Let $f_1$ and $f_2$ be the focis $(\epsilon p, 0)$, $(-\epsilon p, 0)$ of an ellipse $(p\cos(t), q\sin(t))$, where $\epsilon = \sqrt{1 - (q^2/p^2)}$. Then the tangent vector to the ellipse is given by $\gamma'(t) = (-p\sin(t), q\cos(t))$. Let $x$ be any point on the ellipse. We'd like to show that the angle between the line joining $f_1$ and $x$ and the tangent to the ellipse at $x$ is equal to the angle between the line joining $f_2$ and $x$ and the said tangent at $x$.
My source claims that we only need to show that $\frac{(p - f_1)\cdot n}{||p - f_1||} = \frac{(p - f_2)\cdot n}{||p - f_2||}$, where $n = (q\cos(t), p\sin(t))$ is the normal vector to the tangent at $x$. What I don't understand is that why are we using the normal vector to the tangent, when the tangent vector with which we ought to find the angle between vectors $a_i = p - f_i, i \in \{1, 2\}$?
Please see the diagram. Use the fact that the normal vector at any point on ellipse is perpendicular to the tangent line at that point. So if the normal vector bisects $ \angle f_1Pf_2$, $f_1P$ and $f_2P$ will make equal angles with the tangent line.