Proof of the Optical Properties of a Hyperbola
Theorem
For a point $P$ on the hyperbola and the two foci $F_{1}, F_{2}$, if the angles formed at $P$ by the tangent and $\overline{PF_{1}}$, $\overline{PF_{2}}$ are denoted by $\alpha$ and $\beta$ respectively, then $\alpha$ and $\beta$ are equal.
Explanation
Application to telescopes
Proof
I searched extensively, and the method introduced in Proof of the optical property of the parabola is the cleanest; I found no other striking tricks. Among the alternatives I have seen so far, I discovered a proof that requires the fewest auxiliary lemmas and used it to fill in the previously sparse parts1.
Part 1.
$$ {\frac{ x^{2} }{ a^{2} }} - {\frac{ y^{2} }{ b^{2} }} = 1 $$ Without loss of generality (../2720), assume this hyperbola is represented by the equation shown above. If the coordinates of the foci are $c^{2} = a^{2} + b^{2}$ and $F_{1} = (-c, 0)$, $F_{2} = (c, 0)$ and $P = \left( x_{0} , y_{0} \right)$ respectively, 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 $Q = \left( a^{2} / x_{0} , 0 \right)$. From this, 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 in squares is the distance from $Q$ to the focus from $Q$, so it can be expressed as a constant multiple. $$ \begin{align*} \overline{P F_{1}} =& {\frac{ x_{0} }{ a }} \left( c + {\frac{ a^{2} }{ x_{0} }} \right) = {\frac{ x_{0} }{ a }} \overline{Q F_{1}} \\ \overline{P F_{2}} =& {\frac{ x_{0} }{ a }} \left( c - {\frac{ a^{2} }{ x_{0} }} \right) = {\frac{ x_{0} }{ a }} \overline{Q F_{2}} \end{align*} $$
Usually one would finish the proof here by asserting that $\overline{PQ}$ is the angle bisector, but avoiding that auxiliary lemma is the key of this proof.
Part 2.
First, let the tangent at $P$ meet the $x$-axis at $Q$, and let the feet of the perpendiculars from $F_{1}$ and $F_{2}$ to the tangent be $R_{1}$ and $R_{2}$ respectively. The two triangles thus formed, $\triangle{F_{1}QR_{1}}$ and $\triangle{F_{2}QR_{2}}$, are similar, yielding: $$ {\frac{ \overline{R_{1} F_{1}} }{ \overline{R_{2} F_{2}} }} = {\frac{ \overline{Q F_{1}} }{ \overline{Q F_{2}} }} $$ 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 one can conclude that $\alpha$ and $\beta$ are equal.