= Solution
For <reflection between the foci of an ellipse>, assume $s>|A-B|$ and define $\Phi(P)=R_A(P)+R_B(P)$. From part (b),
$$
\nabla\Phi=\mathbf u_A+\mathbf u_B,\qquad\mathbf u_A=\frac{P-A}{|P-A|},\quad\mathbf u_B=\frac{P-B}{|P-B|}.
$$
On the ellipse this sum is normal to the tangent. Thus a unit tangent $\boldsymbol\tau$ obeys $\boldsymbol\tau\cdot\mathbf u_A=-\boldsymbol\tau\cdot\mathbf u_B$. The magnitudes of the projections of the two unit focal directions on the tangent are equal, proving
$$
\boxed{\angle(AP,T)=\angle(BP,T)}
$$
for the corresponding acute, unoriented line angles.
More explicitly, let $\mathbf n=(\mathbf u_A+\mathbf u_B)/|\mathbf u_A+\mathbf u_B|$. The <law of reflection> maps an incoming unit ray $\mathbf u_A$ to
$$
\mathbf u_A-2(\mathbf u_A\cdot\mathbf n)\mathbf n=\mathbf u_A-(\mathbf u_A+\mathbf u_B)=-\mathbf u_B,
$$
which points from $P$ toward $B$. Thus the equal-angle property has the correct incoming/outgoing ray orientation, rather than merely an algebraic bisector assertion.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2013-ellipse-reflected-ray-focusing.png]
{title=Calculated equal-length reflected ray paths between the foci of an ellipse}
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