Question

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

Answer: (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 normal has equal intercepts on the positive $x$-axis and the $y$-axis. Hence, the slope of the normal is $-2\sin \theta =-1$
$\Rightarrow \sin \theta =\frac{1}{2}\Rightarrow \theta =\frac{\pi }{6}$
Y-intercept of the normal $=\frac{5}{\cot \theta }=\frac{5}{\sqrt{3}}$
As it touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow a^2+b^2=\frac{25}{3}$ (Since $c^2=a^2{m}^2+b^2$)