An ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ passes through the vertices of the hyperbola $H : \frac{x^{2}}{49} - \frac{y^{2}}{64} = - 1$ . Let the major and minor axes of the ellipse E coincide with the transverse and conjugate axes of the hyperbola H, respectively. Let the product of the eccentricities of E and H be $\frac{1}{2}$ . If $k$ is the length of the latus rectum of the ellipse E, then the value of 113 $k$ is equal to _______.

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answer is 1552.

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Detailed Solution

$Hyp : \frac{y^{2}}{64} - \frac{x^{2}}{49} = 1$

An ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ passes through the vertices of the hyperbola $H : \frac{x^{2}}{49} - \frac{y^{2}}{64} = - 1$

So b² = 64

$\begin{array}{l} e_{H} = \sqrt{1 + \frac{a^{2}}{b^{2}}} = \sqrt{1 + \frac{49}{64}} \\ Ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \\ e_{E} = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{a^{2}}{64}} \\ b = 8, \sqrt{\frac{1 - a^{2}}{64}} \times \frac{\sqrt{113}}{8} = \frac{1}{2} \Rightarrow \sqrt{64 - a^{2}} \times \sqrt{113} = 32 \\ (64 - a^{2}) = \frac{32^{2}}{113} \\ \Rightarrow a^{2} = 64 - \frac{32^{2}}{113} \\ l = \frac{2 a^{2}}{b} = \frac{2}{8} (64 - \frac{32^{2}}{113}) = \frac{1552}{113} \\ 113 l = 1552 \end{array}$