Let e_1 be the eccentricity of a hyperbola fo | vedclass

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(B) For the first hyperbola,the distance between foci is $2ae_1$ and the distance between directrices is $\frac{2a}{e_1}$.Given $2ae_1 = 2 	imes \fra

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$\frac{\sqrt{10}}{2}$

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(B) For the first hyperbola,the distance between foci is $2ae_1$ and the distance between directrices is $\frac{2a}{e_1}$.Given $2ae_1 = 2 	imes \frac{2a}{e_1}$,which simplifies to $e_1^2 = 2$. Since $e_1 > 1$,we have $e_1 = \sqrt{2}$.For the second hyperbola,the length of the transverse axis is $2a_2$ and the length of the conjugate axis is $2b_2$.Given $2a_2 = 2(2b_2)$,so $a_2 = 2b_2$ or $b_2 = \frac{a_2}{2}$.The eccentricity $e_2$ is given by $e_2 = \sqrt{1 + \frac{b_2^2}{a_2^2}} = \sqrt{1 + \frac{(a_2/2)^2}{a_2^2}} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$.Therefore,$e_1 e_2 = \sqrt{2} 	imes \frac{\sqrt{5}}{2} = \frac{\sqrt{10}}{2}$.Thus,option $B$ is correct.

Let $e_1$ be the eccentricity of a hyperbola for which the distance between its foci is $2$ times the distance between its directrices,and $e_2$ be the eccentricity of another hyperbola for which the length of its transverse axis is twice the length of its conjugate axis. Then $e_1 e_2 =$

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