Proof of the Optical Properties of a Hyperbola
Theorem
For a point $P$ on the hyperbola and the two foci $F_{1}, F_{2}$, let the angles formed at $P$ between the tangent at $P$ and the lines $\overline{PF_{1}}$ and $\overline{PF_{2}}$ be $\alpha$ and $\beta$, respectively. Then $\alpha$ and $\beta$ are equal.
Explanation
In simple terms, light emanating from one focus of a hyperbola, when reflected, propagates as if it came from the other focus.
Applications include building telescopes using a hyperboloid-shaped secondary mirror, and navigation systems that determine a ship’s position by using the time-difference-of-arrival of radio signals sent from two base stations.
Proof
After extensive searching, the method introduced in Proof of the optical property of the parabola turned out to be the cleanest; there were no other clever tricks. Among the alternatives I examined, I found a proof that requires the fewest auxiliary lemmas, and I used it to fill in the missing details1.
Part 1.
$$ {\frac{ x^{2} }{ a^{2} }} - {\frac{ y^{2} }{ b^{2} }} = 1 $$ Without loss of generality, let this hyperbola be represented by the equation shown above. If the coordinates of the foci are $F_{1} = (-c, 0)$ and $F_{2} = (c, 0)$ with respect to $c^{2} = a^{2} + b^{2}$ and $P = \left( x_{0} , y_{0} \right)$, then the equation of the tangent at $P$ is $x_{0} x / a^{2} - y_{0} y / b^{2} = 1$, and substituting $y = 0$ yields $Q = \left( a^{2} / x_{0} , 0 \right)$. Accordingly, the distance from $Q$ to a focus is as follows. $$ \begin{align*} \overline{Q F_{1}} =& c + {\frac{ a^{2} }{ x_{0} }} \\ \overline{Q F_{2}} =& c - {\frac{ a^{2} }{ x_{0} }} \end{align*} $$
$$ \begin{align*} \overline{P F_{1}} =& \sqrt{ \left( x_{0} + c \right)^{2} + y_{0}^{2} } \\ \overline{P F_{2}} =& \sqrt{ \left( x_{0} - c \right)^{2} + y_{0}^{2} } \end{align*} $$ Since $P$ is a point on the hyperbola, $y_{0}^{2} = \left( b/a \right)^{2} \left( x_{0}^{2} - a^{2} \right)$ holds, and using this we can compute the squared lengths of the segments as follows. $$ \begin{align*} & \left( x_{0} \pm c \right)^{2} + y_{0}^{2} \\ =& \left( x_{0} \pm c \right)^{2} + \left( b/a \right)^{2} \left( x_{0}^{2} - a^{2} \right) \\ =& \left( 1 + {\frac{ b^{2} }{ a^{2} }} \right) x_{0}^{2} \pm 2 c x_{0} + c^{2} - b^{2} \\ =& \left( {\frac{ c^{2} }{ a^{2} }} \right) x_{0}^{2} \pm 2 c x_{0} + a^{2} \\ =& \left( {\frac{ c }{ a }} x_{0} \pm a \right)^{2} \\ =& {\frac{ x_{0}^{2} }{ a^{2} }} \left( c \pm {\frac{ a }{ x_{0} }} \right)^{2} \end{align*} $$
The second term grouped under the square is the distance from $Q$ to the focus, so it can be expressed as a constant multiple of that distance. $$ \begin{align*} \overline{P F_{1}} =& {\frac{ x_{0} }{ a }} \left( c + {\frac{ a^{2} }{ x_{0} }} \right) = {\frac{ x_{0} }{ a }} \overline{Q F_{1}} \\ \overline{P F_{2}} =& {\frac{ x_{0} }{ a }} \left( c - {\frac{ a^{2} }{ x_{0} }} \right) = {\frac{ x_{0} }{ a }} \overline{Q F_{2}} \end{align*} $$
Normally one would assert here that $\overline{PQ}$ is the angle bisector and finish the proof, but avoiding the use of such an auxiliary lemma is the core idea of this proof.
Part 2.
Let the tangent at $P$ meet the $x$ axis at $Q$, and let the feet of the perpendiculars from $F_{1}$ and $F_{2}$ onto the tangent be $R_{1}$ and $R_{2}$, respectively. The two triangles $\triangle{F_{1}QR_{1}}$ and $\triangle{F_{2}QR_{2}}$ formed this way are similar, yielding the following. $$ {\frac{ \overline{R_{1} F_{1}} }{ \overline{R_{2} F_{2}} }} = {\frac{ \overline{Q F_{1}} }{ \overline{Q F_{2}} }} $$ By the result obtained in Part 1, the following equality holds. $$ {\frac{ \overline{R_{1} F_{1}} }{ \overline{R_{2} F_{2}} }} = {\frac{ \overline{Q F_{1}} }{ \overline{Q F_{2}} }} = {\frac{ \overline{P F_{1}} }{ \overline{P F_{2}} }} $$
Part 3.
The two triangles $\triangle{P F_{1} R_{1}}$ and $\triangle{P F_{2} R_{2}}$ are similar, and from this we conclude that $\alpha$ and $\beta$ are equal.