Conic section(parabola and ellipse) An ellipse and a parabola have a common focus $S$ and intersect in two real points $P$ and $Q$, of which $P$ is the vertex of the parabola. If $e$ be the eccentricity of the ellipse and $x$, the angle which $SP$ makes with the major axis, prove that 
$$\frac{SQ}{SP}=1+\frac{4e^2\sin^2 x}{(1-e\cos x)^2}$$

Let the parabola be $r=\dfrac{a}{1+\cos (\theta-x)}$, then
$$\theta=x \implies r=\frac{a}{2}=SP$$
Now the the ellipse is
$$r=\frac{a(1+e\cos x)}{2(1+e\cos \theta)}$$
Take $Q$ as $\theta=y$,
$$SQ=\frac{a(1+e\cos x)}{2(1+e\cos y)}=\frac{a}{1+\cos (y-x)}$$
Let $(X,Y)=(\cos x,\cos y)$, then
$$\frac{1+eX}{1+eY} = \frac{2}{1+XY \pm \sqrt{1-X^2}\sqrt{1-Y^2}}$$
Using Mathematica,
\begin{align}
  \frac{1+eX}{1+eY} &= \frac{1+4e^2-2eX-3e^2X^2}{(1-eX)^2} \\
  &= \frac{(1-eX)^2+4e^2(1-X^2)}{(1-eX)^2} \\
  \frac{SQ}{SP} &= 1+\frac{4e^2\sin^2 x}{(1-e\cos x)^2}
\end{align}
