Question

PSP' and QSQ' are two perpendicular focal chords of a conic; prove that $\frac{1}{PS.S{P}^{\prime }}+\frac{1}{QS.S{Q}^{\prime }}$ is constant.

Answer 1

Equation of any conic in polar form is

$\frac{l}{r}=1-e\cos \theta \Rightarrow \frac{1}{r}=\frac{1-e\cos \theta }{l}.....\left(i\right)$

If the vectorical angle of focal chord $PS{P}^{\prime }$ is $\alpha$ then vectorical angle of perpendicular focal chord $QS{Q}^{\prime }$ is $(\alpha +\frac{\pi }{2})$

Now if vectorical angle of $P$ is $\alpha$ then $P^{\prime }$ is $(\pi +\alpha )$

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

$\Rightarrow \frac{1}{SP.S{P}^{\prime }}=\frac{1-e\cos \alpha }{l}\times \frac{1-e\cos (\pi +\alpha )}{l}\Rightarrow \frac{1}{SP.S{P}^{\prime }}=\frac{1-e\cos \alpha }{l}\times \frac{1+e\cos \alpha }{l}\Rightarrow \frac{1}{SP.S{P}^{\prime }}=\frac{1-e^2{\cos }^2\alpha }{l^2}.....\left(ii\right)$

Also the vectorical angle of $Q$ is $(\alpha +\frac{\pi }{2})$ then $Q^{\prime }$ will be $(\alpha +\frac{\pi }{2}+\pi )$

$\Rightarrow \frac{1}{SQ}=\frac{1-e\cos (\alpha +\frac{\pi }{2})}{l}$ and $\frac{1}{S{Q}^{\prime }}=\frac{1-e\cos (\alpha +\frac{\pi }{2}+\pi )}{l}$

$\Rightarrow \frac{1}{SQ.S{Q}^{\prime }}=\frac{1+e\sin \alpha }{l}\times \frac{1-e\sin \alpha }{l}\Rightarrow \frac{1}{SQ.S{Q}^{\prime }}=\frac{1-e^2{\sin }^2\alpha }{l^2}.....\left(iii\right)$

Adding $\left(ii\right)$ and $\left(iii\right)$

$\Rightarrow \frac{1}{SP.S{P}^{\prime }}+\frac{1}{SQ.S{Q}^{\prime }}=\frac{1-e^2{\cos }^2\alpha }{l^2}+\frac{1-e^2{\sin }^2\alpha }{l^2}\Rightarrow \frac{1}{SP.S{P}^{\prime }}+\frac{1}{SQ.S{Q}^{\prime }}=\frac{2-e^2({\cos }^2\alpha +{\sin }^2\alpha )}{l^2}\Rightarrow \frac{1}{SP.S{P}^{\prime }}+\frac{1}{SQ.S{Q}^{\prime }}=\frac{2-e^2}{l^2}$

R.H.S is constant

Hence proved.