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 E₁ be 4. If E₁ and E₂ meet at A, B, C and D, then the area of the quadrilateral ABCD equals:

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6 $\sqrt{6}$

$\frac{12 \sqrt{6}}{5}$

$\frac{18 \sqrt{6}}{5}$

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

answer is D.

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Detailed Solution

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