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Let the hyperbola $H : \frac{ x ^{2}}{ a ^{2}}- y ^{2}=1$ and the ellipse $E: 3 x^{2}+4 y^{2}=12$ 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
$42$
$40$
$36$
$47$
Solution
$\frac{ x ^{2}}{ a ^{2}}-\frac{ y ^{2}}{1}=1$
$e _{ H }=\sqrt{1+\frac{1}{ a ^{2}}} \quad \frac{ x ^{2}}{4}+\frac{ y ^{2}}{3}=1$
$\ell \cdot R .=\frac{2}{ a } \quad \ell R =\frac{2 \times 3}{\frac{2}{1-\frac{3}{4}}}=\frac{1}{2}$
$\frac{2}{ a }=3$
$a =\frac{2}{3}$
$e _{ H }=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$
$12\left( e _{ H }^{2}+ e _{ E }^{2}\right)=12\left(\frac{13}{4}+\frac{1}{4}\right)$
$=\frac{12 \times 14}{4}=42$
