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

[JEE Main 2025, 4 Apr (Shift 1)]

Question

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

[JEE Main 2025, 4 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: b) $24$. $\left(\frac{∓\text{ }{a}^{2}m}{\sqrt{{a}^{2}\text{ }{m}^{2}+{b}^{2}}},\text{ }\frac{\pm {b}^{2}}{\sqrt{{a}^{2}\text{ }{m}^{2}+{b}^{2}}}\right)$ $A\left(\frac{−16}{5},\text{ }\frac{9}{5}\right)\text{ }B\left(\frac{16}{5},\text{ }\frac{−9}{5}\right)$ Point D is $\left(\frac{12}{5},\text{ }\frac{12}{5}\right)$ Area of $ABD=\text{ }\frac{1}{2}\text{ }\left|\begin{matrix}-\frac{16}{5} & \frac{9}{5} & 1 \ \frac{16}{5} & \frac{-9}{5} & 1 \ \frac{12}{5} & \frac{12}{5} & 1\end{matrix}\right|$ = 12 Area of ABCD is 24

Let for two distinct values of p the lines y = x + p touch the ellipse E : \(\frac{{\mathrm{x}}^{2}}{{4}^{2}}+\frac{{\ma…

Practice more JEE Maths PYQs