An ellipse E:x^2a^2+y^2b^2=1 passes through the vertices of the hyperbola H:x^249-y^264=-1…

Question

An ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ passes through the vertices of the hyperbola $H:\dfrac{x^2}{49}-\dfrac{y^2}{64}=-1$. Let the major and minor axes of $E$ coincide with the transverse and conjugate axes of $H$ respectively. Let the product of the eccentricities of $E$ and $H$ be $\dfrac12$. If $l$ is the length of the latus rectum of $E$, then the value of $113l$ is equal to \_\_\_\_\_\_\_\_.

Answer: 1552

Accepted Answer

$H:\dfrac{y^2}{64}-\dfrac{x^2}{49}=1$, vertices $(0,\pm8)$, $e_H=\sqrt{1+\tfrac{49}{64}}=\dfrac{\sqrt{113}}8$.

$E$'s major axis is along $y$ (so $b>a$ in the usual sense --- here $E$'s semi-major is along $y$ with value $\sqrt{b^2}=8$): $e_E=\dfrac{\sqrt{64-a^2}}8$.

$$e_E\cdot e_H=\frac12\ \Longrightarrow\ \frac{\sqrt{113(64-a^2)}}{64}=\frac12\ \Longrightarrow\ 64-a^2=\frac{1024}{113}.$$

$$l=\frac{2a^2}{8}=\frac{2}{8}\left(64-\frac{1024}{113}\right)=\frac{1552}{113}\ \Longrightarrow\ 113l=1552.$$

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