Let E : frac{x^2}{a^2} + frac{y^2}{b^2} = 1, | vedclass

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(C) For ellipse $E$,the foci are $(\pm ae, 0)$,so the distance between foci is $2ae = 2\sqrt{3} \Rightarrow ae = \sqrt{3}$.For hyperbola $H$,the foci

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$8$

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(C) For ellipse $E$,the foci are $(\pm ae, 0)$,so the distance between foci is $2ae = 2\sqrt{3} \Rightarrow ae = \sqrt{3}$.For hyperbola $H$,the foci are $(\pm Ae', 0)$,so the distance between foci is $2Ae' = 2\sqrt{3} \Rightarrow Ae' = \sqrt{3}$.Thus,$ae = Ae' \Rightarrow \frac{e}{e'} = \frac{A}{a}$.Given $\frac{e}{e'} = \frac{1}{3}$,we have $\frac{A}{a} = \frac{1}{3} \Rightarrow a = 3A$.Given $a - A = 2$,substituting $a = 3A$ gives $3A - A = 2$ $\Rightarrow 2A = 2$ $\Rightarrow A = 1$ and $a = 3$.Since $ae = \sqrt{3}$,$3e = \sqrt{3} \Rightarrow e = \frac{1}{\sqrt{3}}$. Then $b^2 = a^2(1 - e^2) = 9(1 - \frac{1}{3}) = 9(\frac{2}{3}) = 6$.Since $Ae' = \sqrt{3}$,$1 \cdot e' = \sqrt{3} \Rightarrow e' = \sqrt{3}$. Then $B^2 = A^2((e')^2 - 1) = 1(3 - 1) = 2$.The length of the latus rectum of $E$ is $L_E = \frac{2b^2}{a} = \frac{2(6)}{3} = 4$.The length of the latus rectum of $H$ is $L_H = \frac{2B^2}{A} = \frac{2(2)}{1} = 4$.The sum of the lengths of their latus rectums is $4 + 4 = 8$.

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