Question

A variable chord of a hyperbola $\frac{x^2}{4}-\frac{y^2}{8}=1$ subtends a right angle at the centre of hyperbola. If the chord of hyperbola touches a fixed circle of radius $R$ which is concentric with hyperbola then find $R ^2$

Answer 1

Let variable chord be $x \cos \alpha+ y \sin \alpha= p$
homogenizing the hyperbola
$\frac{x^2}{4}-\frac{y^2}{8}=\frac{(x \cos \alpha+y \sin \alpha)^2}{p^2}$
Now, coefficient of $x^2+$ coefficient of $y^2=0$
$\frac{\cos ^2 \alpha}{ p ^2}-\frac{1}{4}+\frac{\sin ^2 \alpha}{ p ^2}+\frac{1}{8}=0 \\
\frac{1}{ p ^2}=\frac{1}{8} \Rightarrow p ^2=8$
$\therefore $ required $R ^2=8$