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 $ABCD$ is equal to

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Ellipse

Let for two distinct values of

p

the lines

y = x + p

touch the ellipse

E : \frac{x ^{2}}{4 ^{2}} + \frac{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 to

Answer: (B)

The condition for

y = x + p

to touch the ellipse is

p^{2} = 4^{2} + 3^{2} = 25

, so

p = \pm 5

with contact points

A (- \frac{16}{5}, \frac{9}{5})

and

B (\frac{16}{5}, - \frac{9}{5})

. The line

y = x

meets the ellipse at

C (\frac{12}{5}, \frac{12}{5})

and

D (- \frac{12}{5}, - \frac{12}{5})

. Both chords pass through the centre, so

A C B D

is a parallelogram with diagonals

A B

and

C D

; its area is

\frac{1}{2} A B \times C D = \frac{1}{2} ∣ - 48 ∣ = 24

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Single correct option