JEE Main 2025 — 29 January, Morning Shift — Question 56

$\quad 2 \mathrm{ae}=4$

$2 \mathrm{a}\left(\frac{1}{\sqrt{3}}\right)=4$

$\Rightarrow \mathrm{a}=2 \sqrt{3}$

$\Rightarrow 1-\frac{\mathrm{b}^{2}}{12}=\frac{1}{3}$

$\Rightarrow \mathrm{~b}^{2}=8$

Now $\frac{2 \mathrm{~b}^{2}}{\mathrm{a}} \cdot \frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=\frac{32}{\sqrt{3}}$

$\Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 \mathrm{~A}^{2}}{\mathrm{~B}}=\frac{32}{\sqrt{3}}$

$\Rightarrow A^{2}=2 B$

$1-\frac{\mathrm{A}^{2}}{\mathrm{~B}^{2}}=\frac{1}{3}$

$\Rightarrow 1-\frac{2 \mathrm{~B}}{\mathrm{~B}^{2}}=\frac{1}{3}$

$\Rightarrow \mathrm{~B}=3$ $\Rightarrow A^{2}=6$

$\frac{x^{2}}{12}+\frac{y^{2}}{8}=1$

$\frac{x^{2}}{6}+\frac{y^{2}}{9}=1$

On solving (1) & (2) we get

$$(x, y) \equiv\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right)$$

The four points are vertices of rectangle and its area $=$ $\frac{24 \sqrt{6}}{5}$