Let for two distinct values of p the lines y=x+p touch the ellipse E: x^2 4^2 + y^2 3^2 =1 at the points A and B . Let the line y=x intersect E at the points C and D . Then the area of the quadrilateral A B C D is equal

Correct answer: B. 24 Point of contact are ( a^2 ~m a^2 ~m^2+b^2 , ± b^2 a^2 ~m^2+b^2 ) A((-16)/(5), (9)/(5)) B((16)/(5), (-9)/(5)) Point D is ((12)/(5), (12)/(5)) Area of ABD=(1)/(2)| array ccc -(16)/(5) & (9)/(5) & 1 (16)/(5) & (-9)/(5) & 1 (12)/(5) & (12)/(5) & 1 array | =12 Area of ABCD is =24 option (2)

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Ellipse

Let for two distinct values of p the lines y=x+p touch the ellipse...

Let for two distinct values of

p

the lines

y=x+p

touch the ellipse

\mathrm{E}: \frac{\mathrm{x}^2}{4^2}+\frac{\mathrm{y}^2}{3^2}=1

at the points A and B . Let the line

\mathrm{y}=\mathrm{x}

intersect E at the points C and

D

. Then the area of the quadrilateral

A B C D

is equal to

Options

A $36$
B $24$
C $48$
D $20$

Correct answer

B. $24$

Step-by-step solution

Point of contact are

\left(\frac{\mp \mathrm{a}^2 \mathrm{~m}}{\sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}}, \frac{ \pm \mathrm{b}^2}{\sqrt{\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2}}\right)

\mathrm{A}\left(\frac{-16}{5}, \frac{9}{5}\right) \mathrm{B}\left(\frac{16}{5}, \frac{-9}{5}\right)

Point D is \left(\frac{12}{5}, \frac{12}{5}\right)

<br/>Area of

\mathrm{ABD}=\frac{1}{2}\left|

\begin{array}{ccc}-\frac{16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & \frac{-9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1\end{array}

\right|

<br/>

=12

<br/>Area of ABCD is

=24
option (2)

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