Focal length of an ellipse and related results

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.

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Part I:

After parametrization $P(t)\equiv(ae+r\cos t,r\sin t)$:
$$\frac{(ae+r\cos t)^2}{a^2}+\frac{r^2\sin^2t}{b^2}=1$$
$$b^2a^2e^2+b^2r^2\cos^2t+2ab^2er\cos t+a^2r^2\sin^2t-a^2b^2=0\\
(a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r+(b^2a^2e^2-a^2b^2)=0\\
(a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r-b^4=0\quad(\color{red}{b^2=a^2(1-e^2)})$$
Now: $$r_1=l_1,r_2=-l_2\implies |l_1+l_2|=|r_1-r_2|=\left|\sqrt{(r_1+r_2)^2-4r_1r_2}\right|$$
$$|l_1+l_2|
=\sqrt{\left(\frac{2abe\cos t}{(a^2\sin^2t+b^2\cos^2t)}\right)^2+4\frac{(b^4)}{(a^2\sin^2t+b^2\cos^2t)}}\\=\sqrt{\frac{4a^2b^4e^2\cos^2t}{(a^2\sin^2t+b^2\cos^2t)^2}+4\frac{(b^4)}{(a^2\sin^2t+b^2\cos^2t)}}\\
=\frac{2b^2}{(a^2\sin^2t+b^2\cos^2t)}\sqrt{a^2e^2\cos^2t+a^2\sin^2t+\underbrace{b^2}_{b^2=a^2(1-e^2)}\cos^2t}\\
=\frac{2b^2}{(a^2\sin^2t+b^2\cos^2t)}\sqrt{a^2e^2\cos^2t+a^2\sin^2t+a^2\cos^2t-a^2e^2cos^2t}\\
\fbox{$\huge PQ=\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}$}
$$


Part II

$$(a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r-b^4=0$$

Now:
$$\frac{PS}{SQ}=\frac{l_1}{l_2}=\left|\frac{r_1}{r_2}\right|=\left|\frac{(r_1+r_2)+(r_1-r_2)}{(r_1+r_2)-(r_1-r_2)}\right|=\left|\frac{-\frac{(2ab^2e\cos t)}{(a^2\sin^2t+b^2\cos^2t)}+\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}}{-\frac{(2ab^2e\cos t)}{(a^2\sin^2t+b^2\cos^2t)}-\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}}\right|=\frac{1-e\cos t}{1+e\cos t}$$
Similiarly: $$\frac{PS’}{S’R}=\frac{1+e\cos u}{1-e\cos u}$$
Now take a triangle with vertices $P,S$ and the foot of perpendicular from P to Major Axis, then: $$\cos t=\frac{a\cos\theta-ae}{\sqrt{(a\cos\theta-ae)^2+b^2\sin^2\theta}}=\frac{\cos\theta-e}{1-e\cos\theta}$$
Similiarly:
$$\cos u=\frac{\cos\theta+e}{1+e\cos\theta}$$
Now:
$$\frac{PS}{SQ}=\frac{1-e\cos t}{1+e\cos t}=\frac{1-e\cos\theta-e(\cos\theta-e)}{1-e\cos\theta+e(\cos\theta-e)}=\frac{1-2e\cos\theta+e^2}{1-e^2}$$
Similiarly:
$$\frac{PS’}{S’R}=\frac{1+2e\cos\theta+e^2}{1-e^2}$$
So:
$$\frac{PS}{SQ}+\frac{PS’}{S’R}=\frac{1-2e\cos\theta+e^2}{1-e^2}+\frac{1+2e\cos\theta+e^2}{1-e^2}\\\fbox{$\huge\frac{PS}{SQ}+\frac{PS’}{S’R}=2\left(\frac{1+e^2}{1-e^2}\right)$}$$

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.