Question

Prove that the point of intersection of two perpendicular tangents to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ lies on the circle $x^2+y^2=a^2-b^2$.

Answer 1

Equation of tangent to the ellipse $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

with slope $m$ is given by $y=mx\pm \sqrt{a^2{m}^2-b^2}$

$\Rightarrow y-mx=\pm \sqrt{a^2{m}^2-b^2}$

Now square both sides,

$\Rightarrow (y-mx{)}^2=a^2{m}^2-b^{2}2$

$\Rightarrow y^2+m^2{x}^2-2mxy=a^2{m}^2-b^2$

$\Rightarrow (x^2-a^2)m^2-\left(2xy\right)m+(y^2+b^2)=0$

For tangents to be perpendicular, slope of their tangent should be $-1$

i.e. Product of above quadratic $=-1$

$\Rightarrow \frac{y^2+b^2}{x^2-a^2}=-1$

$\Rightarrow x^2+y^2=a^2-b^2$