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 $a^2+b^2$ is equal to

• A. $5$
• B. $25$
• C. $16$
• D. None of these

Answer 1

The correct option is D None of these

The equation of the normal to the hyperbola
$\frac{x^2}{4}-\frac{y^2}{1}=1$
at $(2\sec \theta ,\tan \theta )$ is $2x\cos \theta +y\cot \theta =5$.
The slope of the normal is
$-2\sin \theta =-1$
or $\sin \theta =\frac{1}{2}$ or $\theta =\frac{\pi }{6}$
Y - intercept of the normal $=\frac{5}{\cot \theta }=\frac{5}{\sqrt{3}}$
As it touches the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
we have $a^2+b^2=\frac{25}{9}$