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(B) For the first hyperbola,the transverse axis is $2a$ and the conjugate axis is $2b$. Given $2a = 2(2b)$,so $a = 2b$. The eccentricity $x$ is given

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$\frac{7}{4}$

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(B) For the first hyperbola,the transverse axis is $2a$ and the conjugate axis is $2b$. Given $2a = 2(2b)$,so $a = 2b$. The eccentricity $x$ is given by $x^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{b^2}{(2b)^2} = 1 + \frac{1}{4} = \frac{5}{4}$.For the second hyperbola,the distance between the foci is $2ae$ and the distance between the directrices is $\frac{2a}{e}$. Given $2ae = 3 	imes \frac{2a}{e}$,which simplifies to $e^2 = 3$. Thus,$y^2 = 3$.Therefore,$y^2 - x^2 = 3 - \frac{5}{4} = \frac{12-5}{4} = \frac{7}{4}$.

Let $x$ be the eccentricity of a hyperbola whose transverse axis is twice its conjugate axis. Let $y$ be the eccentricity of another hyperbola for which the distance between the foci is $3$ times the distance between its directrices. Then $y^2-x^2=$

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