Let the ellipse E_1: frac{x^2}{a^2} + frac{y^ | vedclass

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(D) For $E_1$,$2ae = 4 \implies a(\frac{1}{\sqrt{3}}) = 2 \implies a = 2\sqrt{3}$.Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $\frac{1}{3} = 1 - \frac{b

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

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(D) For $E_1$,$2ae = 4 \implies a(\frac{1}{\sqrt{3}}) = 2 \implies a = 2\sqrt{3}$.Since $e^2 = 1 - \frac{b^2}{a^2}$,we have $\frac{1}{3} = 1 - \frac{b^2}{12} \implies \frac{b^2}{12} = \frac{2}{3} \implies b^2 = 8$.The length of the latus rectum of $E_1$ is $L_1 = \frac{2b^2}{a} = \frac{2(8)}{2\sqrt{3}} = \frac{8}{\sqrt{3}}$.For $E_2$,$e^2 = 1 - \frac{A^2}{B^2} = \frac{1}{3} \implies \frac{A^2}{B^2} = \frac{2}{3} \implies A^2 = \frac{2}{3}B^2$. The length of the latus rectum is $L_2 = \frac{2A^2}{B} = \frac{2(\frac{2}{3}B^2)}{B} = \frac{4B}{3}$.Given $L_1 \cdot L_2 = \frac{32}{\sqrt{3}} \implies \frac{8}{\sqrt{3}} \cdot \frac{4B}{3} = \frac{32}{\sqrt{3}} \implies B = 3$. Then $A^2 = \frac{2}{3}(3^2) = 6$.Equations are $E_1: \frac{x^2}{12} + \frac{y^2}{8} = 1$ and $E_2: \frac{x^2}{6} + \frac{y^2}{9} = 1$.Subtracting the equations: $(\frac{1}{12} - \frac{1}{6})x^2 + (\frac{1}{8} - \frac{1}{9})y^2 = 0 \implies -\frac{1}{12}x^2 + \frac{1}{72}y^2 = 0 \implies y^2 = 6x^2$.Substituting $y^2 = 6x^2$ into $E_1$: $\frac{x^2}{12} + \frac{6x^2}{8} = 1 \implies \frac{x^2}{12} + \frac{3x^2}{4} = 1 \implies \frac{x^2 + 9x^2}{12} = 1 \implies 10x^2 = 12 \implies x^2 = \frac{6}{5} \implies x = \pm \sqrt{\frac{6}{5}}$.Then $y^2 = 6(\frac{6}{5}) = \frac{36}{5} \implies y = \pm \frac{6}{\sqrt{5}}$.The vertices are $(\pm \sqrt{\frac{6}{5}}, \pm \frac{6}{\sqrt{5}})$.Area of rectangle $= 2|x| \cdot 2|y| = 4 \cdot \sqrt{\frac{6}{5}} \cdot \frac{6}{\sqrt{5}} = \frac{24\sqrt{6}}{5}$.

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