Suppose we have an ellipse $\frac{x^2}{p^2}+\frac{y^2}{q^2}=1$ with parametrization $\gamma(t)=(p\cos(t),q\sin(t))$. Let $\vec{p}= \gamma(t_0)=(p\cos(t_0),q\sin(t_0))$ be any point on the ellipse. Let the foci of the ellipse be $\vec{f_{1,2}}=(\pm ep,0)$ where $e=\sqrt{1-\frac{q^2}{p^2}}$. Let $l_1$ be the line joining $\vec{f_1}$ to $\vec{p}$ i.e. $$l_1=(p\cos(t_o)-ep, q\sin(t_0))$$ and let $l_2$ be the line joining $\vec{f_2}$ to $\vec{p}$ i.e. $$l_2=(p\cos(t_0)+ep, q\sin(t_0))$$ I want to show that $l_1$ and $l_2$ make equal angles with the tangent vector $\dot\gamma(t_0)=(-p\cos(t_0),q\sin(t_0)$ at the point $\vec{p}$. In other words I want to show $$\frac{l_1\cdot\dot\gamma(t_0)}{|l_1|\cdot|\dot\gamma(t_0)|}=\frac{l_2\cdot\dot\gamma(t_0)}{|l_2|\cdot|\dot\gamma(t_0)|}$$
After much simplification of the above expression, I get that $$1=-1$$
Since they only differ by a sign, does this show that the lines do indeed form equal angles with the tangent line at $\vec{p}$ but have slopes of opposite sign?
Thanks in advance for any help.
