Question

A normal to the hyperbola $\frac{x^2}{6}-\frac{y^2}{2}$ has equal intercepts on positive $x$ and $y$-axis. If this normal touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then find the value of $\frac{a^2+b^2}{4}$.

Answer 1

The equation of normal at $(\sqrt{6} \sec \theta, \sqrt{2} \tan \theta)$ is $\frac{\sqrt{6} x}{\sec \theta}+\frac{\sqrt{2} y}{\tan \theta}=8$
$\text { Slope }=-1 \Rightarrow \frac{-\sqrt{6}}{\sec \theta} \times \frac{\tan \theta}{\sqrt{2}}=-1 $
$\sin \theta=\frac{1}{\sqrt{3}} $
$N: x+y=4$
Now, $y=-x+4$ is tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Hence, $\left(a^2+b^2\right)=16$