અહી ઉપવલય E _1: frac{ x ^2}{ a ^2}+frac{ y ^2 | vedclass

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Question

$2 ae=4$
$2 a\left(\frac{1}{\sqrt{3}}ight)=4$
$\Rightarrow a=2 \sqrt{3}$
$\Rightarrow 1-\frac{b^2}{12}=\frac{1}{3} \Rightarrow b^2=8$
$	ext { N

Vedclass · Accepted Answer

$\frac{24 \sqrt{6}}{5}$

Vedclass · Answer

$2 ae=4$
$2 a\left(\frac{1}{\sqrt{3}}ight)=4$
$\Rightarrow a=2 \sqrt{3}$
$\Rightarrow 1-\frac{b^2}{12}=\frac{1}{3} \Rightarrow b^2=8$
$	ext { Now } \frac{2 b^2}{a} \cdot \frac{2 A^2}{B}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}ight) \frac{2 A^2}{B}=\frac{32}{\sqrt{3}}$
$\Rightarrow A^2=2 B$
$1-\frac{A^2}{B^2}=\frac{1}{3} \Rightarrow 1-\frac{2 B}{B^2}=\frac{1}{3} \Rightarrow B=3$
$\Rightarrow A^2=6$
$\frac{x^2}{12}+\frac{y^2}{8}=1 \ldots . .(1)$
$\frac{x^2}{6}+\frac{y^2}{9}=1 \ldots .(2)$
On solving $(1) \& (2)$ we get
$(x, y) \equiv\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}ight),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}ight),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}ight),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}ight)$
The four points are vertices of rectangle and its area $= \frac{24 \sqrt{6}}{5}$

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