Question

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

Answer 1

The equation of the normal to the hyperbola $\frac{x^2}{4}-\frac{y^2}{1}=1$ at $(2\sec \theta ,\tan \theta )$ is $2x\cos \theta +y\cot \theta =5...\left(i\right)$

Normal has equal intercepts on positive $x-$and $y-$axes.
Thus slope of the normal is $-2\sin \theta =-1$
$\Rightarrow \sin \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{6},\frac{5\pi }{6}$
But $\theta \ne \frac{5\pi }{6}$ as normal lies in $Q_1$
From $\left(i\right)$, Equation of line will be $y=-x+\frac{5}{\sqrt{3}}$
As it touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$
we have $c^2=a^2{m}^2+b^2$
Here $c=\frac{5}{\sqrt{3}},m=-1$
$\Rightarrow \frac{25}{3}=a^2(-1{)}^2+b^2$
$\Rightarrow \frac{25}{3}=a^2+b^2$
$\Rightarrow 3(a^2+b^2)=25$