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 this normal touches the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,$ then the value $\frac{9}{25}({\mathrm{a}}^2+{\mathrm{b}}^2)=$

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{4}-\frac{{\mathrm{y}}^2}{1}=1 \\ \Rightarrow {\mathrm{a}}^2=4, {\mathrm{b}}^2=1 \\ \mathrm{Equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P}\left(\mathrm{\theta }\right) \mathrm{is} \\ \frac{\mathrm{ax}}{\mathrm{sec\theta }}+\frac{\mathrm{by}}{\mathrm{tan\theta }}={\mathrm{a}}^2+{\mathrm{b}}^2 \\ \Rightarrow \frac{2\mathrm{x}}{\mathrm{sec\theta }}+\frac{\mathrm{y}}{\mathrm{tan\theta }}=5 \\ \Rightarrow 2\mathrm{x} \mathrm{cos\theta }+\mathrm{y} \mathrm{cot\theta }=5 — ① \\ \mathrm{Normal} \mathrm{has} \mathrm{equal} \mathrm{intercefts} \mathrm{on} \mathrm{posotive} \mathrm{x}-\mathrm{axis} \mathrm{and} \mathrm{y}-\mathrm{axis}. \\ \mathrm{Hence} \mathrm{slope} \mathrm{of} \mathrm{normal} \mathrm{is} -2\mathrm{sin\theta }=-1 \\ \Rightarrow \mathrm{sin\theta }=\frac{1}{2} \Rightarrow \mathrm{\theta }\frac{\mathrm{\pi }}{6} \\ ① \Rightarrow 2\mathrm{x}\left(\frac{\sqrt{3}}{1}\right)+\mathrm{y}\left(\sqrt{3}\right)=5 \\ \Rightarrow \sqrt{3}\mathrm{x}+\sqrt{3}\mathrm{y}-5=0 \\ \mathrm{If} \mathrm{it} \mathrm{touches} \mathrm{ellipse} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{then} n^2={\mathrm{a}}^2{\mathrm{l}}^2+{\mathrm{b}}^2{\mathrm{m}}^2 \\ \Rightarrow 25={\mathrm{a}}^2\left(3\right)+{\mathrm{b}}^2\left(3\right) \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=\frac{25}{3} \\ \frac{9}{25}({\mathrm{a}}^2+{\mathrm{b}}^2)=\frac{9}{25}\times \frac{25}{3}=3 \end{gathered}$$