Proof of the Optical Properties of the Ellipse
Theorem
For a point $P$ on an ellipse and the two foci $F_{1}, F_{2}$, let the angles formed by the tangent at $P$ with $\overline{PF_{1}}$ and $\overline{PF_{2}}$ be $\alpha$ and $\beta$ respectively. Then $\alpha$ and $\beta$ are equal.
Explanation
In plain terms, light emitted from one focus of an ellipse is concentrated at the other focus.
Famous applications include the whispering-gallery effect in St. Paul’s Cathedral, where a whisper at one focus can be heard at the other focus, and more practical uses such as medical lithotripters that exploit this property to concentrate energy at a precise point.
Proof
Strategy: There are many ways to prove the optical property of an ellipse; instead of using synthetic geometry or brute-force coordinate computations, we present an interesting proof here 1. The prerequisites to follow this proof are at least undergraduate-level, but the argument is clean and extends to higher-dimensional generalizations such as the ellipsoid. For a planar 2D figure one need not study so much to obtain a proof.
Parameterize a point on the ellipse as $P = P(t)$. By the definition of an ellipse, the sum of the distances from $P$ to $F_{1}$ and from $P$ to $F_{2}$ equals some constant $C$, so we have $$ \left\| P - F_{1} \right\| + \left\| P - F_{2} \right\| = C $$
[Derivative of the norm]: For the distance between a point $\mathbf{a}$ and a vector $\mathbf{x} = \mathbf{x} (t)$, the derivative of the $l^{2}$-norm $\left\| \mathbf{x} - \mathbf{a} \right\|$ is given by $$ {\frac{ d \left\| \mathbf{x} - \mathbf{a} \right\| }{ d t }} = \dot{\mathbf{x}} \cdot {\frac{ \mathbf{x} - \mathbf{a} }{ \left\| \mathbf{x} - \mathbf{a} \right\| }} $$
Using the differentiation of the norm, we obtain the following; $\dot{P}$ is the tangent (in the plane, the direction vector of the tangent line). $$ \dot{P} \cdot {\frac{ P - F_{1} }{ \left\| P - F_{1} \right\| }} + \dot{P} \cdot {\frac{ P - F_{2} }{ \left\| P - F_{2} \right\| }} = 0 $$
[Definition of angle]: For two vectors $\mathbf{u}, \mathbf{v} \in V$, the quantity $\theta$ satisfying the following is defined to be the angle between the two vectors (angle). $$ \cos \theta = {{ \left< \mathbf{u}, \mathbf{v} \right> } \over { \left| \mathbf{u} \right| \left| \mathbf{v} \right| }} $$
[Definition of even function]: A function $f(x)$ satisfying $f(-x) = f(x)$ is called an even function. $$ \begin{align*} & {\frac{ \dot{P} }{ \left\| \dot{P} \right\| }} \cdot {\frac{ P - F_{1} }{ \left\| P - F_{1} \right\| }} + {\frac{ \dot{P} }{ \left\| \dot{P} \right\| }} \cdot {\frac{ P - F_{2} }{ \left\| P - F_{2} \right\| }} = 0 \\ \implies & {\frac{ \left< \dot{P} , P - F_{1} \right> }{ \left\| \dot{P} \right\| \left\| P - F_{1} \right\| }} + {\frac{ \left< \dot{P} , P - F_{2} \right> }{ \left\| \dot{P} \right\| \left\| P - F_{2} \right\| }} = 0 \\ \implies & \cos \left( \alpha \right) + \cos \left( \beta \right) = 0 \\ \implies & \cos \left( \alpha \right) = - \cos \left( \beta \right) \\ \implies & \cos \left( \alpha \right) = \cos \left( \beta \right) \\ \implies & \left| \alpha \right| = \left| \beta \right| \end{align*} $$
Multiplying every term on both sides by $\left\| \dot{P} \right\|^{-1}$ converts them to cosines, from which one sees that the magnitudes of $\alpha$ and $\beta$ are equal.
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