10-2. Parabola, Ellipse, Hyperbola
hard

Let the ellipse, $E _1: \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1, a > b$ and $E _2: \frac{ x ^2}{A^2}+\frac{ y ^2}{B^2}=1, A< B$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$, then the area of the quadrilateral $A B C D$ equals:

A$6 \sqrt{6}$
B$\frac{18 \sqrt{6}}{5}$
C$\frac{12 \sqrt{6}}{5}$
D$\frac{24 \sqrt{6}}{5}$
(JEE MAIN-2025)

Solution

$2 ae=4$
$2 a\left(\frac{1}{\sqrt{3}}\right)=4$
$\Rightarrow a=2 \sqrt{3}$
$\Rightarrow 1-\frac{b^2}{12}=\frac{1}{3} \Rightarrow b^2=8$
$\text { Now } \frac{2 b^2}{a} \cdot \frac{2 A^2}{B}=\frac{32}{\sqrt{3}} \Rightarrow 2\left(\frac{8}{2 \sqrt{3}}\right) \frac{2 A^2}{B}=\frac{32}{\sqrt{3}}$
$\Rightarrow A^2=2 B$
$1-\frac{A^2}{B^2}=\frac{1}{3} \Rightarrow 1-\frac{2 B}{B^2}=\frac{1}{3} \Rightarrow B=3$
$\Rightarrow A^2=6$
$\frac{x^2}{12}+\frac{y^2}{8}=1 \ldots . .(1)$
$\frac{x^2}{6}+\frac{y^2}{9}=1 \ldots .(2)$
On solving $(1) \& (2)$ we get
$(x, y) \equiv\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{6}{\sqrt{5}}\right),\left(\frac{\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right),\left(\frac{-\sqrt{6}}{\sqrt{5}}, \frac{-6}{\sqrt{5}}\right)$
The four points are vertices of rectangle and its area $= \frac{24 \sqrt{6}}{5}$
Standard 11
Mathematics

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