Mathematics · Ellipse

JEE Main 2025 — 4 April, Evening Shift — Question 22

Let for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $E: \frac{x^{2}}{4^{2}}+\frac{y^{2}}{3^{3}}=1$ 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: D

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

$T: y=m x \pm \sqrt{16 m^{2}+9}$

$y=x+p$

$\Rightarrow m=1$

$\Rightarrow p= \pm \sqrt{16+9}$

$= \pm 5$

$T: y=x \pm 5$ will to cut the $E$ at $A\left(-\frac{16}{5}, \frac{9}{5}\right)$

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

Also, $y=x$ will cut the $E$ at $C\left(\frac{12}{5}, \frac{12}{5}\right)$ $D\left(-\frac{12}{5},-\frac{12}{5}\right)$

$A B C D$ in not give in cyclic order

$\therefore$ it does not form any quadrilateral

$\therefore \quad$ No option should match

If order is not considered then

Area $=24$ sq. unit.

Answer key and solution verified before publishing.

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