Question

Prove that perpendicular focal chords of a rectangular hyperbola are equal.

Answer 1

Equation of any conic in polar form is

$\frac{l}{r}=1-e\cos \theta \Rightarrow r=\frac{l}{1-e\cos \theta }$

For rectangular hyperbola $e=\sqrt{2}$

$\Rightarrow r=\frac{l}{1-\sqrt{2}\cos \theta }$

Let $PS{P}^{\prime }$ be a focal chord with vectorical angle of $P$ at $\alpha$ then vectorical angle of $P^{\prime }$ w.r.t to $S$ is $(\pi +\alpha )$

$\Rightarrow SP=\frac{l}{1-\sqrt{2}\cos \alpha }$ and $S{P}^{\prime }=\frac{l}{1-\sqrt{2}\cos (\pi +\alpha )}$

$\Rightarrow PS{P}^{\prime }=\frac{l}{1-\sqrt{2}\cos \alpha }+\frac{l}{1-\sqrt{2}\cos (\pi +\alpha )}\Rightarrow PS{P}^{\prime }=\frac{l}{1-\sqrt{2}\cos \alpha }+\frac{l}{1+\sqrt{2}\cos \alpha }\Rightarrow PS{P}^{\prime }=\frac{2l}{1-2{\cos }^2\alpha }=\frac{-2l}{\cos 2\alpha }$

Lenght of perpendicular $QS{Q}^{\prime }$ is obtained by replacing $\alpha$ by $(\frac{\pi }{2}+\alpha )$

$\Rightarrow QS{Q}^{\prime }=\frac{-2l}{\cos 2(\frac{\pi }{2}+\alpha )}=\frac{-2l}{\cos (\pi +2\alpha )}\Rightarrow QS{Q}^{\prime }=\frac{-2l}{-\cos 2\alpha }=\frac{2l}{\cos 2\alpha }$

Clearly $|PS{P}^{\prime }|=|QS{Q}^{\prime }|$

Hence proved.