JEE Main 2025 — Ellipse Question with Solution

\begin{aligned} &\begin{aligned} & 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 \\ & \text { 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 \mathrm{~A}^2=2 \mathrm{~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 \mathrm{~A}^2=6 \\ & \frac{\mathrm{x}^2}{12}+\frac{\mathrm{y}^2}{8}=1 \ldots . .(1) \\ & \frac{\mathrm{x}^2}{6}+\frac{\mathrm{y}^2}{9}=1 \ldots . .(2) \end{aligned}\\ &\text { On solving (1) & (2) we get } \end{aligned}

$$\begin{array}{rl}& (x,y)\equiv \left(\frac{\sqrt{6}}{\sqrt{5}},\frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}},\frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}},\frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}},\frac{-6}{\sqrt{5}}\right)\\ & \text{ The four points are vertices of rectangle and its area }=\\ & \frac{24\sqrt{6}}{5}\end{array}$$