Question

The tangents at two points, P and Q, of a conic meet in T, and S is the focus ; prove that if the conic be a parabola, then $S{T}^2=SP.SQ.$

Answer 1

Equation of parabola in polar form is

$\frac{2a}{r}=1-\cos \theta$

$\Rightarrow \frac{2a}{r}=2{\sin }^2\frac{\theta }{2}\Rightarrow r=a{\csc }^2\frac{\theta }{2}$

Let the vectorical angle of $P$ and $Q$ be $\alpha$ and $\beta$ respectively

$\Rightarrow SP=a{\csc }^2\frac{\alpha }{2}$ and $SQ=a{\csc }^2\frac{\beta }{2}........\left(i\right)$

Tangents at $P$ and $Q$ intersect at $T$ , let its polar coordinates be $(r^{\prime },\varphi )$

$\Rightarrow \frac{2a}{r}=\cos (\varphi -\alpha )-\cos \varphi ......\left(ii\right)\Rightarrow \frac{2a}{r}=\cos (\varphi -\beta )-\cos \varphi ......\left(iii\right)$

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

$\Rightarrow \cos (\varphi -\alpha )=\cos (\varphi -\beta )\Rightarrow \varphi -\alpha =\varphi -\beta \Rightarrow \varphi =\frac{\alpha +\beta }{2}$

substituting in $\left(ii\right)$

$\Rightarrow \frac{2a}{r^{\prime }}=\cos (\frac{\alpha +\beta }{2}-\alpha )-\cos \frac{\alpha +\beta }{2}\Rightarrow \frac{2a}{r^{\prime }}=\cos \frac{\alpha -\beta }{2}-\cos \frac{\alpha +\beta }{2}\Rightarrow \frac{2a}{r^{\prime }}=2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\Rightarrow \frac{a}{ST}=\sin \frac{\alpha }{2}\sin \frac{\beta }{2}\Rightarrow ST=a\csc \frac{\alpha }{2}\csc \frac{\beta }{2}\Rightarrow {ST}^2=a^2{\csc }^2\frac{\alpha }{2}{\csc }^2\frac{\beta }{2}\Rightarrow {ST}^2=\left(a{\csc }^2\frac{\alpha }{2}\right)\left(a{\csc }^2\frac{\beta }{2}\right)$

using $\left(i\right)$

$\Rightarrow ST=SP.SQ$

Hence proved.