A normal to the hyperbola  $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{1}=1$has equal intercepts on positive x-and y-axes. If the normal touches the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ and if ${\mathrm{a}}^2+{\mathrm{b}}^2$ is k, then $\sqrt{3k}$ is equal to

Normal to $\frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{1}=1$ is $\frac{2\mathrm{x}}{\sec \mathrm{\theta }}+\frac{\mathrm{y}}{\tan \mathrm{\theta }}=5$,

Given

 $$\begin{gathered} \frac{5}{2}\sec \mathrm{\theta }=5\tan \mathrm{\theta } \\ \Rightarrow \sec \mathrm{\theta }=2\tan \mathrm{\theta } \\ \therefore \mathrm{\theta }=\frac{\mathrm{\pi }}{6}\text{, } \end{gathered}$$

Now equation of normal is  $\sqrt{3}\mathrm{x}+\sqrt{3}\mathrm{y}=5$

$\Rightarrow y=-x+\frac{5}{\sqrt{3}}$

But this touches $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

then

$$\begin{gathered} \frac{25}{3}=(a^2+b^2) \\ \Rightarrow a^2+b^2=\frac{25}{3} \\ By comparision k=\frac{25}{3} \\ Now \sqrt{3k}=5 \end{gathered}$$