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(A) Let the hyperbola $H$ be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Distance between foci is $2ae = 26$,so $ae = 13$. Distance between directrices i

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$\frac{13}{12}$

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(A) Let the hyperbola $H$ be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Distance between foci is $2ae = 26$,so $ae = 13$. Distance between directrices is $\frac{2a}{e} = \frac{50}{13}$,so $\frac{a}{e} = \frac{25}{13}$,which means $a = \frac{25e}{13}$. Substituting $a$ in $ae = 13$: $(\frac{25e}{13})e = 13 \implies e^2 = \frac{169}{25} \implies e = \frac{13}{5}$. Now,$a = \frac{25}{13} 	imes \frac{13}{5} = 5$. Since $b^2 = a^2(e^2 - 1)$,we have $b^2 = 25(\frac{169}{25} - 1) = 169 - 25 = 144$,so $b = 12$. The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. Its eccentricity $e'$ satisfies $e'^2 = 1 + \frac{a^2}{b^2} = 1 + \frac{25}{144} = \frac{169}{144}$. Thus,$e' = \sqrt{\frac{169}{144}} = \frac{13}{12}$.

If the distance between the foci of a hyperbola $H$ is $26$ and the distance between its directrices is $\frac{50}{13}$,then the eccentricity of the conjugate hyperbola of the hyperbola $H$ is

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