Question

Let the hyperbola $H:\frac{x^2}{a^2}-y^2=1$ and the ellipse $E:3x^2+4y^2=12$ be such that the length of latus rectum of $H$ is equal to the length of latus rectum of $E$. If $e_H$ and $e_E$ are the eccentricities of $H$ and $E$ respectively, then the value of $12\left(e_H^2+e_E^2\right)$ is equal to ______

Answer 1

Answer: 42

$\frac{x^2}{a^2}-\frac{y^2}{1}=1$
$\frac{x^2}{4}+\frac{y^2}{3}=1$
$e_H=\sqrt{1+\frac{1}{a^2}}$
$e_E=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
$\ell \cdot R.=\frac{2}{a}\ell R=\frac{2\times 3}{2}=3$
$\frac{2}{a}=3$
$a=\frac{2}{3}$
$e_H=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$
$12\left(e_H^2+e_E^2\right)=12\left(\frac{13}{4}+\frac{1}{4}\right)$
$=\frac{12\times 14}{4}=42$