The eccentricity of an ellipse E with centre | vedclass

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(D) For ellipse $E$, eccentricity $e = \frac{\sqrt{3}}{2}$ and directrix $x = \frac{a}{e} = \frac{4\sqrt{6}}{3}$.Thus, $a = \frac{4\sqrt{6}}{3} \cdot

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$\frac{8}{7}$

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(D) For ellipse $E$, eccentricity $e = \frac{\sqrt{3}}{2}$ and directrix $x = \frac{a}{e} = \frac{4\sqrt{6}}{3}$.Thus, $a = \frac{4\sqrt{6}}{3} \cdot \frac{\sqrt{3}}{2} = 2\sqrt{2}$.Since $b^2 = a^2(1 - e^2)$, we have $b^2 = 8(1 - 3/4) = 2$, so $b = \sqrt{2}$.For hyperbola $H$, eccentricity $e_H = a = 2\sqrt{2}$.The length of the latus rectum of $H$ is $\frac{2b_H^2}{a_H} = 2b = 2\sqrt{2}$.Also, for hyperbola $H$, $b_H^2 = a_H^2(e_H^2 - 1) = a_H^2(8 - 1) = 7a_H^2$.Substituting this into the latus rectum equation: $\frac{2(7a_H^2)}{a_H} = 2\sqrt{2} \implies 14a_H = 2\sqrt{2} \implies a_H = \frac{\sqrt{2}}{7}$.The distance between the foci of $H$ is $2a_H e_H = 2 \cdot \frac{\sqrt{2}}{7} \cdot 2\sqrt{2} = \frac{8}{7}$.

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