JEE Main 2022 (Online) 24th June Evening ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Hyperbola
Answer$42$
Solution
Ellipse $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$: latus rectum $=\dfrac{2\cdot3}{2}=3$, $e_E^2=1-\dfrac34=\dfrac14$. Hyperbola: latus rectum $=\dfrac{2\cdot1}{a}=3\Rightarrow a=\tfrac23$, $e_H^2=1+\dfrac{1}{a^2}=\dfrac{13}{4}$. $12\left(\dfrac{13}{4}+\dfrac14\right)=42$.