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

Answer: 8

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 $y^2=0$
$$\begin{gathered} \frac{{\cos }^2\alpha }{p^2}-\frac{1}{4}+\frac{{\sin }^2\alpha }{p^2}+\frac{1}{8}=0 \\ <br/>\frac{1}{p^2}=\frac{1}{8}\Rightarrow p^2=8 \end{gathered}$$
$\therefore$ required $R^2=8$