JEE Main 2024 (Online) 9th April Evening ShiftQuestion 1
Let the foci of a hyperbola $H$ coincide with the foci of the ellipse $E:\frac{(x-1)^{2}}{100}+\frac{(y-1)^{2}}{75}=1$ and the eccentricity of the hyperbola $H$ be the reciprocal of the eccentricity of the ellipse $E$. If the length of the transverse axis of $H$ is $\alpha$ and the length of its conjugate axis is $\beta$, then $3\alpha^{2}+2\beta^{2}$ is equal to
Hyperbola
Solution
Ellipse: $a_E^2=100$, $b_E^2=75$, $c=5$, $e_E=\tfrac12$. Hyperbola: $e=2$, $ae=5\Rightarrow a=\dfrac52$. $\alpha=2a=5$; $b^2=a^2(e^2-1)=\dfrac{75}{4}$, $\beta^2=4b^2=75$. $3\alpha^2+2\beta^2=75+150=225$.