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

$$\frac{SQ}{SP}=1+\frac{4e^2\sin^2 x}{(1-e\cos x)^2}$$

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    $\begingroup$ Try drawing the figure. It might help. And also, you are supposed to show your attempt on the question. $\endgroup$ – Manish Kundu Feb 9 '18 at 17:06
  • $\begingroup$ Let equation of a ellipse be x^2 / a^2 + y^2 / b^2=1 having focus (a e, 0). Suppose vertex of parabola is P(x', y'). The equation of the parabola according to the problem is (y-y')^2 = 4a e (x-x'). Since P(x', y') lies on the ellipse so x'^2 / a^2 + y'^2 / b^2=1. $\endgroup$ – Bal G Srivastava Feb 9 '18 at 17:15
  • $\begingroup$ Please edit that into the question instead of adding a comment. $\endgroup$ – amd Feb 9 '18 at 18:30
  • $\begingroup$ Since you’ve fixed the ellipse in standard position, why are you also assuming that the parabola’s axis is parallel to the $x$-axis? $\endgroup$ – amd Feb 9 '18 at 22:05

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}

  • $\begingroup$ Thanks a lot... $\endgroup$ – Bal G Srivastava Feb 10 '18 at 13:40

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