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. 
 A: 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 $13n/72$.
A: If we set $P_i=(x_i,y_i)=\left(3\cos\theta_i,2\sin\theta_i\right)$ we have 
$$\vartheta_0+\frac{\pi(i-1)}{n}=\arctan\frac{y_1}{x_i}=\arctan\left(\frac{2}{3}\tan\theta_i\right)$$
from which:
$$ \tan\theta_i = \frac{3}{2}\tan\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) $$
and
$$ OP_i^2 = 9\cos^2\theta_i+4\sin^2\theta_i = 4+5\cos^2\theta_i=\frac{9+4 \tan^2\theta_i}{1+\tan^2\theta_i}$$
gives that the wanted sum is:
$$ \sum_{i=1}^{n}\frac{1+\tan^2\theta_i}{9+4\tan^2\theta_i}=\sum_{i=1}^{n}\frac{1+\frac{9}{4}\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) }{9+9\tan^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) } $$
or:
$$ \frac{1}{36}\sum_{i=1}^{n}\left[4\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) +9\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right) \right]$$
that can be computed by exploiting the duplication formulas for $\sin$/$\cos$, De Moivre's formula and geometric series, or simply by noticing that it does not really depend on $\vartheta_0$ by differentiation with respect to $\vartheta_0$:
$$\forall\, n,\vartheta_0,\qquad \sum_{i=1}^{n}\sin^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)=
\sum_{i=1}^{n}\cos^2\left(\vartheta_0+\frac{\pi(i-1)}{n}\right)=\color{red}{\frac{n}{2}}\tag{1}$$
gives:

$$ \sum_{i=1}^{n}\frac{1}{OP_i^2} = \color{red}{\frac{13n}{72}}.\tag{2}$$

