Let for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $\mathrm{E}:…
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
$36$
$24$
$48$
$20$
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 $\left(\frac{12}{5}, \frac{12}{5}\right)$ Area of $\mathrm{ABD}=\frac{1}{2}\left|\begin{array}{ccc}-\frac{16}{5} & \frac{9}{5} & 1 \\ \frac{16}{5} & \frac{-9}{5} & 1 \\ \frac{12}{5} & \frac{12}{5} & 1\end{array}\right|$ $=12$ Area of ABCD is $=24$ option (2)