TITLE: Focal length of an ellipse and related results
QUESTION [8 upvotes]: There are 2 questions(part of same question but I divided it into two):

Q1. Prove that the length of the focal chord of the ellipse $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ which is inclined to the major axis at an angle $\theta$ is $$\frac {2ab^2}{a^2\sin^2\theta+b^2\cos^2\theta}$$ I tried to solve this using the parametric form of a line, i.e., $(x,y)=(ae+r\cos\theta,r\sin\theta)$, plugging this into the given equation to find $r_1-r_2$ which is giving a different solution.

Q2. If $PSQ$ and $PS'R$ are two chords of an ellipse through its two focii S and S', then prove that $$\frac {PS}{SQ}+\frac {PS'}{S'R}=2\left(\frac {1+e^2}{1-e^2}\right)$$ Here I fixed $P \equiv (a\cos\theta,b\sin\theta)$ and found out the parametric angles of $Q$ and $R$ using the result $$\tan\theta_1/2*\tan\theta_2/2=\left(\frac {e-1}{e+1}\right)^{\pm1}$$, $+$ when chord passes through $S$ and $-$ when chord passes through $S'$. But it gives a very ugly expression. So how do I do it  in an easy way? Hints are welcome.

REPLY [2 votes]: First part
Using Newton polar form focal ray length of radius vector, total length is much easier to calculate.
$$  =\frac{p}{1- e \,cos \theta }+   \frac{p}{1+ e \,cos  \theta },\, e^2 = 1- (b/a)^2, p = b^2/a $$
and simplify to obtain the required result.
Second part
...also along similar lines with quotients instead of focal ray part sum.