JEE Main 2022 (Online) 28th July Morning ShiftQuestion 1
For the hyperbola $H:x^2-y^2=1$ and the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b>0$, let the (1) eccentricity of $E$ be reciprocal of the eccentricity of $H$, and (2) the line $y=\sqrt{\frac52}x+K$ be a common tangent of $E$ and $H$. Then $4(a^2+b^2)$ is equal to ________.
Hyperbola
Answer$3$
Solution
$e_H=\sqrt2$, so $e_E=\frac1{\sqrt2}$ and $b^2=\frac{a^2}2$. Tangent $y=mx+K$ to $x^2-y^2=1$: $K^2=m^2-1=\frac32$. Tangent to the ellipse: $K^2=a^2m^2+b^2=3a^2$, so $a^2=\frac12$, $b^2=\frac14$ and $4(a^2+b^2)=3$.