate choice in this context.</p>
<p>The slope of the line PQ is $m_{PQ} = \frac{-6 - 54}{1 - 81} = \frac{-60}{-80} = \frac{3}{4}$. The slope of the line perpendicular to PQ is $m_{\perp PQ} = -1/m_{PQ} = -\frac{4}{3}$. Thus, the equation of the line through M and perpendicular to PQ is $(y - 9) = -\frac{4}{3}(x - 21)$, or $4x + 3y = 111$.</p>
<p>Now, we check which of the given points do NOT lie on this line.</p>
<p>Option A : $(6, 29)$
Substitute these values into the equation :</p>
<p>$4\times6 + 3\times29 = 111?$</p>
<p>$24 + 87 = 111?$</p>
<p>$111 = 111$</p>
<p>So, option A does lie on the line.</p>
<p>Option B : $(-3, 43)$
Substitute these values into the equation :</p>
<p>$4\times(-3) + 3\times43 = 111?$</p>
<p>$-12 + 129 = 111?$</p>
<p>$117 = 111$</p>
<p>So, option B does <b>NOT</b> lie on the line.</p>
<p>Option C : $(3, 33)$
Substitute these values into the equation :</p>
<p>$4\times3 + 3\times33 = 111?$</p>
<p>$12 + 99 = 111?$</p>
<p>$111 = 111$</p>
<p>So, option C does lie on the line.</p>
<p>Option D : $(-6, 45)$
Substitute these values into the equation :</p>
<p>$4\times(-6) + 3\times45 = 111?$</p>
<p>$-24 + 135 = 111?$</p>
<p>$111 = 111$</p>
<p>So, option D does lie on the line.</p>
<p>Therefore, $(-3, 43)$ which does not lie on the line passing through M and perpendicular to the line PQ.</p>
| mcq | jee-main-2023-online-13th-april-morning-shift |
1lgzyfu4u | maths | parabola | chord-of-parabola | <p>Let $$R$$ be the focus of the parabola $$y^{2}=20 x$$ and the line $$y=m x+c$$ intersect the parabola at two points $$P$$ and $$Q$$.
<br/><br/>Let the point $$G(10,10)$$ be the centroid of the triangle $$P Q R$$. If $$c-m=6$$, then $$(P Q)^{2}$$ is :</p> | [{"identifier": "A", "content": "317"}, {"identifier": "B", "content": "325"}, {"identifier": "C", "content": "346"}, {"identifier": "D", "content": "296"}] | ["B"] | null | $$
y^2=20 x, y=m x+\mathrm{c}
$$
<br/><br/>Put value of $x$
<br/><br/>$$
\begin{aligned}
& y^2=20\left(\frac{y-c}{m}\right) \\\\
& \Rightarrow y^2-\frac{20}{m} y+\frac{20}{m} c=0 .......(i)
\end{aligned}
$$
<br/><br/>Since, centroid $=(10,10)$
<br/><br/>$$
\begin{aligned}
& \text { So, } \frac{y_1+y_2+0}{3}=10 \\\\
& \Rightarrow y_1+y_2=30
\end{aligned}
$$
<br/><br/>From (1),
<br/><br/>$$
\text { Sum of roots }=\frac{20}{m}=30 \Rightarrow m=\frac{2}{3}
$$
<br/><br/>Also, $c-m=6 \Rightarrow c=6+\frac{2}{3}=\frac{20}{3}$
<br/><br/>Now, the equation is :
<br/><br/>$$
\begin{aligned}
& y^2-\frac{20}{2} \times 3 y+\frac{20}{2} \times 3 \times \frac{20}{3}=0 \\\\
& \Rightarrow y^2-30 y+200=0 \\\\
& \Rightarrow y^2-20 y-10 y+200=0 \\\\
& \Rightarrow(y-20)(y-10)=0 \\\\
& \Rightarrow y=10,20 \Rightarrow x=5, x=20 \\\\
& \therefore P \equiv(5,10), Q \equiv(20,20) \\\\
& \text { So, }(P Q)^2=(20-5)^2+(20-10)^2 \\\\
& =225+100=325
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
jaoe38c1lsfl7tt9 | maths | parabola | chord-of-parabola | <p>Let $$P(\alpha, \beta)$$ be a point on the parabola $$y^2=4 x$$. If $$P$$ also lies on the chord of the parabola $$x^2=8 y$$ whose mid point is $$\left(1, \frac{5}{4}\right)$$, then $$(\alpha-28)(\beta-8)$$ is equal to _________.</p> | [] | null | 192 | <p>Parabola is $$x^2=8 y$$</p>
<p>Chord with mid point $$\left(\mathrm{x}_1, \mathrm{y}_1\right)$$ is $$\mathrm{T}=\mathrm{S}_1$$</p>
<p>$$\begin{aligned}
& \therefore \mathrm{xx}_1-4\left(\mathrm{y}+\mathrm{y}_1\right)=\mathrm{x}_1^2-8 \mathrm{y}_1 \\
& \therefore\left(\mathrm{x}_1, \mathrm{y}_1\right)=\left(1, \frac{5}{4}\right) \\
& \Rightarrow \mathrm{x}-4\left(\mathrm{y}+\frac{5}{4}\right)=1-8 \times \frac{5}{4}=-9
\end{aligned}$$</p>
<p>$$\therefore x-4 y+4=0$$ ...... (i)</p>
<p>$$(\alpha, \beta)$$ lies on (i) & also on $$y^2=4 x$$</p>
<p>$$\begin{aligned}
& \therefore \alpha-4 \beta+4=0 \text{ .... (ii)} \\
& \& ~\beta^2=4 \alpha \text{ .... (iii)}
\end{aligned}$$</p>
<p>Solving (ii) & (iii)</p>
<p>$$\begin{aligned}
& \beta^2=4(4 \beta-4) \Rightarrow \beta^2-16 \beta+16=0 \\
& \therefore \beta=8 \pm 4 \sqrt{3} \text { and } \alpha=4 \beta-4=28 \pm 16 \sqrt{3} \\
& \therefore(\alpha, \quad \beta) \quad=\quad(28+16 \sqrt{3}, 8+4 \sqrt{3}) \quad \& \\
& (28-16 \sqrt{3}, 8-4 \sqrt{3}) \\
& \therefore(\alpha-28)(\beta-8)=( \pm 16 \sqrt{3})( \pm 4 \sqrt{3}) \\
& =192
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-evening-shift |
lv0vxdso | maths | parabola | chord-of-parabola | <p>Let the length of the focal chord PQ of the parabola $$y^2=12 x$$ be 15 units. If the distance of $$\mathrm{PQ}$$ from the origin is $$\mathrm{p}$$, then $$10 \mathrm{p}^2$$ is equal to __________.</p> | [] | null | 72 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwk9u5jo/5c19a59f-8f6e-4ba6-8e8c-779764eeedda/afcaae30-1992-11ef-bba0-ebdcce35fbed/file-1lwk9u5jp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwk9u5jo/5c19a59f-8f6e-4ba6-8e8c-779764eeedda/afcaae30-1992-11ef-bba0-ebdcce35fbed/file-1lwk9u5jp.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) 4th April Morning Shift Mathematics - Parabola Question 7 English Explanation"></p>
<p>$$\begin{aligned}
& A B=15 \Rightarrow\left(3\left(t^2-\frac{1}{t^2}\right)\right)+\left(6\left(t+\frac{1}{t}\right)\right)^2=225 \\
& \Rightarrow 9\left(t^2-\frac{1}{t^2}\right)+36\left(t+\frac{1}{t}\right)^2=225
\end{aligned}$$</p>
<p>$$
\begin{aligned}
\Rightarrow & \left.\left.\left(t+\frac{1}{t}\right)^2 \right[\,\left(t-\frac{1}{t}\right)+4\right]=25 \\
& \left(t+\frac{1}{t}\right)^2\left(t+\frac{1}{t}\right)^2=25 \Rightarrow\left(t+\frac{1}{t}\right)^4=25 \\
\Rightarrow & t+\frac{1}{t}= \pm \sqrt{5} \Rightarrow\left(t-\frac{1}{t}\right)= \pm 1
\end{aligned}$$</p>
<p>Equation of $$A B:(y-6 t)=\left(\frac{2 t}{t^2-1}\right)\left(x-3 t^2\right)$$</p>
<p>$$\Rightarrow$$ Distance from $$y-6 t=m x-3 m t^2$$</p>
<p>$$\Rightarrow p=\frac{\left|3 m t^2-6 t\right|}{\sqrt{1+m^2}}=\frac{\left|\left(\frac{6 t}{t^2-1}\right)\right|}{\sqrt{5}}=\frac{6}{\sqrt{5}}$$</p>
<p>$$\left[ m=\frac{2 t}{t^2-1}=\frac{}{t-\frac{1}{t}}= \pm 2 \Rightarrow m^2=4\right]$$</p>
<p>$$\Rightarrow \quad 10 p^2=\frac{10 \times 36}{5} \Rightarrow 72$$</p> | integer | jee-main-2024-online-4th-april-morning-shift |
lv2er40w | maths | parabola | chord-of-parabola | <p>Let $$P Q$$ be a chord of the parabola $$y^2=12 x$$ and the midpoint of $$P Q$$ be at $$(4,1)$$. Then, which of the following point lies on the line passing through the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ ?</p> | [{"identifier": "A", "content": "$$(3,-3)$$\n"}, {"identifier": "B", "content": "$$\\left(\\frac{1}{2},-20\\right)$$\n"}, {"identifier": "C", "content": "$$(2,-9)$$\n"}, {"identifier": "D", "content": "$$\\left(\\frac{3}{2},-16\\right)$$"}] | ["B"] | null | <p>$$y^2=12 x$$</p>
<p>Chord $$P Q$$ having mid-point $$(x_1, y_1)=(4,1)$$ equation of chord $$P Q$$</p>
<p>$$\begin{aligned}
& T=S_1 \\
& y y_1-12 \frac{\left(x+x_1\right)}{2}=y_1^2-12 x_1 \\
& y-6(x+4)=1-12 \times 4 \\
& y-6 x-24=-47 \\
& y-6 x+23=0
\end{aligned}$$</p>
<p>From option (4) $$x=\frac{1}{2}$$ & $$y=-20$$</p>
<p>$$-20-6 \times \frac{1}{2}+23=0$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv7v3k4p | maths | parabola | chord-of-parabola | <p>Suppose $$\mathrm{AB}$$ is a focal chord of the parabola $$y^2=12 x$$ of length $$l$$ and slope $$\mathrm{m}<\sqrt{3}$$. If the distance of the chord $$\mathrm{AB}$$ from the origin is $$\mathrm{d}$$, then $$l \mathrm{~d}^2$$ is equal to _________.</p> | [] | null | 108 | <p>Equation of focal chord</p>
<p>$$y-0=\tan \theta .(x-3)$$</p>
<p>Distance from origin</p>
<p>$$\begin{aligned}
& d=\left|\frac{-3 \tan \theta}{\sqrt{1+\tan ^2 \theta}}\right| \\
& I=4 \times 3 \operatorname{cosec}^2 \theta \\
& I. d^2=\frac{9 \tan ^2 \theta}{1+\tan ^2 \theta} \times 12 \operatorname{cosec}^2 \theta \\
& =\frac{108 \operatorname{cosec}^2 \theta}{1+\cot ^2 \theta}=108
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-morning-shift |
9taDurH0zt2wYzwP | maths | parabola | common-tangent | Two common tangents to the circle $${x^2} + {y^2} = 2{a^2}$$ and parabola $${y^2} = 8ax$$ are : | [{"identifier": "A", "content": "$$x = \\pm \\left( {y + 2a} \\right)$$ "}, {"identifier": "B", "content": "$$y = \\pm \\left( {x + 2a} \\right)$$ "}, {"identifier": "C", "content": "$$x = \\pm \\left( {y + a} \\right)$$ "}, {"identifier": "D", "content": "$$y = \\pm \\left( {x + a} \\right)$$ "}] | ["B"] | null | Any tangent to the parabola $${y^2} = 8ax$$ is
<br><br>$$y = mx + {{2a} \over m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>If $$(i)$$ is a tangent to the circle, $${x^2} + {y^2} = 2{a^2}$$ then,
<br><br>$$\sqrt {2a} = \pm {{2a} \over {m\sqrt {{m^2} + 1} }}$$
<br><br>$$ \Rightarrow {m^2}\left( {1 + {m^2}} \right) = 2$$
<br><br>$$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$
<br><br>$$ \Rightarrow m = \pm 1.$$
<br><br>So from $$(i),$$ $$y = \pm \left( {x + 2a} \right).$$ | mcq | aieee-2002 |
NtIfshmqF4cgY2Iy | maths | parabola | common-tangent | <b>Given :</b> A circle, $$2{x^2} + 2{y^2} = 5$$ and a parabola, $${y^2} = 4\sqrt 5 x$$.
<br/><b>Statement-1 :</b> An equation of a common tangent to these curves is $$y = x + \sqrt 5 $$.
<p><b>Statement-2 :</b> If the line, $$y = mx + {{\sqrt 5 } \over m}\left( {m \ne 0} \right)$$ is their common tangent, then $$m$$ satiesfies $${m^4} - 3{m^2} + 2 = 0$$.</p> | [{"identifier": "A", "content": "Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}, {"identifier": "B", "content": "Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "C", "content": "Statement-1 is true; Statement-2 is false."}, {"identifier": "D", "content": "Statement-1 is false Statement-2 is true."}] | ["B"] | null | Let common tangent be
<br><br>$$y = mx + {{\sqrt 5 } \over m}$$
<br><br>Since, perpendicular distance from center of the circle to
<br><br>the common tangent is equal to radius of the circle,
<br><br>therefore $${{{{\sqrt 5 } \over m}} \over {\sqrt {1 + {m^2}} }} = \sqrt {{5 \over 2}} $$
<br><br>On squaring both the side, we get
<br><br>$${m^2}\left( {1 + {m^2}} \right) = 2$$
<br><br>$$ \Rightarrow {m^4} + {m^2} - 2 = 0$$
<br><br>$$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$$
<br><br>$$ \Rightarrow m = \pm 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ ( as $$m \ne \pm \sqrt 2 $$ )
<br><br>$$y = \pm \left( {x + \sqrt 5 } \right),$$ both statements are correct as $$m = \pm 1$$
<br><br>satisfies the given equation of statement - $$2.$$ | mcq | jee-main-2013-offline |
FxrHLpBJBnUdwvlp1iuzC | maths | parabola | common-tangent | If the common tangents to the parabola, x<sup>2</sup> = 4y and the circle, x<sup>2</sup> + y<sup>2</sup> = 4 intersect at the point P, then the distance of P from the origin, is : | [{"identifier": "A", "content": "$$\\sqrt 2 + 1$$"}, {"identifier": "B", "content": "2(3 + 2 $$\\sqrt 2 $$)"}, {"identifier": "C", "content": "2($$\\sqrt 2 $$ + 1)"}, {"identifier": "D", "content": "3 + 2$$\\sqrt 2 $$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267100/exam_images/dqtfzq9ngy0k8tl3a08o.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2017 (Online) 8th April Morning Slot Mathematics - Parabola Question 104 English Explanation">
<br>Tangent to x<sup>2</sup> + y<sup>2</sup> = 4 is
<br><br>y = mx $$ \pm $$ 2$$\sqrt {1 + {m^2}} $$
<br><br>Also, x<sup>2</sup> = 4y
<br><br>x<sup>2</sup> = 4mx + 8$$\sqrt {1 + {m^2}} $$
<br><br> or x<sup>2</sup> = 4mx $$-$$ 8$$\sqrt {1 + {m^2}} $$
<br><br>For D = 0
<br><br>we have; 16m<sup>2</sup> + 4.8$$\sqrt {1 + {m^2}} $$ = 0
<br><br>$$ \Rightarrow $$ m<sup>2</sup> + 2$$\sqrt {1 + {m^2}} $$ = 0
<br><br>$$ \Rightarrow $$ m<sup>2</sup> = $$-$$ 2$$\sqrt {1 + {m^2}} $$
<br><br>$$ \Rightarrow $$ m<sup>4</sup> = 4 + 4m<sup>2</sup>
<br><br>$$ \Rightarrow $$ m<sup>4</sup> $$-$$ 4m<sup>2</sup> $$-$$ 4 = 0
<br><br>$$ \Rightarrow $$ m<sup>2</sup> = $${{4 \pm \sqrt {16 + 16} } \over 2}$$
<br><br>$$ \Rightarrow $$ m<sup>2</sup> = $${{4 \pm 4\sqrt 2 } \over 2}$$
<br><br>$$ \Rightarrow $$ m<sup>2</sup> = 2 + 2$$\sqrt 2 $$ | mcq | jee-main-2017-online-8th-april-morning-slot |
2P3MCSR3dlMypVZh7Vinq | maths | parabola | common-tangent | Two parabolas with a common vertex and with axes along x-axis and $$y$$-axis, respectively intersect each other in the first quadrant. If the length of the latus rectum of each parabola is $$3$$, then the equation of the common tangent to the two parabolas is : | [{"identifier": "A", "content": "4(x + y) + 3 = 0"}, {"identifier": "B", "content": "3(x + y) + 4 = 0"}, {"identifier": "C", "content": "8(2x + y) + 3 = 0"}, {"identifier": "D", "content": "x + 2y + 3 = 0"}] | ["A"] | null | As origin is the only common point to x-axis and y-axis, so origin is the common vertex
<br><br>Let the equation of two of parabolas be y<sup>2</sup> = 4ax and x<sup>2</sup> = 4by
<br><br>Now latus rectum of both parabolas = 3
<br><br>$$\therefore\,\,\,$$ 4a = 4b = 3
<br><br>$$ \Rightarrow $$$$\,\,\,$$ a = b = $${3 \over 4}$$
<br><br>$$\therefore\,\,\,$$ Two parabolas are y<sup>2</sup> = 3x and x<sup>2</sup> = 3y
<br><br>Suppose y = mx + c is in the common tangent.
<br><br>$$\therefore\,\,\,$$ y<sup>2</sup> = 3x $$ \Rightarrow $$ (mx + c)<sup>2</sup> = 3x $$ \Rightarrow $$ m<sup>2</sup>x<sup>2</sup> + (2mc $$-$$ 3) x + c<sup>2</sup> = 0
<br><br>As, the tangent touches at one point only
<br><br>So, b<sup>2</sup> $$-$$ 4ac = 0
<br><br>$$ \Rightarrow $$ (2mc $$-$$ 3)<sup>2</sup> $$-$$ 4m<sup>2</sup>c<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ 4m<sup>2</sup>c<sup>2</sup> + 9 $$-$$ 12mc $$-$$ 4m<sup>2</sup>c<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ c = $${9 \over {12m}}$$ = $${3 \over {4m}}$$ . . . .(i)
<br><br>$$\therefore\,\,\,$$ x<sup>2</sup> = 3y $$ \Rightarrow $$ x<sup>2</sup> = 3 (mx + c ) $$ \Rightarrow $$ x<sup>2</sup> $$-$$ 3mx $$-$$ 3c = 0
<br><br>Again, b<sup>2</sup> $$-$$ 4ac = 0
<br><br>$$ \Rightarrow $$ 9m<sup>2</sup> $$-$$ 4(1) ($$-$$3c) = 0
<br><br>$$ \Rightarrow $$ 9m<sup>2</sup> = $$-$$ 12c . . . . .(ii)
<br><br>From (i) and (ii)
<br><br>m<sup>2</sup> = $${{ - 4c} \over 3}$$ = $${{ - 4} \over 3}$$ $$\left( {{3 \over {4m}}} \right)$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ m<sup>3</sup> = $$-$$ 1 $$ \Rightarrow $$ m = $$-$$ 1 $$ \Rightarrow $$ c = $${{ - 3} \over 4}$$
<br><br>Hence, y = mx + c = $$-$$ x $$-$$ $${3 \over 4}$$
<br><br>$$ \Rightarrow $$ 4 (x + y) + 3 = 0 | mcq | jee-main-2018-online-15th-april-morning-slot |
2ZGfQ5EYzWZNuxj1U6Eng | maths | parabola | common-tangent | Equation of a common tangent to the circle, x<sup>2</sup> + y<sup>2</sup> – 6x = 0 and the parabola, y<sup>2</sup> = 4x is : | [{"identifier": "A", "content": "$$2\\sqrt 3 $$y = 12x + 1"}, {"identifier": "B", "content": "$$\\sqrt 3 $$y = x + 3"}, {"identifier": "C", "content": "$$2\\sqrt 3 $$y = -x - 12"}, {"identifier": "D", "content": "$$\\sqrt 3 $$y = 3x + 1"}] | ["B"] | null | We know,
<br><br>Equation of tangent to the parabola y<sup>2</sup> = 4ax is,
<br><br>y = mx + $${a \over m}$$
<br><br>$$ \therefore $$ Equation of tangent to the parabola y<sup>2</sup> = 4x is,
<br><br>y = mx + $${1 \over m}$$
<br><br>$$ \Rightarrow $$ m<sup>2</sup>x $$-$$ ym + 1 = 0
<br><br>This tangent is also the tangent to the circle x<sup>2</sup> + y<sup>2</sup> $$-$$ 6x = 0
<br><br>So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.
<br><br>Here center is at (3, 0) of the circle and radius = 3
<br><br>$$ \therefore $$ $$\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$$
<br><br>$$ \Rightarrow $$ (3m<sup>2</sup> + 1)<sup>2</sup> = 9(m<sup>4</sup> + m<sup>2</sup>)
<br><br>$$ \Rightarrow $$ 9m<sup>4</sup> + 6m<sup>2</sup> + 1 = 9m<sup>4</sup> + 9m<sup>2</sup>
<br><br>$$ \Rightarrow $$ 3m<sup>2</sup> = 1
<br><br>$$ \Rightarrow $$ m = $$ \pm $$ $${1 \over {\sqrt 3 }}$$
<br><br>So, possible tangents are
<br><br>y = $${1 \over {\sqrt 3 }}$$x + $$\sqrt 3 $$
<br><br>$$ \Rightarrow $$ $$\sqrt 3 $$y = x + 3
<br><br>or y = $$-$$ $${x \over {\sqrt 3 }}$$ $$-$$ $$\sqrt 3 $$
<br><br>$$ \Rightarrow $$ $$\sqrt 3 y$$ = $$-$$ x $$-$$ 3 | mcq | jee-main-2019-online-9th-january-morning-slot |
rYubXCr5SvNnWKqiFu3rsa0w2w9jx25iuwn | maths | parabola | common-tangent | If the line ax + y = c, touches both the curves x<sup>2</sup>
+ y<sup>2</sup>
= 1 and y<sup>2</sup>
= 4$$\sqrt 2 $$x , then |c| is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$\\sqrt 2 $$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | Tangent to the curve y<sup>2</sup> = 4$$\sqrt 2$$x is y = mx + $${{\sqrt 2 } \over m}$$<br><br>
It is tangent to the circle x<sup>2</sup> + <sup>y2</sup> = 1<br><br>
$$ \therefore $$ $$\left| {{{\sqrt 2 /m} \over {\sqrt {1 + {m^2}} }}} \right| = 1 \Rightarrow m = \pm 1$$<br><br>
$$ \therefore $$ tangent are y = x + $$\sqrt 2$$ & y = – x – $$\sqrt 2$$<br><br>
Compare with y = – ax + c<br><br>
$$ \Rightarrow a = \pm 1 $$ & $$ c = \pm \sqrt 2 $$
| mcq | jee-main-2019-online-10th-april-evening-slot |
hEggXEf1frYDmowbRp3rsa0w2w9jx5c8n12 | maths | parabola | common-tangent | Let P be the point of intersection of the common tangents to the parabola y<sup>2</sup>
= 12x and the hyperbola
8x<sup>2</sup>
– y<sup>2</sup>
= 8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS'
in a ratio :
| [{"identifier": "A", "content": "14 : 13"}, {"identifier": "B", "content": "13 : 11"}, {"identifier": "C", "content": "5 : 4"}, {"identifier": "D", "content": "2 : 1"}] | ["C"] | null | Let equation of common tangent is y = mx + $${3 \over m}$$<br><br>
$$ \therefore $$ $${\left( {{3 \over m}} \right)^2}$$ = 1, m<sup>2</sup> - 8<br><br>
$$ \Rightarrow {m^4} - 8{m^2} - 9 = 0$$<br><br>
$$ \Rightarrow {m^2} = 9 \Rightarrow m = \pm 3$$<br><br>
$$ \therefore $$ equation of common tangents are y = 3x + 1 & y = -3x - 1<br><br>
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264098/exam_images/o0bfsfl0vlazf4qy2pvy.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Morning Slot Mathematics - Parabola Question 86 English Explanation">
$$ \therefore $$ $${{PS} \over {PS}} = {{3 + {1 \over 3}} \over { - {1 \over 3} + 3}} = {5 \over 4}$$ | mcq | jee-main-2019-online-12th-april-morning-slot |
7CKGmlezauHN5uhw6S3rsa0w2w9jxad6eun | maths | parabola | common-tangent | The equation of common tangent to the curves y<sup>2</sup>
= 16x and xy = –4, is :
| [{"identifier": "A", "content": "x \u2013 y + 4 = 0"}, {"identifier": "B", "content": "x + y + 4 = 0"}, {"identifier": "C", "content": "x \u2013 2y + 16 = 0"}, {"identifier": "D", "content": "2x \u2013 y + 2 = 0"}] | ["A"] | null | Let the equation of tangent to parabola<br><br>
y<sup>2</sup> = 16x is y = mx + $${4 \over m}$$ ...... (1)<br><br>
It is given that tangent to xy = -4 .........(2)<br><br>
Solving (1) and (2) we get<br><br>
$$x\left( {mx + {4 \over m}} \right) + 4 = 0$$<br><br>
$$ \Rightarrow m{x^2} + {4 \over m}x + 4 = 0$$<br><br>
Now for tangent D = 0 $$ \Rightarrow {{16} \over {{m^2}}} - 16m = 0$$<br><br>
$$ \Rightarrow {m^3} = 1$$ $$ \Rightarrow $$ m = 1<br><br>
Now putting value of m in Equation (1) <br><br>
y = x + 4 or <b>x – y + 4 = 0</b> | mcq | jee-main-2019-online-12th-april-evening-slot |
3X6EHkJyHixQzo6fqxjgy2xukfg6l5e2 | maths | parabola | common-tangent | If the common tangent to the parabolas, <br/>y<sup>2</sup> = 4x and x<sup>2</sup> = 4y also touches the circle, x<sup>2</sup> + y<sup>2</sup> = c<sup>2</sup>,<br/> then c is equal to : | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "B", "content": "$${1 \\over {2\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}] | ["A"] | null | $$y = mx + {1 \over m}$$ (tangent at y<sup>2</sup>
= 4x)
<br><br>y = mx – m<sup>2</sup>
(tangent at x<sup>2</sup>
= 4y)
<br><br>$${1 \over m} = - {m^2}$$ (for common tangent)
<br><br>m<sup>3</sup>
= – 1
<br><br>$$ \Rightarrow $$ m = - 1
<br><br>$$ \therefore $$ Equation of tangent
<br><br>y = –x –1
<br><br>x + y + 1 = 0
<br><br>This line touches circle whose center at (0, 0),
<br><br>$$ \therefore $$ apply p( Distance from center of the circle of the line ) = r ( Radius of the circle )
<br><br>c = $$\left| {{{0 + 0 + 1} \over {\sqrt 2 }}} \right| = {1 \over {\sqrt 2 }}$$ | mcq | jee-main-2020-online-5th-september-morning-slot |
rvy2Lk1McfAFgqhIjD1klt9nkb3 | maths | parabola | common-tangent | A line is a common tangent to the circle (x $$-$$ 3)<sup>2</sup> + y<sup>2</sup> = 9 and the parabola y<sup>2</sup> = 4x. If the two points of contact (a, b) and (c, d) are distinct and lie in the first quadrant, then 2(a + c) is equal to _________. | [] | null | 9 | Circle : (x $$-$$ 3)<sup>2</sup> + y<sup>2</sup> = 9<br><br>Parabola : y<sup>2</sup> = 4x<br><br>Let tangent y = mx + $${a \over m}$$<br><br>y = mx + $${1 \over m}$$<br><br>m<sup>2</sup>x $$-$$ my + 1 = 0<br><br>the above line is also tangent to circle<br><br>(x $$-$$ 3)<sup>2</sup> + y<sup>2</sup> = 9<br><br>$$\therefore$$ $$ \bot $$ from (3, 0) = 3
<br><br>$$\left| {{{3{m^2} - 0 + 1} \over {\sqrt {{m^2} + {m^4}} }}} \right| = 3$$
<br><br>(3m<sup>2</sup> + 1)<sup>2</sup> = 9(m<sup>2</sup> + m<sup>4</sup>)<br><br>$$6{m^2} + 1 + 9{m^4} = 9{m^2} + 9{m^4}$$<br><br>$$3{m^2} = 1$$<br><br>$$m = \pm {1 \over {\sqrt 3 }}$$<br><br>$$ \therefore $$ tangent is<br><br>$$y = {1 \over {\sqrt 3 }}x + \sqrt 3 $$<br><br>(it will be used)<br><br>or<br><br>$$y = - {1 \over {\sqrt 3 }}x - \sqrt 3 $$<br><br>(rejected)<br><br>$$m = {1 \over {\sqrt 3 }}$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264738/exam_images/lvif1mocwwwtckxww1qh.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Parabola Question 68 English Explanation"><br><br>For parabola <br><br>$$\left( {{a \over {{m^2}}},{{2a} \over m}} \right) \equiv (3,2\sqrt 3 )$$ = (c, d)<br><br>for circle $$y = {1 \over {\sqrt 3 }}x + \sqrt 3 $$<br><br>&<br><br>$${(x - 3)^2} + {y^2} = 9$$<br><br>Solving,<br><br>$${(x - 3)^2} + {\left( {{1 \over {\sqrt 3 }}x + \sqrt 3 } \right)^2} = 9$$<br><br>$${x^2} + 9 - 6x + {1 \over 3}{x^2} + 3 + 2x = 9$$<br><br>$${4 \over 3}{x^2} - 4x + 3 = 0$$<br><br>$$4{x^2} - 12x + 9 = 0$$<br><br>$$4{x^2} - 6x - 6x + 9 = 0$$<br><br>$$2x(2x - 3) - 3(2x - 3) = 0$$<br><br>$$(2x - 3)(2x - 3) = 0$$<br><br>$$x = {3 \over 2}$$<br><br>$$ \therefore $$ $$y = {1 \over {\sqrt 3 }}\left( {{3 \over 2}} \right) + \sqrt 3 $$<br><br>$$y = {{\sqrt 3 } \over 2} + \sqrt 3 $$<br><br>$$y = {{3\sqrt 3 } \over 2}$$<br><br>$$(a,b) \equiv \left( {{3 \over 2},{{3\sqrt 3 } \over 2}} \right)$$<br><br>$$2(a + c) = 2\left( {{3 \over 2} + 3} \right)$$<br><br>$$ = 2\left( {{3 \over 2} + {6 \over 2}} \right) = 9$$ | integer | jee-main-2021-online-25th-february-evening-slot |
4NuKcpiQ85rmuFolIW1kmkn9dip | maths | parabola | common-tangent | Let L be a tangent line to the parabola y<sup>2</sup> = 4x $$-$$ 20 at (6, 2). If L is also a tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over b} = 1$$, then the value of b is equal to : | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "16"}, {"identifier": "D", "content": "11"}] | ["B"] | null | Parabola y<sup>2</sup> = 4x $$-$$ 20<br><br>Tangent at P(6, 2) will be <br><br>$$2y = 4\left( {{{x + 6} \over 2}} \right) - 20$$<br><br>2y = 2x + 12 $$-$$ 20<br><br>2y = 2x $$-$$ 8<br><br>y = x $$-$$ 4<br><br>x $$-$$ y $$-$$ 4 = 0 ....... (1)<br><br>This is also tangent to ellipse $${{{x^2}} \over 2} + {{{y^2}} \over b} = 1$$<br><br>Apply c<sup>2</sup> = a<sup>2</sup>m<sup>2</sup> + b<sup>2</sup><br><br>($$-$$4)<sup>2</sup> = (2)(1) + b<br><br>b = 14 | mcq | jee-main-2021-online-17th-march-evening-shift |
1ktkea35m | maths | parabola | common-tangent | A tangent line L is drawn at the point (2, $$-$$4) on the parabola y<sup>2</sup> = 8x. If the line L is also tangent to the circle x<sup>2</sup> + y<sup>2</sup> = a, then 'a' is equal to ___________. | [] | null | 2 | tangent of y<sup>2</sup> = 8x is y = mx + $${2 \over m}$$<br><br>P(2, $$-$$4) $$\Rightarrow$$ $$-$$4 = 2m + $${2 \over m}$$<br><br>$$\Rightarrow$$ m + $${1 \over m}$$ = $$-$$2 $$\Rightarrow$$ m = $$-$$1<br><br>$$\therefore$$ tangent is y = $$-$$x $$-$$2<br><br>$$\Rightarrow$$ x + y + 2 = 0 ...... (1)<br><br>(1) is also tangent to x<sup>2</sup> + y<sup>2</sup> = a<br><br>So, $${2 \over {\sqrt 2 }} = \sqrt a \Rightarrow \sqrt a = \sqrt 2 $$<br><br>$$\Rightarrow$$ a = 2 | integer | jee-main-2021-online-31st-august-evening-shift |
1l58ah6pn | maths | parabola | common-tangent | <p>Let the common tangents to the curves $$4({x^2} + {y^2}) = 9$$ and $${y^2} = 4x$$ intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If e and l respectively denote the eccentricity and the length of the latus rectum of this ellipse, then $${l \over {{e^2}}}$$ is equal to ______________.</p> | [] | null | 4 | <p>Let y = mx + c is the common tangent</p>
<p>So $$c = {1 \over m} = \pm \,{3 \over 2}\sqrt {1 + {m^2}} \Rightarrow {m^2} = {1 \over 3}$$</p>
<p>So equation of common tangents will be $$y = \pm \,{1 \over {\sqrt 3 }}x \pm \,\sqrt 3 $$, which intersects at Q($$-$$3, 0)</p>
<p>Major axis and minor axis of ellipse are 12 and 6.</p>
<p>So eccentricity</p>
<p>$${e^2} = 1 - {1 \over 4} = {3 \over 4}$$ and length of latus rectum $$ = {{2{b^2}} \over a} = 3$$</p>
<p>Hence, $${l \over {{e^2}}} = {3 \over {3/4}} = 4$$</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l5aisjco | maths | parabola | common-tangent | <p>If $$y = {m_1}x + {c_1}$$ and $$y = {m_2}x + {c_2}$$, $${m_1} \ne {m_2}$$ are two common tangents of circle $${x^2} + {y^2} = 2$$ and parabola y<sup>2</sup> = x, then the value of $$8|{m_1}{m_2}|$$ is equal to :</p> | [{"identifier": "A", "content": "$$3 + 4\\sqrt 2 $$"}, {"identifier": "B", "content": "$$ - 5 + 6\\sqrt 2 $$"}, {"identifier": "C", "content": "$$ - 4 + 3\\sqrt 2 $$"}, {"identifier": "D", "content": "$$7 + 6\\sqrt 2 $$"}] | ["C"] | null | <p>Let tangent to $${y^2} = x$$ be</p>
<p>$$y = mx + {1 \over {4m}}$$</p>
<p>For it being tangent to circle.</p>
<p>$$\left| {{{{1 \over 4}m} \over {\sqrt {1 + {m^2}} }}} \right| = \sqrt 2 $$</p>
<p>$$ \Rightarrow 32{m^4} + 32{m^2} - 1 = 0$$</p>
<p>$$ \Rightarrow {m^2} = {{ - 32 \pm \sqrt {{{(32)}^2} + 4(32)} } \over {64}}$$</p>
<p>$$ \Rightarrow 8{m_1}{m_2} = - 4 + 3\sqrt 2 $$</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5c0w908 | maths | parabola | common-tangent | <p>Let x<sup>2</sup> + y<sup>2</sup> + Ax + By + C = 0 be a circle passing through (0, 6) and touching the parabola y = x<sup>2</sup> at (2, 4). Then A + C is equal to ___________.</p> | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "88/5"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "$$-$$8"}] | ["A"] | null | For tangent to parabola $y=x^{2}$ at $(2,4)$
<br/><br/>
$$
\left.\frac{d y}{d x}\right|_{(2,4)}=4
$$
<br/><br/>
Equation of tangent is
$$
y-4=4(x-2)
$$
<br/><br/>
$\Rightarrow 4 x-y-4=0$
<br/><br/>
Family of circle can be given by
<br/><br/>
$(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0$
<br/><br/>
As it passes through $(0,6)$
<br/><br/>
$2^{2}+2^{2}+\lambda(-10)=0$
<br/><br/>
$\Rightarrow \lambda=\frac{4}{5}$
<br/><br/>
Equation of circle is
<br/><br/>
$$
\begin{aligned}
&(x-2)^{2}+(y-4)^{2}+\frac{4}{5}(4 x-y-4)=0 \\\\
&\Rightarrow \left(x^{2}+y^{2}-4 x-8 y+20\right)+\left(\frac{16}{5} x-\frac{4}{5} y-\frac{16}{5}\right)=0 \\\\
&A=-4+\frac{16}{5}, C=20-\frac{16}{5}
\end{aligned}
$$
<br/><br/>
So, $A+C=16$ | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6hyskfa | maths | parabola | common-tangent | <p>The equation of a common tangent to the parabolas $$y=x^{2}$$ and $$y=-(x-2)^{2}$$ is</p> | [{"identifier": "A", "content": "$$y=4(x-2)$$"}, {"identifier": "B", "content": "$$y=4(x-1)$$"}, {"identifier": "C", "content": "$$y=4(x+1)$$"}, {"identifier": "D", "content": "$$y=4(x+2)$$"}] | ["B"] | null | <p>Equation of tangent of slope $$m$$ to $$y$$ $$= x^2$$</p>
<p>$$y = mx - {1 \over 4}{m^2}$$</p>
<p>Equation of tangent of slope $$m$$ to $$y = - {(x - 2)^2}$$</p>
<p>$$y = m(x - 2) + {1 \over 4}{m^2}$$</p>
<p>If both equation represent the same line</p>
<p>$${1 \over 4}{m^2} - 2m = - {1 \over 4}{m^2}$$</p>
<p>$$m = 0,\,4$$</p>
<p>So, equation of tangent</p>
<p>$$y = 4x - 4$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6np9p51 | maths | parabola | common-tangent | <p>Two tangent lines $$l_{1}$$ and $$l_{2}$$ are drawn from the point $$(2,0)$$ to the parabola $$2 \mathrm{y}^{2}=-x$$. If the lines $$l_{1}$$ and $$l_{2}$$ are also tangent to the circle $$(x-5)^{2}+y^{2}=r$$, then 17r is equal to ___________.</p> | [] | null | 9 | <p>Given : $${y^2} = {{ - x} \over 2}$$</p>
<p>$$\eqalign{
& T \equiv y = mx - {1 \over {8m}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow (2,0) \cr} $$</p>
<p>$$ \Rightarrow {m^2} = {1 \over {16}} \Rightarrow m = \, \pm \,{1 \over 4}$$</p>
<p>Tangents are $$y = {1 \over 4}x - {1 \over 2},\,y = {{ - x} \over 4} + {1 \over 2}$$</p>
<p>$$4y = x - 2$$ and $$4y + x = 2$$</p>
<p>If these are also tangent to circle then $${d_c} = r$$</p>
<p>$$ \Rightarrow \left| {{{5 - 2} \over {\sqrt {17} }}} \right| = \sqrt r \Rightarrow r = {\left( {{3 \over {\sqrt {17} }}} \right)^2}$$</p>
<p>$$ \Rightarrow 17r = 17\,.\,{9 \over {17}} = 9$$</p> | integer | jee-main-2022-online-28th-july-evening-shift |
ldqwkapa | maths | parabola | common-tangent | Let $A$ be a point on the $x$-axis. Common tangents are drawn from $A$ to the curves $x^2+y^2=8$ and $y^2=16 x$. If one of these tangents touches the two curves at $Q$ and $R$, then $(Q R)^2$ is equal to : | [{"identifier": "A", "content": "76"}, {"identifier": "B", "content": "81"}, {"identifier": "C", "content": "64"}, {"identifier": "D", "content": "72"}] | ["D"] | null | <p>Let a tangent on $${y^2} = 16x$$ be $$y = mx + {4 \over m}$$</p>
<p>For common to $${x^2} + {y^2} = 8$$</p>
<p>$${4 \over m} = 2\sqrt 2 (1 + {m^2})$$</p>
<p>$$ \Rightarrow {2 \over {{m^2}}} = 1 + {m^2} \Rightarrow m = \, \pm 1$$</p>
<p>Taking one of the tangent $$y = x + 4$$</p>
<p>Point of tangency with $${y^2} = 4x$$</p>
<p>$${x^2} + 8 + 16 = 4x \Rightarrow x = 4$$ & $$y = 8$$</p>
<p>$$\therefore$$ $$Q(4,8)$$</p>
<p>and for $${x^2} + {y^2} = 8$$</p>
<p>$$2{x^2} + 8x + 8 = 0$$</p>
<p>$${x^2} + 4x + 4 = 8 \Rightarrow x = - 2,y = 2 \Rightarrow R = ( - 2,2)$$</p>
<p>$${(QR)^2} = {6^2} + {6^2} = 72$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldv2j1e5 | maths | parabola | common-tangent | <p>The distance of the point $$(6,-2\sqrt2)$$ from the common tangent $$\mathrm{y=mx+c,m > 0}$$, of the curves $$x=2y^2$$ and $$x=1+y^2$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{3}$$"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$$\\frac{14}{3}$$"}, {"identifier": "D", "content": "5$$\\sqrt3$$"}] | ["B"] | null | $$
\begin{aligned}
& y^2=\frac{x}{2} \Rightarrow \text { tangent } y=m x+\frac{1}{8 m} \\\\
& y^2=x-1 \Rightarrow \text { tangent } y=m(x-1)+\frac{1}{4 m} \\\\
& \text { For common tangent } \frac{1}{8 m}=-m+\frac{1}{4 m} \\\\
& \Rightarrow 1=-8 m^2+2 \\\\
& \because m>0 \Rightarrow m=\frac{1}{2 \sqrt{2}} \\\\
& \Rightarrow \text { Common tangent is } y=\frac{x}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}} \\\\
& \Rightarrow x-2 \sqrt{2} y+1=0
\end{aligned}
$$<br/><br/>
Distance of point $(6,-2 \sqrt{2})$ from common tangent $=5$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1lgxwffpa | maths | parabola | common-tangent | <p>Let a common tangent to the curves $${y^2} = 4x$$ and $${(x - 4)^2} + {y^2} = 16$$ touch the curves at the points P and Q. Then $${(PQ)^2}$$ is equal to __________.</p> | [] | null | 32 | Tangent of slope $m$ to the parabola
<br/><br/>$y^2=4 x$ is given by $y=m x+\frac{1}{m}$ and <br/><br/>Tangent of slope $m$ to the circle $(x-4)^2+y^2=16$ is given by
<br/><br/>$$
y=m(x-4) \pm 4 \sqrt{1+m^2}
$$
<br/><br/>For common tangent
<br/><br/>$$
\begin{aligned}
& \frac{1}{m}=-4 m \pm 4 \sqrt{1+m^2} \\\\
& \Rightarrow \left(\frac{1}{m}+4 m\right)^2=\left( \pm 4 \sqrt{1+m^2}\right)^2
\end{aligned}
$$
<br/><br/>On squaring both sides, we get
<br/><br/>$$
\begin{aligned}
& =\frac{1}{m^2}+16 m^2+8=16+16 m^2 \\\\
& \Rightarrow \frac{1}{m^2} =8 \Rightarrow m= \pm \frac{1}{2 \sqrt{2}}
\end{aligned}
$$
<br/><br/>Then, the point of contact on parabola is $(8,4 \sqrt{2})$
<br/><br/>Length of tangent $P Q$ from $(8,4 \sqrt{2})$ on the circle is
<br/><br/>$$
\begin{array}{ll}
&\Rightarrow P Q=\sqrt{(8-4)^2+(4 \sqrt{2})^2-16} \\\\
&\Rightarrow P Q=\sqrt{16+32-16} \\\\
&\Rightarrow P Q=\sqrt{32} \Rightarrow(P Q)^2=32
\end{array}
$$ | integer | jee-main-2023-online-10th-april-morning-shift |
lsaq1mxh | maths | parabola | common-tangent | Let the line $\mathrm{L}: \sqrt{2} x+y=\alpha$ pass through the point of the intersection $\mathrm{P}$ (in the first quadrant) of the circle $x^2+y^2=3$ and the parabola $x^2=2 y$. Let the line $\mathrm{L}$ touch two circles $\mathrm{C}_1$ and $\mathrm{C}_2$ of equal radius $2 \sqrt{3}$. If the centres $Q_1$ and $Q_2$ of the circles $C_1$ and $C_2$ lie on the $y$-axis, then the square of the area of the triangle $\mathrm{PQ}_1 \mathrm{Q}_2$ is equal to ___________. | [] | null | 72 | <p>$x^2+y^2=3$ and $x^2=2 y$</p>
$y^2+2 y-3=0 $
<br/><br/>$\Rightarrow(y+3)(y-1)=0$
<br/><br/>$y=-3$ (Rejected) or $y=1$
<br/><br/>For $\mathrm{y}=1, \mathrm{x}=\sqrt{2} \Rightarrow P(\sqrt{2}, 1)$
<br/><br/>$p$ lies on the line
<br/><br/>$$
\begin{aligned}
& \sqrt{2} x+y=\alpha \\\\
& \sqrt{2}(\sqrt{2})+1=\alpha \\\\
& \alpha=3
\end{aligned}
$$
<br/><br/>For circle $\mathrm{C}_1$
<br/><br/>$\mathrm{Q}_1$ lies on $\mathrm{y}$ axis
<br/><br/>Let $\mathrm{Q}_1(0, \alpha)$ coordinates
<br/><br/>$\mathrm{R}_1=2 \sqrt{3}$ (Given
<br/><br/>Line $\mathrm{L}$ act as tangent
<br/><br/>Apply P $=r$ (condition of tangency)
<br/><br/>$\begin{aligned} & \Rightarrow\left|\frac{\alpha-3}{\sqrt{3}}\right|=2 \sqrt{3} \\\\ & \Rightarrow|\alpha-3|=6\end{aligned}$
<br/><br/>$$ \therefore $$ $\alpha-3=6$
<br/><br/>$\Rightarrow \alpha=9$
<br/><br/>$\begin{gathered}\text { or } \alpha-3=-6 \\\\ \Rightarrow \alpha=-3\end{gathered}$
<br/><br/>$\begin{aligned} & \triangle P Q_1 Q_2=\frac{1}{2}\left|\begin{array}{ccc}\sqrt{2} & 1 & 1 \\ 0 & 9 & 1 \\ 0 & -3 & 1\end{array}\right| \\\\ & =\frac{1}{2}(\sqrt{2}(12))=6 \sqrt{2} \\\\ & \left(\triangle P Q_1 Q_2\right)^2=72\end{aligned}$ | integer | jee-main-2024-online-1st-february-morning-shift |
lvc57b5a | maths | parabola | common-tangent | <p>Let $$C$$ be the circle of minimum area touching the parabola $$y=6-x^2$$ and the lines $$y=\sqrt{3}|x|$$. Then, which one of the following points lies on the circle $$C$$ ?</p> | [{"identifier": "A", "content": "$$(1,2)$$\n"}, {"identifier": "B", "content": "$$(2,2)$$\n"}, {"identifier": "C", "content": "$$(1,1)$$\n"}, {"identifier": "D", "content": "$$(2,4)$$"}] | ["D"] | null | <p>Let centre be (0, k)</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd1ocxr/d26061ca-43e1-4f4a-a0fc-77af60f91e91/491160f0-1599-11ef-acc0-2dadf5616f59/file-1lwd1ocxs.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd1ocxr/d26061ca-43e1-4f4a-a0fc-77af60f91e91/491160f0-1599-11ef-acc0-2dadf5616f59/file-1lwd1ocxs.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 3 English Explanation"></p>
<p>Now radius is $$r=6-k$$</p>
<p>Also, $$6-k=\left|\frac{k}{2}\right|$$</p>
<p>$$\begin{aligned}
& \Rightarrow 6-k=\frac{k}{2} \\
& \Rightarrow 12-2 k=k \\
& \Rightarrow k=4
\end{aligned}$$</p>
<p>Radius, $$r=6-4=2$$</p>
<p>So circle will be</p>
<p>$$\begin{aligned}
& (x)^2+(y-k)^2=4 \\
& x^2+(y-4)^2=4
\end{aligned}$$</p>
<p>$$(2,4)$$ satisfies this equation.</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
evpXasRCTfr92zs0 | maths | parabola | locus | Let $$P$$ be the point $$(1, 0)$$ and $$Q$$ a point on the parabola $${y^2} = 8x$$. The locus of mid point of $$PQ$$ is : | [{"identifier": "A", "content": "$${y^2} - 4x + 2 = 0$$ "}, {"identifier": "B", "content": "$${y^2} + 4x + 2 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 4y + 2 = 0$$"}, {"identifier": "D", "content": "$${x^2} - 4y + 2 = 0$$"}] | ["A"] | null | $$P = \left( {1,0} \right)\,\,Q = \left( {h,k} \right)$$ Such that $${k^2} = 8h$$
<br><br>Let $$\left( {\alpha ,\beta } \right)$$ be the midpoint of $$PQ$$
<br><br>$$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$$
<br><br>$$ \therefore $$ $$2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$$
<br><br>$${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$$
<br><br>$$ \Rightarrow {y^2} - 4x + 2 = 0.$$ | mcq | aieee-2005 |
QFprF6TCLSKFhBp8 | maths | parabola | locus | If two tangents drawn from a point $$P$$ to the parabola $${y^2} = 4x$$ are at right angles, then the locus of $$P$$ is | [{"identifier": "A", "content": "$$2x+1=0$$"}, {"identifier": "B", "content": "$$x=-1$$ "}, {"identifier": "C", "content": "$$2x-1=0$$ "}, {"identifier": "D", "content": "$$x=1$$ "}] | ["B"] | null | The locus of perpendicular tangents is directrix
<br><br>i.e., $$x=-1$$ | mcq | aieee-2010 |
2KV25TyV87muR2Xt | maths | parabola | locus | Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $${{x^2} = 8y}$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, then locus of $$P$$ is : | [{"identifier": "A", "content": "$${y^2} = 2x$$ "}, {"identifier": "B", "content": "$${{x^2} = 2y}$$ "}, {"identifier": "C", "content": "$${{x^2} = y}$$"}, {"identifier": "D", "content": "$${y^2} = x$$ "}] | ["B"] | null | <p>Let the coordinates of Q and P be (x<sub>1</sub>, y<sub>1</sub>) and (h, k) respectively.</p>
<p>$$\because$$ Q lies on x<sup>2</sup> = 8y,</p>
<p>$$\therefore$$ x$$_1^2$$ = 8y ....... (1)</p>
<p>Again, P divides OQ internally in the ratio 1 : 3.</p>
<p>$$\therefore$$ $$h = {{{x_1} + 0} \over 4} = {{{x_1}} \over 4}$$ or x<sub>1</sub> = 4h and</p>
<p>$$k = {{{y_1} + 0} \over 4} = {{{y_1}} \over 4}$$ or y<sub>1</sub> = 4k</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxw9mg20/83fce52e-2630-4574-a56c-ffe9c399a8da/aef7f480-6b3f-11ec-8608-9b519146e1b7/file-1kxw9mg21.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxw9mg20/83fce52e-2630-4574-a56c-ffe9c399a8da/aef7f480-6b3f-11ec-8608-9b519146e1b7/file-1kxw9mg21.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2015 (Offline) Mathematics - Parabola Question 110 English Explanation"> </p>
<p>Now putting x<sub>1</sub> and y<sub>1</sub> in (1) we get,</p>
<p>16h<sup>2</sup> = 32k or, h<sup>2</sup> = 2k</p>
<p>$$\therefore$$ the locus of P is given by, x<sup>2</sup> = 2y.</p> | mcq | jee-main-2015-offline |
8r6EX9Sd1qXSMJx8mk7k9k2k5gypkn5 | maths | parabola | locus | The locus of a point which divides the line
segment joining the point (0, –1) and a point on
the parabola, x<sup>2</sup> = 4y, internally in the ratio
1 : 2, is : | [{"identifier": "A", "content": "9x<sup>2</sup> \u2013 3y = 2"}, {"identifier": "B", "content": "4x<sup>2</sup> \u2013 3y = 2"}, {"identifier": "C", "content": "x<sup>2</sup> \u2013 3y = 2"}, {"identifier": "D", "content": "9x<sup>2</sup> \u2013 12y = 8"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266223/exam_images/w7aqioqqabs5dzmfr7a7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 8th January Morning Slot Mathematics - Parabola Question 82 English Explanation">
<br>Take point P(2t, t<sup>2</sup>
) on parabola x<sup>2</sup>
= 4y
<br><br>h = $${{2t + 0} \over 3}$$ and k = $${{{t^2} - 2} \over 3}$$
<br><br>$$ \Rightarrow $$ t = $${{3h} \over 2}$$ and 3k + 2 = t<sup>2</sup>
<br><br>$$ \therefore $$ 3k + 2 = $${{9{h^2}} \over 4}$$
<br><br>$$ \Rightarrow $$ 9x<sup>2</sup> – 12y = 8 | mcq | jee-main-2020-online-8th-january-morning-slot |
Zs5uWLenVyBwNLYaxY1klrhcogv | maths | parabola | locus | The locus of the mid-point of the line segment joining the focus of the parabola y<sup>2</sup> = 4ax to a
moving point of the parabola, is another parabola whose directrix is : | [{"identifier": "A", "content": "x = 0"}, {"identifier": "B", "content": "x = - $${a \\over 2}$$"}, {"identifier": "C", "content": "x = a"}, {"identifier": "D", "content": "x = $${a \\over 2}$$"}] | ["A"] | null | Given, equation of parabola $$\Rightarrow$$ y<sup>2</sup> = 4ax<br><br>Focus = S(a, 0)<br><br>Let any point on the parabola be P(at<sup>2</sup>, 2at).<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxof1w7l/840d5c1f-ce31-4701-93d3-208583e5d683/a5091c10-66ee-11ec-9866-0df874a38238/file-1kxof1w7m.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxof1w7l/840d5c1f-ce31-4701-93d3-208583e5d683/a5091c10-66ee-11ec-9866-0df874a38238/file-1kxof1w7m.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2021 (Online) 24th February Morning Shift Mathematics - Parabola Question 72 English Explanation"><br>and let the mid-point of PS be M(h, k).<br><br>$$\therefore$$ $$h{{a{t^2} + a} \over 2};k = {{2at + 0} \over 2}$$<br><br>$$ \Rightarrow {t^2} = {{2h - a} \over a};t = {k \over a}$$<br><br>$$ \Rightarrow {t^2} = {{{k^2}} \over {{a^2}}}$$<br><br>Now, $${{2h - a} \over a} = {{{k^2}} \over {{a^2}}}$$<br><br>$$ \Rightarrow 2h - a = {{{k^2}} \over a} \Rightarrow {k^2} = a(2h - a)$$<br><br>$$\therefore$$ Locus of (h, k) is $${y^2} = a(2x - a)$$<br><br>$${y^2} = 2a\left( {x - {a \over 2}} \right)$$<br><br>$$\therefore$$ The directrix of this parabola is <br><br>$$x - {a \over 2} = - {a \over 2} \Rightarrow x = 0$$ | mcq | jee-main-2021-online-24th-february-morning-slot |
1krrulg3k | maths | parabola | locus | Let P be a variable point on the parabola $$y = 4{x^2} + 1$$. Then, the locus of the mid-point of the point P and the foot of the perpendicular drawn from the point P to the line y = x is : | [{"identifier": "A", "content": "$${(3x - y)^2} + (x - 3y) + 2 = 0$$"}, {"identifier": "B", "content": "$$2{(3x - y)^2} + (x - 3y) + 2 = 0$$"}, {"identifier": "C", "content": "$${(3x - y)^2} + 2(x - 3y) + 2 = 0$$"}, {"identifier": "D", "content": "$$2{(x - 3y)^2} + (3x - y) + 2 = 0$$"}] | ["B"] | null | Given, parabola $$y = 4{x^2} + 1$$<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l09usogd/0e8d490d-75bf-4d50-a3ef-e9477504eeb8/1b3947d0-9a51-11ec-b1c5-39b9b722e9af/file-1l09usoge.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l09usogd/0e8d490d-75bf-4d50-a3ef-e9477504eeb8/1b3947d0-9a51-11ec-b1c5-39b9b722e9af/file-1l09usoge.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 20th July Evening Shift Mathematics - Parabola Question 62 English Explanation"><br>Let R(a, b) be mid-point of line joining point P and Q where PQ is perpendicular to line y = x.<br><br>Let coordinates of P be P(x, y), Q(q, q) and R(a, b) then,<br><br>$$a = {{x + q} \over 2}$$ and $$b = {{y + q} \over 2}$$<br><br>Now, slope of line y = x is m<sub>1</sub> = 1<br><br>Slope of line PQ be<br><br>$${{b - q} \over {a - q}} = {m_2}$$ (say)<br><br>$$\because$$ Line y = x and PQ are perpendicular to each other,<br><br>m<sub>1</sub> . m<sub>2</sub> = $$-$$1<br><br>$$ \Rightarrow {{b - q} \over {a - q}} = - 1 \Rightarrow b - q = q - a$$<br><br>$$ \Rightarrow q = {{b + a} \over 2}$$<br><br>$$\therefore$$ $$a = {{x + q} \over 2} = {{x + \left( {{{b + a} \over 2}} \right)} \over 2} = {{2x + b + a} \over 4}$$<br><br>$$ \Rightarrow x = {{4a - b - a} \over 2} = {{3a - b} \over 2}$$<br><br>and $$b = {{y + q} \over 2} = {{y + \left( {{{b + a} \over 2}} \right)} \over 2} = {{2y + b + a} \over 4}$$<br><br>$$ \Rightarrow y = {{3b - a} \over 2}$$<br><br>Put (x, y) in equation of parabola as P(x, y) is variable point on parabola<br><br>$${{3b - a} \over 2} = 4{\left( {{{3a - b} \over 2}} \right)^2} + 1$$<br><br>$${{(3b - a)} \over 2} = {(3a - b)^2} + 1$$<br><br>$$ \Rightarrow (3b - a) = 2{(3a - b)^2} + 2$$<br><br>Replace (a, b) as (x, y) $$\Rightarrow$$ (3y $$-$$ x) = 2(3x $$-$$ y)<sup>2</sup> + 2<br><br>or $$2{(3x - y)^2} + (x - 3y) + 2 = 0$$ | mcq | jee-main-2021-online-20th-july-evening-shift |
1ktfzhxju | maths | parabola | locus | If two tangents drawn from a point P to the <br/>parabola y<sup>2</sup> = 16(x $$-$$ 3) are at right angles, then the locus of point P is : | [{"identifier": "A", "content": "x + 3 = 0"}, {"identifier": "B", "content": "x + 1 = 0"}, {"identifier": "C", "content": "x + 2 = 0"}, {"identifier": "D", "content": "x + 4 = 0"}] | ["B"] | null | Locus is directrix of parabola<br><br>x $$-$$ 3 + 4 = 0 $$\Rightarrow$$ x + 1 = 0. | mcq | jee-main-2021-online-27th-august-evening-shift |
ldqwyq2x | maths | parabola | locus | The parabolas : $a x^2+2 b x+c y=0$ and $d x^2+2 e x+f y=0$ intersect on the line $y=1$. If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in G.P., then : | [{"identifier": "A", "content": "$\\frac{d}{a}, \\frac{e}{b}, \\frac{f}{c}$ are in A.P."}, {"identifier": "B", "content": "$\\frac{d}{a}, \\frac{e}{b}, \\frac{f}{c}$ are in G.P."}, {"identifier": "C", "content": "$d, e, f$ are in A.P."}, {"identifier": "D", "content": "$d, e, f$ are in G.P."}] | ["A"] | null | <p>Let point of intersection be ($$\alpha,1$$)</p>
<p>$$\alpha x^2+2b\alpha+c=0$$ ..... (i)</p>
<p>and $$d\alpha^2+2e\alpha+f=0$$ .... (ii)</p>
<p>$$\Rightarrow a\alpha^2+2\sqrt{ac}\alpha+c=0$$ ($$\because$$ $$b^2=ac$$)</p>
<p>$${\left( {\sqrt a \alpha + \sqrt c } \right)^2} = 0$$</p>
<p>$$\alpha = - \sqrt {{c \over a}} $$</p>
<p>Put the value of $$\alpha$$ in (ii),</p>
<p>$$d{c \over a} - 2e\sqrt {{c \over a}} + f = 0$$</p>
<p>$${d \over a} - {{2e} \over {\sqrt {ac} }} + {f \over c} = 0$$</p>
<p>$${d \over a} + {f \over c} = 2{e \over b}$$</p>
<p>$${d \over a},{e \over b},{f \over c}$$ are in A.P.</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldu41vqc | maths | parabola | locus | <p>The equations of two sides of a variable triangle are $$x=0$$ and $$y=3$$, and its third side is a tangent to the parabola $$y^2=6x$$. The locus of its circumcentre is :</p> | [{"identifier": "A", "content": "$$4{y^2} - 18y - 3x - 18 = 0$$"}, {"identifier": "B", "content": "$$4{y^2} + 18y + 3x + 18 = 0$$"}, {"identifier": "C", "content": "$$4{y^2} - 18y + 3x + 18 = 0$$"}, {"identifier": "D", "content": "$$4{y^2} - 18y - 3x + 18 = 0$$"}] | ["C"] | null | Third side of triangle
<br><br>
$t y=x+\frac{3}{2} t^{2}$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef4vkb3/455513c9-54ce-4923-8412-8892bb374d49/794a91f0-b263-11ed-a6d1-894f1caef997/file-1lef4vkb4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lef4vkb3/455513c9-54ce-4923-8412-8892bb374d49/794a91f0-b263-11ed-a6d1-894f1caef997/file-1lef4vkb4.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 - Parabola Question 23 English Explanation 1"><br>
$$
\begin{aligned}
& \therefore H \equiv(0,3) \quad G \equiv\left(t-\frac{3 t}{2}\right)
\end{aligned}
$$<br><br>
Let O (h, k)<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lef4wwy4/94c63b59-5b17-404e-a886-f1440050aed4/9edd19b0-b263-11ed-a126-5dfa1a9d5fb8/file-1lef4wwy5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lef4wwy4/94c63b59-5b17-404e-a886-f1440050aed4/9edd19b0-b263-11ed-a126-5dfa1a9d5fb8/file-1lef4wwy5.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 - Parabola Question 23 English Explanation 2"><br>
$$
\begin{aligned}
& \Rightarrow \frac{2 h}{3}=t-\frac{t^{2}}{2} \& \frac{2 k+3}{3}=2+\frac{t}{2} \\\\
& \Rightarrow 4 h=6 t-3 t^{2} \& 4 k=6+3 t \\\\
& \Rightarrow 4 h=2(4 k-6)-3\left(\frac{(4 k-6)^{2}}{9}\right) \\\\
& \Rightarrow 4 h=6 k-9-\left(4 k^{2}+9-12 k\right) \\\\
& \Rightarrow 4 k^{2}-18 k+3 h+18=0 \\\\
& \Rightarrow 4 y^{2}-18 y+3 x+18=0
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-evening-shift |
dn3PoBQdOLPq820v | maths | parabola | normal-to-parabola | The normal at the point$$\left( {bt_1^2,2b{t_1}} \right)$$ on a parabola meets the parabola again in the point $$\left( {bt_2^2,2b{t_2}} \right)$$, then : | [{"identifier": "A", "content": "$${t_2} = {t_1} + {2 \\over {{t_1}}}$$ "}, {"identifier": "B", "content": "$${t_2} = -{t_1} - {2 \\over {{t_1}}}$$"}, {"identifier": "C", "content": "$${t_2} = -{t_1} + {2 \\over {{t_1}}}$$"}, {"identifier": "D", "content": "$${t_2} = {t_1} - {2 \\over {{t_1}}}$$"}] | ["B"] | null | Equation of the normal to a parabola $${y^2} = 4bx$$ at point
<br><br>$$\left( {bt_1^2,2b{t_1}} \right)$$ is $$y = - {t_1}x + 2b{t_1} + bt_1^3$$
<br><br>As given, it also passes through $$\left( {bt_2^2,2b{t_2}} \right)$$ then
<br><br>$$2b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3$$
<br><br>$$2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)$$
<br><br>$$ = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)$$
<br><br>$$ \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)$$
<br><br>$$ \Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}$$
<br><br>$$ \Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}$$ | mcq | aieee-2003 |
rF6OFC4tBXIjjAbO | maths | parabola | normal-to-parabola | Let $$P$$ be the point on the parabola, $${{y^2} = 8x}$$ which is at a minimum distance from the centre $$C$$ of the circle, $${x^2} + {\left( {y + 6} \right)^2} = 1$$. Then the equation of the circle, passing through $$C$$ and having its centre at $$P$$ is: | [{"identifier": "A", "content": "$${{x^2} + {y^2} - {x \\over 4} + 2y - 24 = 0}$$ "}, {"identifier": "B", "content": "$${{x^2} + {y^2} - 4x + 9y + 18 = 0}$$ "}, {"identifier": "C", "content": "$${{x^2} + {y^2} - 4x + 8y + 12 = 0}$$"}, {"identifier": "D", "content": "$${{x^2} + {y^2} - x + 4y - 12 = 0}$$"}] | ["C"] | null | Minimum distance $$ \Rightarrow $$ perpendicular distance
<br><br>$$E{q^n}$$ of normal at $$p\left( {2{t^2},\,4t} \right)$$
<br><br>$$y = - tx + 4t + 2{t^3}$$
<br><br>It passes through $$C\left( {0, - 6} \right) \Rightarrow {t^3} + 2t + 3 = 0$$
<br><br>$$ \Rightarrow t = - 1$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264958/exam_images/exuu3huqxlmuv2m9xipb.webp" loading="lazy" alt="JEE Main 2016 (Offline) Mathematics - Parabola Question 109 English Explanation">
<br><br>Center of new circle $$ = P\left( {2t{}^2,4t} \right) = P\left( {2, - 4} \right)$$
<br><br>Radius $$ = PC = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} = 2\sqrt 2 $$
<br><br>$$\therefore$$ Equation of the circle is
<br><br>$${\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2}$$
<br><br>$$ \Rightarrow {x^2} + y{}^2 - 4x + 8y + 12 = 0$$ | mcq | jee-main-2016-offline |
IDKgSlFmiu6IKjtRmTg5G | maths | parabola | normal-to-parabola | P and Q are two distinct points on the parabola, y<sup>2</sup> = 4x, with parameters t and t<sub>1</sub> respectively. If the normal at P passes through Q, then the minimum value of $$t_1^2$$ is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["D"] | null | t<sub>1</sub> = $$-$$ t $$-$$ $${2 \over t}$$
<br><br>$$t_1^2$$ = t<sup>2</sup> + $${4 \over {{t^2}}}$$ + 4
<br><br>t<sup>2</sup> + $${4 \over {{t^2}}}$$ $$ \ge $$ 2$$\sqrt {{t^2}.{4 \over {{t^2}}}} = 4$$
<br><br>Minimum value of $$t_1^2$$ = 8 | mcq | jee-main-2016-online-10th-april-morning-slot |
aYPG8LoQabTT8eGQ2ua7k | maths | parabola | normal-to-parabola | If y = mx + c is the normal at a point on the parabola y<sup>2</sup> = 8x whose focal distance is 8 units, then $$\left| c \right|$$ is equal to : | [{"identifier": "A", "content": "$$2\\sqrt 3 $$ "}, {"identifier": "B", "content": "$$8\\sqrt 3 $$"}, {"identifier": "C", "content": "$$10\\sqrt 3 $$"}, {"identifier": "D", "content": "$$16\\sqrt 3 $$"}] | ["C"] | null | c = $$-$$ 29m $$-$$ 9m<sup>3</sup>
<br><br>a = 2
<br><br>Given (at<sup>2</sup> $$-$$ a)<sup>2</sup> + 4a<sup>2</sup>t<sup>2</sup> = 64
<br><br>$$ \Rightarrow $$ (a(t<sup>2</sup> + 1)) = 8
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + 1 = 4 $$ \Rightarrow $$ t<sup>2</sup> = 3
<br><br>$$ \Rightarrow $$ t = $$\sqrt 3 $$
<br><br>$$ \therefore $$ c = 2at(2 + t<sup>2</sup>)
<br><br>= $$2\sqrt 3 \left( 5 \right)$$
<br><br>$$\left| c \right|$$ = 10$$\sqrt 3 $$ | mcq | jee-main-2017-online-9th-april-morning-slot |
YjXpdEyhqPAl1IA4 | maths | parabola | normal-to-parabola | Tangent and normal are drawn at P(16, 16) on the parabola y<sup>2</sup> = 16x, which intersect the axis of the
parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and $$\angle $$CPB =
$$\theta $$, then a value of tan$$\theta $$ is : | [{"identifier": "A", "content": "$${4 \\over 3}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["C"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263693/exam_images/be9bgy7e2oqgaj3q4lr8.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Parabola Question 108 English Explanation 1">
<br><br>As equation of tangent PA at (x<sub>1</sub>, y<sub>1</sub>) on the parabola y<sup>2</sup> = 4ax,
<br><br>yy<sub>1</sub> = 2a (x + x<sub>1</sub>)
<br><br>here (x<sub>1</sub>, y<sub>1</sub>) = ( 16, 16)
<br><br>y . 16 = 2.4 (x + 16)
<br><br>$$ \Rightarrow $$ 2y = x + 16 .....(1)
<br><br>At pont A value of y = 0
<br><br>putting y = 0 in equation (1) we get,
<br><br>0 = x + 16
<br><br>$$ \Rightarrow $$ x = $$-$$ 16
<br><br>$$\therefore\,\,\,$$ Coordinate of point A = ($$-$$ 16, 0)
<br><br>Slope of line P A :
<br><br>As$$\,\,\,\,$$ 2y = x + 16
<br><br>$$\therefore\,\,\,$$ y = $${1 \over 2}\,$$ x + 8
<br><br>$$\therefore\,\,\,$$ Slope (m) = $${1 \over 2}\,$$
<br><br>Let slope of perpendicular line PB passing through point p(16, 16) = m'
<br><br>$$\therefore\,\,\,$$ m m' = $$-$$ 1
<br><br>$$ \Rightarrow $$ $${1 \over 2}$$ $$ \times $$ m' = $$-$$ 1
<br><br>$$ \Rightarrow $$ ' = $$-$$ 2
<br><br>As Equation of normal PB, when slope is m,
<br><br>y = mx $$-$$ 2am $$-$$am<sup>3</sup>
<br><br>Here m = m' = $$-$$ 2 and a = 4
<br><br>$$\therefore\,\,\,$$ y = $$-$$2x $$-$$ 2(4) ($$-$$2) $$-$$ 4 . ($$-$$2)<sup>3</sup>
<br><br>$$ \Rightarrow $$ y = $$-$$ 2x + 16 + 32
<br><br>$$ \Rightarrow $$ y = $$-$$ 2x + 48 ..... (2)
<br><br>At point B, y = 0
<br><br>puttig y = 0 at equation (2) we get,
<br><br>0 = $$-$$ 2x + 48
<br><br>$$ \Rightarrow $$x = 24
<br><br>$$\therefore\,\,\,$$ Coordinate of point B = (24, 0)
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263531/exam_images/y0zkxse1bq2gklpforj2.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Parabola Question 108 English Explanation 2">
<br><br>A circle is passing through point P, A and B, and C is the center of the circle.
<br><br>So, AC and BC are the radius.
<br><br>Then AC = BC, So C is the middle point of line AB.
<br><br>$$\therefore\,\,\,$$ C = $$\left( {{{24 - 16} \over 2},{{0 + 0} \over 2}} \right)$$ = (4, 0)
<br><br>$$\angle $$CPB = $$\theta $$ and we have to find tan $$\theta $$.
<br><br>Slope of line PC = $${{16 - 0} \over {16 - 4}}$$ = $${4 \over 3}$$ =m<sub>1</sub>
<br><br>and we know slope of line PB = $$-$$2 = m<sub>2</sub>
<br><br>$$\therefore\,\,\,$$ tan $$\theta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = $$\left| {{{{4 \over 3} + 2} \over {1 - \left( {{4 \over 3}} \right)\left( 2 \right)}}} \right|$$
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = $$\left| {{{{{10} \over 3}} \over { - {5 \over 3}}}} \right|$$
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = $$\left| { - 2} \right|$$
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = 2
| mcq | jee-main-2018-offline |
xHGKs3J3uPk1BI64yCR9N | maths | parabola | normal-to-parabola | If the parabolas y<sup>2</sup> = 4b(x – c) and y<sup>2</sup> = 8ax have a common normal, then which on of the following is a valid choice for the ordered triad (a, b, c)? | [{"identifier": "A", "content": "(1, 1, 3)"}, {"identifier": "B", "content": "(1, 1, 0)"}, {"identifier": "C", "content": "$$\\left( {{1 \\over 2},2,0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},2,3} \\right)$$"}] | ["A"] | null | Normal to the two given curves are
<br><br>y = m(x – c) – 2bm – bm<sup>3</sup>,
<br><br>y = mx – 4am – 2am<sup>3</sup>
<br><br>If they have a common normal, then
<br><br>(c + 2b)m + bm<sup>3</sup>
= 4am + 2am<sup>3</sup>
<br><br>$$ \Rightarrow $$ (4a – c – 2b) m = (b – 2a)m<sup>3</sup>
<br><br>$$ \Rightarrow $$ (4a – c – 2b) = (b – 2a)m<sup>2</sup>
<br><br>$$ \Rightarrow $$ m<sup>2</sup> = $${c \over {2a - b}} - 2$$ $$>$$ 0
<br><br>By checking all options we found (A) is right option.
<br><br><b>Note :</b> We get that all the options are correct for m = 0
i.e., when common normal is x-axis. But may be in question they want common
normal other than x – axis, hence answer is (A). | mcq | jee-main-2019-online-10th-january-morning-slot |
R82TmCwmYj5NHZ0Vju18hoxe66ijvwvfhji | maths | parabola | normal-to-parabola | The area (in sq. units) of the smaller of the two
circles that touch the parabola, y<sup>2 </sup> = 4x at the point
(1, 2) and the x-axis is :- | [{"identifier": "A", "content": "$$4\\pi \\left( {3 +\\sqrt 2 } \\right)$$"}, {"identifier": "B", "content": "$$8\\pi \\left( {2 - \\sqrt 2 } \\right)$$"}, {"identifier": "C", "content": "$$8\\pi \\left( {3 - 2\\sqrt 2 } \\right)$$"}, {"identifier": "D", "content": "$$4\\pi \\left( {2 - \\sqrt 2 } \\right)$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264518/exam_images/csbm6jz6ovhaimejwhpu.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Parabola Question 88 English Explanation">
<br><br>Equation of tangent to the
parabola y<sup>2</sup> = 4x at P(1, 2),
<br><br>T = 0
<br><br>2y = $$4\left( {{{x + 1} \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ y = x + 1
<br><br>Equation of normal of the tangent at point P(1, 2)
<br><br>y - 2 = (-1)(x - 1)
<br><br>$$ \Rightarrow $$ y - 2 = - x + 1
<br><br>$$ \Rightarrow $$ x + y - 3 = 0
<br><br>This normal also passes through the center (h, r) of the circle.
<br><br>$$ \therefore $$ h + k - 3 = 0
<br><br>$$ \Rightarrow $$ h = 3 - r
<br><br>So center is (3 - r, r)
<br><br>From picture you can see,
<br><br>PC = r
<br><br>$$ \Rightarrow $$ (PC)<sup>2</sup> = r<sup>2</sup>
<br><br>$$ \Rightarrow $$ (3 - r - 1)<sup>2</sup> + (r - 2)<sup>2</sup> = r<sup>2</sup>
<br><br>$$ \Rightarrow $$ 4 + r<sup>2</sup> - 4r + r<sup>2</sup> + 4 - 4r = r<sup>2</sup>
<br><br>$$ \Rightarrow $$ r<sup>2</sup> - 8r + 8 = 0
<br><br>$$ \therefore $$ r = $${{8 \pm \sqrt {64 - 32} } \over 2}$$
<br><br>$$ \Rightarrow $$ r = $${{8 \pm \sqrt {32} } \over 2}$$
<br><br>$$ \Rightarrow $$ r = $${{8 \pm 4\sqrt 2 } \over 2}$$
<br><br>$$ \therefore $$ r = 4 + $${2\sqrt 2 }$$ and 4 - $${2\sqrt 2 }$$
<br><br>If r = 4 + $${2\sqrt 2 }$$ then center of the circle is
<br><br>(-1 - $${2\sqrt 2 }$$, 4 + $${2\sqrt 2 }$$). From the diagram you can see both the x coordinate and y coordinate of the circle should be positive but here x coordinate is negative.
<br><br>So possible value of radius r = 4 - $${2\sqrt 2 }$$
<br><br>Then area of the circle
<br><br>= $$\pi $$r<sup>2</sup>
<br><br>= $$\pi $$(4 - $${2\sqrt 2 }$$)<sup>2</sup>
<br><br>= $$\pi $$(16 + 8 - $${16\sqrt 2 }$$)
<br><br>= $$8\pi \left( {3 - 2\sqrt 2 } \right)$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
qrVANARoXQ1CZObJIUjgy2xukf49t0l9 | maths | parabola | normal-to-parabola | If the tangent to the curve, y = e<sup>x</sup>
at a point
(c, e<sup>c</sup>) and the normal to the parabola, y<sup>2</sup> = 4x
at the point (1, 2) intersect at the same point on
the x-axis, then the value of c is ________ .
| [] | null | 4 | For $$y = {e^x}$$<br><br>$${{dy} \over {dx}} = {e^x}$$<br><br>$${\left. {{{dy} \over {dx}}} \right|_{x = c}} = {e^c}$$<br><br>Tangent is $$y - {e^c} = {e^c}(x - c)$$<br><br>Put y = 0, x = c$$ - $$1.........(i)<br><br>For y<sup>2</sup> = 4x<br><br>$$2y{{dy} \over {dx}} = 4 \Rightarrow {\left. {{{ - dx} \over {dy}}} \right|_{y = 2}} = - 1$$<br><br>Normal is $$y - 2 = - 1(x - 1)$$<br><br>Put y = 0, x = 3 ...........(ii)<br><br>From (i) and (ii); $$c - 1$$ = 3<br><br>$$ \Rightarrow $$ c = 4 | integer | jee-main-2020-online-3rd-september-evening-slot |
52WEiaRzrURiv0g4XY1kmhw1mcp | maths | parabola | normal-to-parabola | If the three normals drawn to the parabola, y<sup>2</sup> = 2x pass through the point (a, 0) a $$\ne$$ 0, then 'a' must be greater than : | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$$-$$1"}, {"identifier": "D", "content": "$$-$$$${1 \\over 2}$$"}] | ["B"] | null | Let the equation of the normal is <br><br>y = mx $$-$$ 2am $$-$$ am<sup>3</sup><br><br>here 4a = 2 $$ \Rightarrow $$ a = $${1 \over 2}$$<br><br>y = mx $$-$$ m $$-$$ $${1 \over 2}$$m<sup>3</sup><br><br>It passing through A(a, 0) then<br><br>0 = am $$-$$ m $$-$$ $${1 \over 2}$$m<sup>3</sup><br><br>m = 0, a $$-$$ 1 $$-$$ $${1 \over 2}$$m<sup>2</sup> = 0<br><br>m<sup>2</sup> = 2(a $$-$$ 1) > 0<br><br>$$ \Rightarrow $$ a > 1 | mcq | jee-main-2021-online-16th-march-morning-shift |
1krpzucdd | maths | parabola | normal-to-parabola | Let the tangent to the parabola S : y<sup>2</sup> = 2x at the point P(2, 2) meet the x-axis at Q and normal at it meet the parabola S at the point R. Then the area (in sq. units) of the triangle PQR is equal to : | [{"identifier": "A", "content": "$${{25} \\over 2}$$"}, {"identifier": "B", "content": "$${{35} \\over 2}$$"}, {"identifier": "C", "content": "$${{15} \\over 2}$$"}, {"identifier": "D", "content": "25"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266392/exam_images/kfqvzltdhtj2yuoqxjth.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 20th July Morning Shift Mathematics - Parabola Question 63 English Explanation"><br><br>Tangent at P : y(2) = 2(1/2) (x + 2)<br><br>$$\Rightarrow$$ 2y = x + 2<br><br>$$\therefore$$ Q = ($$-$$2, 0)<br><br>Normal at P : y $$-$$ 2 = $$-$$ $${{(2)} \over {2.{1 \over 2}}}(x - 2)$$<br><br>$$\Rightarrow$$ y $$-$$ 2 = $$-$$ 2(x $$-$$ 2)<br><br>$$\Rightarrow$$ y = 6 $$-$$ 2x<br><br>$$\therefore$$ Solving with y<sup>2</sup> = 2x $$\Rightarrow$$ R$$\left( {{9 \over 2} - 3} \right)$$<br><br>$$\therefore$$ Ar ($$\Delta$$PQR) = $${1 \over 2}\left| {\matrix{
2 & 2 & 1 \cr
{ - 2} & 1 & 1 \cr
{{9 \over 2}} & 3 & { - 1} \cr
} } \right|$$<br><br>$$ = {{25} \over 2}$$ sq. units. | mcq | jee-main-2021-online-20th-july-morning-shift |
1ktely03s | maths | parabola | normal-to-parabola | A tangent and a normal are drawn at the point P(2, $$-$$4) on the parabola y<sup>2</sup> = 8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to : | [{"identifier": "A", "content": "$$-$$16"}, {"identifier": "B", "content": "$$-$$18"}, {"identifier": "C", "content": "$$-$$12"}, {"identifier": "D", "content": "$$-$$20"}] | ["A"] | null | <p>Given, parabola</p>
<p>$${y^2} = 8x$$ ...... (i)</p>
<p>Equation of tangent at $$P(2, - 4)$$ is</p>
<p>$$ - 4y = 4(x + 2)$$</p>
<p>or, $$x + y + 2 = 0$$ ..... (ii)</p>
<p>and Equation of normal to the parabola is</p>
<p>$$x - y + C = 0$$</p>
<p>$$\therefore$$ Normal passes through $$(2, - 4)$$</p>
<p>$$\therefore$$ $$C = - 6$$</p>
<p>Normal : $$x - y = 6$$ ..... (iii)</p>
<p>Equation of directrix of parabola</p>
<p>$$x = - 2$$ ..... (iv)</p>
<p>Point of intersection of tangent and normal with directrix are $$x = - 2$$ at $$A( - 2,0)$$ and $$B( - 2, - 8)$$ respectively.</p>
<p>$$Q(a,b)$$ and $$P(2, - 4)$$ are given and AQBP is a square.</p>
<p>Mid-point of AB = Mid-point of PQ</p>
<p>$$ \Rightarrow ( - 2, - 4) = \left( {{{a + 2} \over 2},{{b - 4} \over 2}} \right) \Rightarrow a = - 6,b = - 4$$</p>
<p>$$ \Rightarrow 2a + b = - 16$$</p> | mcq | jee-main-2021-online-27th-august-morning-shift |
1kto6jqrb | maths | parabola | normal-to-parabola | Consider the parabola with vertex $$\left( {{1 \over 2},{3 \over 4}} \right)$$ and the directrix $$y = {1 \over 2}$$. Let P be the point where the parabola meets the line $$x = - {1 \over 2}$$. If the normal to the parabola at P intersects the parabola again at the point Q, then (PQ)<sup>2</sup> is equal to : | [{"identifier": "A", "content": "$${{75} \\over 8}$$"}, {"identifier": "B", "content": "$${{125} \\over {16}}$$"}, {"identifier": "C", "content": "$${{25} \\over 2}$$"}, {"identifier": "D", "content": "$${{15} \\over 2}$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwopy91l/b12d9de1-4cc2-4e41-a53c-978e58e1dcd9/0cc11390-534d-11ec-9cbb-695a838b20fb/file-1kwopy91m.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwopy91l/b12d9de1-4cc2-4e41-a53c-978e58e1dcd9/0cc11390-534d-11ec-9cbb-695a838b20fb/file-1kwopy91m.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 50vh" alt="JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Parabola Question 56 English Explanation"><br>$$\left( {y - {3 \over 4}} \right) = {\left( {x - {1 \over 2}} \right)^2}$$ .... (1)<br><br>For $$x = - {1 \over 2}$$<br><br>$$y - {3 \over 4} = 1 \Rightarrow y = {7 \over 4} \Rightarrow P\left( { - {1 \over 2},{7 \over 4}} \right)$$<br><br>Now, $$y' = 2\left( {x - {1 \over 2}} \right)$$<br><br>At $$x = - {1 \over 2}$$<br><br>$$ \Rightarrow {m_T} = - 2,{m_N} = {1 \over 2}$$<br><br>Equation of normal is <br><br>$$y - {7 \over 4} = {1 \over 2}\left( {x + {1 \over 2}} \right)$$<br><br>$$y = {x \over 2} + 2$$<br><br>Now put y in equation (1)<br><br>$${x \over 2} + 2 - {3 \over 4} = {\left( {x - {1 \over 2}} \right)^2}$$<br><br>$$ \Rightarrow x = 2$$ & $$ - {1 \over 2}$$<br><br>$$\Rightarrow$$ Q(2, 3)<br><br>Now, $${(PQ)^2} = {{125} \over {16}}$$<br><br>Option (b) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l589slgl | maths | parabola | normal-to-parabola | <p>Let the normal at the point on the parabola y<sup>2</sup> = 6x pass through the point (5, $$-$$8). If the tangent at P to the parabola intersects its directrix at the point Q, then the ordinate of the point Q is :</p> | [{"identifier": "A", "content": "$$-$$3"}, {"identifier": "B", "content": "$$-$$$${{9} \\over 4}$$"}, {"identifier": "C", "content": "$$-$$$${{5} \\over 2}$$"}, {"identifier": "D", "content": "$$-$$2"}] | ["B"] | null | <p>Let P(at<sup>2</sup>, 2at) where a = $${3 \over 2}$$</p>
<p>T : yt = x + at<sup>2</sup> So point Q is $$\left( { - a,\,at - {a \over t}} \right)$$</p>
<p>N : y = $$-$$tx + 2at + at<sup>3</sup> passes through (5, $$-$$8)</p>
<p>$$-$$8 = $$-$$5t + 3t + $${3 \over 2}$$t<sup>3</sup></p>
<p>$$\Rightarrow$$ 3t<sup>3</sup> $$-$$ 4t + 16 = 0</p>
<p>$$\Rightarrow$$ (t + 2) (3t<sup>2</sup> $$-$$ 6t + 8) = 0</p>
<p>$$\Rightarrow$$ t = 2</p>
<p>So ordinate of point Q is $$-$$$${9 \over 4}$$.</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1ldr6xehk | maths | parabola | normal-to-parabola | <p>If $$\mathrm{P}(\mathrm{h}, \mathrm{k})$$ be a point on the parabola $$x=4 y^{2}$$, which is nearest to the point $$\mathrm{Q}(0,33)$$, then the distance of $$\mathrm{P}$$ from the directrix of the parabola $$\quad y^{2}=4(x+y)$$ is equal to :</p> | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "4"}] | ["C"] | null | <p>Equation of normal</p>
<p>$$y = - tx + 2.{1 \over {16}}t + {1 \over {16}}{t^3}$$</p>
<p>$$33 = {t \over 8} + {{{t^3}} \over {16}}$$</p>
<p>$$ \Rightarrow {t^3} + 2t = 528$$</p>
<p>$$t = 8$$</p>
<p>$$(a{t^2},2at) = (4,1)$$</p>
<p>Distance from $$x = - 2$$</p> | mcq | jee-main-2023-online-30th-january-morning-shift |
1lgsw758t | maths | parabola | normal-to-parabola | <p>Let the tangent to the parabola $$\mathrm{y}^{2}=12 \mathrm{x}$$ at the point $$(3, \alpha)$$ be perpendicular to the line $$2 x+2 y=3$$. Then the square of distance of the point $$(6,-4)$$ from the normal to the hyperbola $$\alpha^{2} x^{2}-9 y^{2}=9 \alpha^{2}$$ at its point $$(\alpha-1, \alpha+2)$$ is equal to _________.</p> | [] | null | 116 | $\because \mathrm{P}(3, \alpha)$ lies on $\mathrm{y}^2=12 \mathrm{x}$
<br/><br/>$\Rightarrow \alpha= \pm 6$
<br/><br/>$$
\text { But, }\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|_{(3, \alpha)}=\frac{6}{\alpha}=1 \Rightarrow \alpha=6(\alpha=-6 \text { reject })
$$
<br/><br/>Now, hyperbola $\frac{x^2}{9}-\frac{y^2}{36}=1$, normal at
<br/><br/>$$
\begin{aligned}
& \mathrm{Q}(\alpha-1, \alpha+2) \text { is } \frac{9 \mathrm{x}}{5}+\frac{36 y}{8}=45 \\\\
& \Rightarrow 2 x+5 y-50=0
\end{aligned}
$$
<br/><br/>Now, distance of $(6,-4)$ from $2 x+5 y-50=0$ is equal to
<br/><br/>$$
\begin{aligned}
& \left|\frac{2(6)-5(4)-50}{\sqrt{2^2+5^2}}\right|=\frac{58}{\sqrt{29}} \\\\
& \Rightarrow \text { Squareof distance }=116
\end{aligned}
$$ | integer | jee-main-2023-online-11th-april-evening-shift |
lsbki99q | maths | parabola | normal-to-parabola | If the shortest distance of the parabola $y^2=4 x$ from the centre of the circle $x^2+y^2-4 x-16 y+64=0$ is $\mathrm{d}$, then $\mathrm{d}^2$ is equal to : | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "36"}] | ["C"] | null | <p>Equation of normal to parabola</p>
<p>$$\mathrm{y=m x-2 m-m^3}$$</p>
<p>this normal passing through center of circle $$(2,8)$$</p>
<p>$$\begin{aligned}
& 8=2 \mathrm{~m}-2 \mathrm{~m}-\mathrm{m}^3 \\
& \mathrm{~m}=-2
\end{aligned}$$</p>
<p>So point $$\mathrm{P}$$ on parabola $$\Rightarrow\left(\mathrm{am}^2,-2 \mathrm{am}\right)=(4,4)$$</p>
<p>And $$\mathrm{C}=(2,8)$$</p>
<p>$$\begin{aligned}
& \mathrm{PC}=\sqrt{4+16}=\sqrt{20} \\
& \mathrm{~d}^2=20
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
1lgypxtlo | maths | parabola | pair-of-tangents | <p>The ordinates of the points P and $$\mathrm{Q}$$ on the parabola with focus $$(3,0)$$ and directrix $$x=-3$$ are in the ratio $$3: 1$$. If $$\mathrm{R}(\alpha, \beta)$$ is the point of intersection of the tangents to the parabola at $$\mathrm{P}$$ and $$\mathrm{Q}$$, then $$\frac{\beta^{2}}{\alpha}$$ is equal to _______________.</p> | [] | null | 16 | $$
\begin{aligned}
& \text { Give parabola is : } y^2=12 x \quad(\because a=3) \\\\
& \text { So, } \mathrm{P} \equiv\left(a t_1^2, 2 a t_1\right) \\\\
& \mathrm{Q} \equiv\left(a t_2^2, 2 a t_2\right) \\\\
& \text { So, point } \mathrm{R}(\alpha, \beta) \equiv\left(a t_1 t_2, a\left(t_1+t_2\right)\right) \\\\
& \equiv((3 t)(3 t), 3(t+3 t))=\left(9 t^2, 12 t\right) \\\\
& \therefore \frac{\beta^2}{\alpha}=\frac{144 t^2}{9 t^2}=16
\end{aligned}
$$ | integer | jee-main-2023-online-8th-april-evening-shift |
AspiOe4ZyUJUoiRTgB09N | maths | parabola | position-of-point-and-chord-joining-of-two-points | Let A(4, $$-$$ 4) and B(9, 6) be points on the parabola, y<sup>2</sup> = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $$\Delta $$ACB is maximum. Then, the area (in sq. units) of $$\Delta $$ACB, is : | [{"identifier": "A", "content": "$$31{1 \\over 4}$$"}, {"identifier": "B", "content": "$$30{1 \\over 2}$$"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "$$31{3 \\over 4}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264184/exam_images/dtyf9rtine6siiyeobjo.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Parabola Question 98 English Explanation">
<br><br>$$\Delta ABC = {1 \over 2}\left| {\matrix{
4 & { - 4} & 1 \cr
9 & 6 & 1 \cr
{{t^2}} & {2t} & 1 \cr
} } \right|$$
<br><br>D = 60 + 10t $$-$$ 10t<sup>2</sup>
<br><br>$${{d\Delta } \over {dt}} = 0 \Rightarrow t = {1 \over 2}$$
<br><br>$${{{d^2}\Delta } \over {d{t^2}}} = - 20 < 0$$
<br><br>$$ \therefore $$ max at $$t = {1 \over 2}$$
<br><br>max area $$\Delta = 65 - {5 \over 2}$$
<br><br>$$ = {{125} \over 2} = 31{1 \over 4}$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
aqj3lPSrYhSaqBB1SDp7y | maths | parabola | position-of-point-and-chord-joining-of-two-points | If the area of the triangle whose one vertex is at the vertex of the parabola, y<sup>2</sup> + 4(x – a<sup>2</sup>) = 0 and the othertwo vertices are the points of intersection of the parabola and y-axis, is 250 sq. units, then a value of 'a' is : | [{"identifier": "A", "content": "$$5\\sqrt 5 $$"}, {"identifier": "B", "content": "$${\\left( {10} \\right)^{2/3}}$$"}, {"identifier": "C", "content": "$$5\\left( {{2^{1/3}}} \\right)$$"}, {"identifier": "D", "content": "5"}] | ["D"] | null | Vertex is (a<sup>2</sup>, 0)
<br><br>y<sup>2</sup> $$=$$ $$-$$(x $$-$$ a<sup>2</sup>) and x $$=$$ 0 $$ \Rightarrow $$ (0, $$ \pm $$ 2a)
<br><br>Area of triangle is $$ = {1 \over 2}.$$4a.(a<sup>2</sup>) = 250
<br><br>$$ \Rightarrow $$ a<sup>3</sup> = 125 or a = 5 | mcq | jee-main-2019-online-11th-january-evening-slot |
ElZ6UyxucQlC2uTZyf0kk | maths | parabola | position-of-point-and-chord-joining-of-two-points | The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x<sup>2</sup> such that the rectangle lies inside the parabola, is :
| [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "20$$\\sqrt 2 $$"}, {"identifier": "C", "content": "18$$\\sqrt 3 $$"}, {"identifier": "D", "content": "32"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267052/exam_images/mnslvvuc5ia6lxqcjf12.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Parabola Question 94 English Explanation">
<br>f (a) = 2a(12 $$-$$ a)<sup>2</sup>
<br><br>f '(a) = 2(12 $$-$$ 3a<sup>2</sup>)
<br><br>Maximum at a = 2
<br><br>maximum area = f(2) = 32 | mcq | jee-main-2019-online-12th-january-morning-slot |
INEmnq65GsBAbnWypAlve | maths | parabola | position-of-point-and-chord-joining-of-two-points | Let P(4, –4) and Q(9, 6) be two points on the parabola, y<sup>2</sup> = 4x and let x be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of $$\Delta $$PXQ is maximum. Then this maximum area (in sq. units) is : | [{"identifier": "A", "content": "$${{625} \\over 4}$$"}, {"identifier": "B", "content": "$${{125} \\over 4}$$"}, {"identifier": "C", "content": "$${{75} \\over 2}$$"}, {"identifier": "D", "content": "$${{125} \\over 2}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266547/exam_images/ql9pnjqathvtux4uc0ar.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Parabola Question 93 English Explanation">
<br>y<sup>2</sup> = 4x
<br><br>2yy' = 4
<br><br>y ' = $${1 \over t} = 2,$$ $$t = {1 \over 2}$$
<br><br>Area = $${1 \over 2}\left| {\matrix{
{{1 \over 4}} & 1 & 1 \cr
9 & 6 & 1 \cr
4 & { - 4} & 1 \cr
} } \right| = {{225} \over 4}$$ | mcq | jee-main-2019-online-12th-january-morning-slot |
b93Qm4pBuW9a6iMQF5jgy2xukez5j28t | maths | parabola | position-of-point-and-chord-joining-of-two-points | The area (in sq. units) of an equilateral triangle
inscribed in the parabola y<sup>2</sup> = 8x, with one of
its vertices on the vertex of this parabola, is : | [{"identifier": "A", "content": "$$256\\sqrt 3 $$"}, {"identifier": "B", "content": "$$64\\sqrt 3 $$"}, {"identifier": "C", "content": "$$128\\sqrt 3 $$"}, {"identifier": "D", "content": "$$192\\sqrt 3 $$"}] | ["D"] | null | Let A = (2t<sup>2</sup>
, 4t)
<br><br>B = (2t<sup>2</sup>
, -4t) (by symmetry as equilateral
triangle)
<br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263641/exam_images/fiy4gcor4i4z43cmqerx.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Evening Slot Mathematics - Parabola Question 79 English Explanation">
<br><br>tan 30<sup>o</sup> = $$\frac{4t}{2t^{2}} $$ = $$\frac{2}{t} $$
<br><br>$$ \Rightarrow $$ t = 2$$\sqrt{3} $$
<br><br> AB = 8t = 16$$\sqrt{3} $$
<br><br> Area of equilateral = $$\frac{1}{2} \times 16\sqrt{3} \times 24$$
<br><br>= $$192\sqrt 3 $$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
L76uaJ1Go15PoOPVJXjgy2xukf0pks9k | maths | parabola | position-of-point-and-chord-joining-of-two-points | Let P be a point on the parabola, y<sup>2</sup>
= 12x and
N be the foot of the perpendicular drawn from
P on the axis of the parabola. A line is now
drawn through the mid-point M of PN, parallel
to its axis which meets the parabola at Q. If the
y-intercept of the line NQ is $${4 \over 3}$$,
then : | [{"identifier": "A", "content": "MQ = $${1 \\over 3}$$"}, {"identifier": "B", "content": "PN = 4"}, {"identifier": "C", "content": "PN = 3"}, {"identifier": "D", "content": "MQ = $${1 \\over 4}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265153/exam_images/uwoptt2srzgkzl7wy0ci.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - Parabola Question 78 English Explanation">
<br><br>Let P = (3t<sup>2</sup>
, 6t) ; N = (3t<sup>2</sup>
, 0)
<br><br>M = (3t<sup>2</sup>
, 3t)
<br><br>Let point Q = (h, 3t) which lies on the parabola y<sup>2</sup> = 12x
<br><br>$$ \therefore $$ 9t<sup>2</sup> = 12h
<br><br>$$ \Rightarrow $$ h = $${{3{t^2}} \over 4}$$
<br><br>Equation of NQ
<br><br>y - 0 = $${{3t} \over {{{3{t^2}} \over 4} - 3{t^2}}}\left( {x - 3{t^2}} \right)$$
<br><br>$$ \Rightarrow $$ y = $${{ - 4} \over {3t}}\left( {x - 3{t^2}} \right)$$
<br><br>For y-intercept of NQ, x = 0
<br><br>$$ \therefore $$ y = 4t
<br><br>Given y-intercept of the line NQ is $${4 \over 3}$$
<br><br>$$ \therefore $$ 4t = $${4 \over 3}$$
<br><br>$$ \Rightarrow $$ t = $${1 \over 3}$$
<br><br>$$ \therefore $$ MQ = $${{9{t^2}} \over 4}$$ = $${1 \over 4}$$
<br><br>PN = 6t = 2 | mcq | jee-main-2020-online-3rd-september-morning-slot |
ir2l5yqzikZtHZQSfz1klrlp86w | maths | parabola | position-of-point-and-chord-joining-of-two-points | If P is a point on the parabola y = x<sup>2</sup> + 4 which is closest to the straight line y = 4x $$-$$ 1, then the co-ordinates of P are : | [{"identifier": "A", "content": "($$-$$2, 8)"}, {"identifier": "B", "content": "(2, 8)"}, {"identifier": "C", "content": "(1, 5)"}, {"identifier": "D", "content": "(3, 13)"}] | ["B"] | null | Given, curve y = x<sup>2</sup> + 4<br><br>and, line y = 4x $$-$$ 1<br><br>Here, y = x<sup>2</sup> + 4<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxoae8y6/3b0cfa8b-90f4-4e91-ba41-db601a79f96b/6ef08ee0-66dc-11ec-8eea-2b236925a651/file-1kxoae8y7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxoae8y6/3b0cfa8b-90f4-4e91-ba41-db601a79f96b/6ef08ee0-66dc-11ec-8eea-2b236925a651/file-1kxoae8y7.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2021 (Online) 24th February Evening Shift Mathematics - Parabola Question 71 English Explanation"><br>$$\therefore$$ $${{dy} \over {dx}} = 2x$$ ..... (i)<br><br>and y = 4x $$-$$ 1<br><br>$${{dy} \over {dx}} = 4$$ ..... (ii)<br><br>Let the required point be P(x<sub>1</sub>, y<sub>1</sub>).<br><br>$$\therefore$$ $${\left. {{{dy} \over {dx}}} \right|_P} = 2{x_1}$$ ..... (iii)<br><br>$$\because$$ Slopes will be equal.<br><br>$$\therefore$$ 2x<sub>1</sub> = 4 [from Eqs. (ii) and (iii)]<br><br>$$ \Rightarrow {x_1} = {4 \over 2} = 2$$<br><br>Now, the given point P(x<sub>1</sub>, y<sub>1</sub>) lies on curve y = x<sup>2</sup> + 4,<br><br>$$\therefore$$ y<sub>1</sub> = x$$_1^2$$ + 4<br><br>$$\Rightarrow$$ y<sub>1</sub> = 2<sup>2</sup> + 4 = 8<br><br>Hence, required coordinate of P = (2, 8) | mcq | jee-main-2021-online-24th-february-evening-slot |
CA3h2nvK6MgPA2sGlp1klt7a0gw | maths | parabola | position-of-point-and-chord-joining-of-two-points | The shortest distance between the line x $$-$$ y = 1 and the curve x<sup>2</sup> = 2y is : | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${1 \\over 2{\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}] | ["B"] | null | Shortest distance must be along common normal<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264020/exam_images/ipwtluvs8xeh3hc4yv0v.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Parabola Question 69 English Explanation">
<br><br>Let a point on curve has x coordinate = h
<br><br>Then y coordinate :
<br><br>h<sup>2</sup> = 2y
<br><br>$$ \Rightarrow $$ y = $${{{h^2}} \over 2}$$
<br><br>So point is = (h, $${{{h^2}} \over 2}$$)
<br><br>m<sub>1</sub> (slope of line x $$-$$ y = 1) = 1
<br><br>$$ \Rightarrow $$ slope of perpendicular line = $$-$$1
<br><br>Slope of the perpendicular line on the curve x<sup>2</sup> = 2y,
<br><br>$${m_2} = {{2x} \over 2} = x \Rightarrow {m_2} = h $$
<br><br>$$ \therefore $$ Slope of normal = $$ - {1 \over h}$$<br><br>$$ - {1 \over h}$$ = $$-$$1 $$ \Rightarrow $$ h = 1<br><br>So point is $$\left( {1,{1 \over 2}} \right)$$<br><br>$$D = \left| {{{1 - {1 \over 2} - 1} \over {\sqrt {1 + 1} }}} \right| = {1 \over {2\sqrt 2 }}$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
1krrwj01w | maths | parabola | position-of-point-and-chord-joining-of-two-points | If the point on the curve y<sup>2</sup> = 6x, nearest to the point $$\left( {3,{3 \over 2}} \right)$$ is ($$\alpha$$, $$\beta$$), then 2($$\alpha$$ + $$\beta$$) is equal to _____________. | [] | null | 9 | Let, $$P \equiv \left( {{3 \over 2}{t^2},3t} \right)$$ is on the curve.<br><br>Normal at point P<br><br>$$tx + y = 3t + {3 \over 2}{t^3}$$<br><br>Passes through $$\left( {3,{3 \over 2}} \right)$$<br><br>$$ \Rightarrow 3t + {3 \over 2} = 3t + {3 \over 2}{t^3}$$<br><br>$$ \Rightarrow {t^3} = 1 \Rightarrow t = 1$$<br><br>$$P \equiv \left( {{3 \over 2},3} \right) = (\alpha ,\beta )$$ <br><br>$$2(\alpha + \beta ) = 2\left( {{3 \over 2} + 3} \right) = 9$$ | integer | jee-main-2021-online-20th-july-evening-shift |
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