 Solution

Let $r_A=|\mathbf{P}-\mathbf{A}|$, $r_B=|\mathbf{P}-\mathbf{B}|$, and define the outward-from-focus unit vectors ${\mathbf{u}}_A=(\mathbf{P}-\mathbf{A})/r_A$ and ${\mathbf{u}}_B=(\mathbf{P}-\mathbf{B})/r_B$. The ellipse is the nondegenerate level curve $r_A+r_B=s$, with $s>|\mathbf{A}-\mathbf{B}|$. The previous gradient result givesIf $\mathbf{t}$ is a unit tangent at $P$, orthogonality to the level-curve gradient impliesTherefore $|\mathbf{t}\cdot {\mathbf{u}}_A|=|\mathbf{t}\cdot {\mathbf{u}}_B|$. These absolute dot products are the cosines of the acute angles made by the two focal lines with the unoriented tangent line. Thus those angles are equal.

Equivalently, the outward unit normal is $\mathbf{n}=({\mathbf{u}}_A+{\mathbf{u}}_B)/|{\mathbf{u}}_A+{\mathbf{u}}_B|$, the bisector of the two unit focal directions. Its denominator is nonzero for a nondegenerate ellipse. Using $|{\mathbf{u}}_A|=|{\mathbf{u}}_B|=1$, the specular reflection formula becomesThis algebra proves the equal-angle statement with the incident and reflected ray directions correctly oriented, rather than confusing the line $BP$ with the outgoing direction $PB$.