$\text{ A normal to the hyperbola }\frac{x^2}{6}-\frac{y^2}{2}=1\text{ has equal intercepts on positive x and y axes. If this normal touches}$

$\text{ the ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\text{ then the value of }\left(a^2+b^2\right)\text{ is }$

$\text{The equation of normal to hyperbola}$

$\begin{array}{l}\frac{x^2}{6}-\frac{y^2}{2}=1\text{ at }(\sqrt{6}\sec \theta ,\sqrt{2}\tan \theta )\text{ is }\\ \frac{\sqrt{6}x}{\sec \theta }+\frac{\sqrt{2}y}{\tan \theta }=8\end{array}$

$\begin{array}{l}\text{ Slope }=-1\\ \Rightarrow \frac{-\sqrt{6}}{\sec \theta }\times \frac{\tan \theta }{\sqrt{2}}=-1\\ \Rightarrow \sin \theta =\frac{1}{\sqrt{3}}\end{array}$

$\begin{array}{l}\text{ Thus, equation of normal is }x+y=4\\ \text{ So, }y=-x+4\text{ is tangent to }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ \text{ Hence, }\left(a^2+b^2\right)=16\end{array}$