$\text{ A normal to the hyperbola }\frac{x^2}{4}-\frac{y^2}{1}=1\text{ has equal intercepts on the positive }x\text{ - and }y\text{ -axes. If this normal touches the }$

$\text{ ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\text{ then }{a}^2+b^2\text{ is equal to }$

$\text{The equation of the normal to the hyperbola}\frac{x^2}{4}-\frac{y^2}{1}=1$

$\begin{array}{l}\text{ at }(2\sec \theta ,\tan \theta )\text{ is }2x\cos \theta +y\cot \theta =5\text{ . }\\ \text{ The slope of the normal is }\end{array}$

$\begin{array}{l}-2\sin \theta =-1\\ \text{ or }\sin \theta =\frac{1}{2}\text{ or }\theta =\frac{\pi }{6}\\ Y\text{ -intercept of the normal }=\frac{5}{\cot \theta }=\frac{5}{\sqrt{3}}\end{array}$

$\begin{array}{l}Y\text{ -intercept of the normal }=\frac{5}{\cot \theta }=\frac{5}{\sqrt{3}}\\ \text{ As it touches the ellipse }\end{array}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\text{ we have }{a}^2+b^2=\frac{25}{3}\text{ (Using }{a}^2{m}^2+b^2=c^2\text{ ) }$