 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),On the ellipse this sum is normal to the tangent. Thus a unit tangent $\mathbf{\tau }$ obeys $\mathbf{\tau }\cdot {\mathbf{u}}_A=-\mathbf{\tau }\cdot {\mathbf{u}}_B$. The magnitudes of the projections of the two unit focal directions on the tangent are equal, provingfor 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 ${\mathbf{u}}_A$ towhich points from $P$ toward $B$. Thus the equal-angle property has the correct incoming/outgoing ray orientation, rather than merely an algebraic bisector assertion.