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

Answer: 4

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, $\left(a^2+b^2\right)=16$