JEE Main 2022 — Hyperbola Question with Solution

Given $H:x^2-y^2=1,E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$e_H=\sqrt{2} \& e_E=\frac{1}{e_H}=\frac{1}{\sqrt{2}}$

For hyperbola  $e^2=1-\frac{b^2}{a^2}=\frac{1}{2}\Rightarrow \frac{b^2}{a^2}=\frac{1}{2}$

Also given that the common tangent of $H \& E$ is $y=\sqrt{\frac{5}{2}}x+k$ $\left(\mathrm{i}.\mathrm{e}.m=\sqrt{\frac{5}{2}}\right)$

We know that the condition for common tangent of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and hyperbola $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$ is $a^2{m}^2+b^2=A^2{m}^2-B^2$

Now, for common tangency: $\frac{5}{2}{a}^2+b^2=\frac{5}{2}-1$

$\frac{5}{2}+\frac{b^2}{a^2}=\frac{3}{2a^2}\Rightarrow a^2=\frac{1}{2}$

$\therefore a^2+b^2=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$

$\Rightarrow 4\left(a^2+b^2\right)=3$