Let the ellipse E_1: (x^2)/(a^2) + (y^2)/(b^2) = 1, a b and E_2: (x^2)/(A^2) + (y^2)/(B^…

Question

Let the ellipse $E_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a > b$ and $E_2: \frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& 2 \mathrm{ae}=4 \
& 2 \mathrm{a}\left(\frac{1}{\sqrt{3}}\right)=4 \
& \Rightarrow \mathrm{a}=2 \sqrt{3} \
& \Rightarrow 1-\frac{\mathrm{b}^2}{12}=\frac{1}{3} \Rightarrow \mathrm{~b}^2=8 \
& \text { Now } \frac{2 \mathrm{~b}^2}{\mathrm{a}} \cdot \frac{2 \mathrm{~A}^2}{\mathrm{~B}}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 \mathrm{~A}^2}{\mathrm{~B}}=\frac{32}{\sqrt{3}} \
& \Rightarrow \mathrm{~A}^2=2 \mathrm{~B} \
& 1-\frac{\mathrm{A}^2}{\mathrm{~B}^2}=\frac{1}{3} \Rightarrow 1-\frac{2 \mathrm{~B}}{\mathrm{~B}^2}=\frac{1}{3} \Rightarrow \mathrm{~B}=3 \
& \Rightarrow \mathrm{~A}^2=6 \
& \frac{\mathrm{x}^2}{12}+\frac{\mathrm{y}^2}{8}=1 \ldots . .(1) \
& \frac{\mathrm{x}^2}{6}+\frac{\mathrm{y}^2}{9}=1 \ldots . .(2)
\end{aligned}\
\&\text { On solving (1) \& (2) we get }
\end{aligned}$$

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

Solutions · 1