Let each of the two ellipses E_1: frac{x^2}{a | vedclass

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(B) For $E_1$,the distance between foci is $2ae = 8$. Given $e = \frac{4}{5}$,we have $2a(\frac{4}{5}) = 8 \Rightarrow a = 5$.Using $b^2 = a^2(1 - e^2

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$\frac{32}{5}$

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(B) For $E_1$,the distance between foci is $2ae = 8$. Given $e = \frac{4}{5}$,we have $2a(\frac{4}{5}) = 8 \Rightarrow a = 5$.Using $b^2 = a^2(1 - e^2)$,we get $b^2 = 25(1 - \frac{16}{25}) = 25(\frac{9}{25}) = 9$.The length of the latus rectum $\ell_1 = \frac{2b^2}{a} = \frac{2 	imes 9}{5} = \frac{18}{5}$.For $E_2$,$A The length of the latus rectum $\ell_2 = \frac{2A^2}{B} = \frac{2(9/25)B^2}{B} = \frac{18}{25}B$.Given $2\ell_1^2 = 9\ell_2$,we have $2(\frac{18}{5})^2 = 9(\frac{18}{25}B) \Rightarrow 2 	imes \frac{324}{25} = \frac{162}{25}B$.Solving for $B$,$B = \frac{2 	imes 324}{162} = 4$.The distance between the foci of $E_2$ is $2Be = 2 	imes 4 	imes \frac{4}{5} = \frac{32}{5}$.

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