Question

The number of points on hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ from where mutually perpendicular tangents can be drawn to the circle $x^2+y^2=a^2$ is

Answer 1

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Rightarrow$Hyperbola
$x^2+y^2=a^2\Rightarrow$Circle
Required points will lie on the intersection of hyperbola with director circle
Equation of director circle: $x^2+y^2=a^2-b^2⟶1$
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{x^2}{a^2}=\frac{b^2+y^2}{b^2}$
$x^2=\frac{a^2}{b^2}[b^2+y^2]$
Put value of $x^2$ in equation 1
$\frac{a^2}{b^2}[b^2+y^2]+y^2=a^2-b^2$
$a^2{b}^2+a^2{y}^2+b^2{y}^2=a^2{b}^2-b^4$
$y^2=\frac{-b^4}{a^2+b^2}$
$y=\pm i{b}^2\left[\frac{1}{\sqrt{a^2+b^2}}\right]$
$\therefore$ We get 2 points