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 2)]

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YESPYQ · Accepted Answer

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

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