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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 $H$. Let the product of the eccentricities of $E$ and $H$ be $\frac{1}{2}$. If $l$ is the length of the latus rectum of the ellipse $E$, then the value of $113\,l$ is equal to $....$
$1500$
$1552$
$1000$
$1553$
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 $H: \frac{x^{2}}{49}-\frac{y^{2}}{64}=-1$.
$S_{0} b^{2}=64$
$e_{H}=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{1+\frac{49}{64}}$
Ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$e_{E}=\sqrt{1-\frac{a^{2}}{b^{2}}}=\sqrt{1-\frac{a^{2}}{64}}$
$b=8, \sqrt{\frac{1-a^{2}}{64}} \times \frac{\sqrt{113}}{8}=\frac{1}{2} \Rightarrow \sqrt{64-a^{2}} \times \sqrt{113}=32$
$\left(64-a^{2}\right)=\frac{32^{2}}{113}$
$\Rightarrow a^{2}=64-\frac{32^{2}}{113}$
$l=\frac{2 a^{2}}{b}=\frac{2}{8}\left(64-\frac{32^{2}}{113}\right)=\frac{1552}{113}$
$113\,l=1552$
