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.
Best Answer
You have chosen an orientation of your tangent. Your expressions are the cosines of the angles between the tangent (with given orientation) and the vectors joining the foci to $p$. If you draw a picture you will convince yourself that those angles have to be supplementary in order for your property to be true. Therefore, what you need is that their cosines be opposite, not equal. And that is what you got.