A conic touches the ellipse $1/r=1+e\cos\theta$ at the point $\theta$= α. They have a common focus at the pole and have the same eccentricity. Show that the length of the latus rectum is $$\frac{2l(1-e^2)}{e^2+2e\cos\alpha+1}$$
My attempt: Since they have the same focus and they touch each other at $\theta=\alpha$ then equation of the tangent for the conic section at alpha will be $l/r =e\cos\theta+\cos(\theta-\alpha)$ and equation of the tangent for the ellipse will be $l/r =e\cos\theta+\cos(\theta-\alpha)${equation of ellipse is $l/r=1+e\cos\theta$}.
Am I on the correct path? What will be my next step? I have to prove that length of latus rectum is $\frac{2l(1-a^2)}{e^2+2e\cos\alpha+1}$.
If $P$ is the tangency points of the ellipses, then the normal at $P$ must be the same for both ellipses, and this is also the bisector of the angle of vertex $P$ subtending the foci. As the ellipses have a focus ($F$) in common, that means that the other foci ($G$ and $H$ in figure below) must be aligned with $P$.
But the ellipses are also similar (because they have the same eccentricity), hence triangles $FPG$ and $FPH$ are similar and: $$ PH:PF=PF:PG. $$ From that, it is not difficult to find the required expression for the latus rectum $2l'$ of the second ellipse. Setting: $2a=PF+PG$, $2a'=PF+PH$, $r=PF$, from the above proportion one gets: $$ 2a'={2ar\over2a-r}. $$ Hence: $$ 2l'=(1-e^2)2a'=(1-e^2){2ar\over2a-r}. $$ Inserting here $a=l/(1-e^2)$ and $r=l/(1+e\cos\alpha)$ one gets the desired result.