Question
The eccentricity of an ellipse E with centre at the origin $O$ is $\frac{\sqrt{3}}{2}$ and its directrices are $x = \pm \frac{4\sqrt{6}}{3}$. Let $H:\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E , and whose length of latus rectum is equal to the length of minor axis of E . Then the distance between the foci of H is :
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Solution
Sol. [Key Concept: From $e = \frac{\sqrt{3}}{2}$ and the directrix get $a = 2\sqrt{2,}\ b^{2} = 2$ for $E$; the hyperbola's eccentricity equals $E$'s semi-major axis and its latus rectum equals $E$'s minor axis, fixing $a,b$ for the foci distance.]
$$e = \frac{\sqrt{3}}{2}$$
$$\frac{a}{e} = \frac{4\sqrt{6}}{3}$$
$${a = 2\sqrt{2} }{(ae)^{2} = a^{2} - b^{2} }{b^{2} = 2 }$$so $E$ : $\frac{x^{2}}{8} + \frac{y^{2}}{2} = 1$
Now $e_{H} = 2\sqrt{2} \Rightarrow {\text{ }b}^{2} = 7a^{2}$
$${\frac{2{\text{ }b}^{2}}{a} = 2\sqrt{2} \Rightarrow a = \frac{\sqrt{2}}{7} }{\left( ae_{H} \right)^{2} = a^{2} + b^{2} = 8a^{2} = \frac{16}{49} }{{ae}_{H} = \frac{4}{7} }$$Distance between focii $= 2{ae}_{H} = \frac{8}{7}$
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