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 t

Question

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

Professor PC Thomas Classes · Answer

(42) : The given equation of hyperbola and ellipse can be written as,\ $H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{1^{2}}=1$ and $E: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$\ Now, $\frac{2(1)^{2}}{a}=\frac{2(3)}{2}$\ $\left[\begin{array}{l}\because 	ext { Length of latus rectum of both } \ 	ext { hyperbola and ellipse are equal }\end{array}ight]$\ $\Rightarrow a=\frac{2}{3}$ So, $e_{H}^{2}=1+\frac{b^{2}}{a^{2}}=1+\frac{9}{4}=\frac{13}{4}$\ and $e_{E}^{2}=1-\frac{b^{2}}{a^{2}}=1-\frac{3}{4}=\frac{1}{4}$\ So, $12\left(e_{H}^{2}+e_{E}^{2}ight)=12 \cdot \frac{14}{4}=42$\