Points $P_i$ on an ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ Consider an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with $O$ as the origin. $n$ points denoted as $P_1,P_2,\cdots$ are taken on the ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ where $i\in(1,n-1)$. Find the value of: $$\sum_{i=1}^n\frac{1}{OP_i^2}$$
I took $P_1$ as $(3,0)$. Let other points be $P_i(3\cos\theta_i,2\sin\theta_i)$.
$$\tan\frac{\pi}{n}=\frac{2}{3}\tan\theta_2$$
$$\tan\frac{2\pi}{n}=\frac{2}{3}\tan\theta_3$$
and so on.
$$OP_1^2=a^2$$
$$OP_2^2=9\cos^2\theta_2+4\sin^2\theta_2=9\cos^2\theta_2\left(1+\tan^2\frac{\pi}{n}\right)$$
$$OP_3^2=9\cos^2\theta_3+4\sin^2\theta_3=9\cos^2\theta_3\left(1+\tan^2\frac{2\pi}{n}\right)$$
and so on.
But I am not able to compute the required sum. 

Better to set up a system of two equations:
$$
3\cos\theta_2=r_2\cos(\pi/n)
\quad\hbox{and}\quad 
2\sin\theta_2=r_2\sin(\pi/n),
$$
where $r_2=OP_2$.
From that you get, after eliminating $\theta_2$:
$$
{1\over r_2^2}={\cos^2(\pi/n)\over9}+{\sin^2(\pi/n)\over4},
$$
and in general:
$$
{1\over r_{k+1}^2}={\cos^2(k\pi/n)\over9}+{\sin^2(k\pi/n)\over4}=
{1\over4}-{5\over36}\cos^2{k\pi\over n}.
$$
Using $\cos^2\alpha=(1/2)(\cos2\alpha+1)$ we can rewrite that formula as:
$$
{1\over r_{k+1}^2}=
{13\over72}-{5\over72}\cos{2k\pi\over n},
$$
so that
$$
\sum_{k=0}^{n-1}{1\over r_{k+1}^2}=
{13\over72}n-{5\over72}\sum_{k=0}^{n-1}\cos{2k\pi\over n}.
$$
But the last sum vanishes by symmetry, so the result is $13n/72$.
