The Reflective Property of the Ellipse
The normal to an ellipse bisects the focal angle: a gradient proof and Heron's shortest-path proof.
1 Statement
Theorem 1.1 (Reflective property). Let \(E\) be an ellipse with foci \(F_1\), \(F_2\). At every point \(P \in E\), the normal line to \(E\) bisects the angle \(F_1 P F_2\). Consequently, a ray from \(F_1\) that reflects off \(E\) at \(P\) passes through \(F_2\).
2 Setting
Definition 2.1 (Ellipse). Let \(F_1, F_2 \in \R ^2\) and \(a > c \geq 0\), where \(2c = \abs {F_1 - F_2}\). The ellipse with foci \(F_1\), \(F_2\) and major axis \(2a\) is the level set
Definition 2.2 (Law of reflection). A ray travelling in the unit direction \(d\) that meets a curve with unit normal \(n\) leaves in the direction \(d' = d - 2 (d \cdot n) n\), the mirror image of \(d\) in the tangent line.
The circle is the case \(c = 0\), and no point needs separate treatment: unlike a proof using slopes, nothing goes wrong where the tangent or a focal line is vertical.
3 Idea of the proof
The gradient of a distance \(\abs {X - F}\) is the unit vector pointing away from \(F\). So \(\nabla g\) is a sum of two unit vectors, and a sum of two unit vectors bisects the angle between them.
4 First proof: the gradient
Lemma 4.1 (\(E\) is a smooth curve). Neither focus lies on \(E\). At every \(X \in E\), \(g\) is differentiable with
Proof. By the triangle inequality, \(g(F_1) = g(F_2) = 2c < 2a\), so the foci are not on \(E\). Away from \(F_j\), the map \(X \mapsto \abs {X - F_j}\) is differentiable with gradient \(u_j\). If \(u_1 + u_2 = 0\), then \(X\) lies on the segment \(F_1F_2\), so \(g(X) = 2c < 2a\) and \(X \notin E\).
By Lemma 4.1 and the implicit function theorem, \(E\) is a smooth curve near each of its points, with normal direction \(\nabla g = u_1 + u_2\).
Proof of Theorem 1.1. Let \(P \in E\) and set \(u_j\) as in Lemma 4.1, so \(u_1\) points from \(F_1\) to \(P\) and \(u_2\) from \(F_2\) to \(P\). Since \(\abs {u_1} = \abs {u_2} = 1\),
so \(u_1\) and \(u_2\) make equal angles with the normal \(u_1 + u_2\). Equivalently, \(-u_1\) and \(-u_2\), which point from \(P\) to the foci, make equal angles with the inward normal. This is the first claim.
For the second, the incoming ray travels in direction \(d = u_1\). Put \(s = 1 + u_1 \cdot u_2 > 0\), so that \(\abs {u_1 + u_2}^2 = 2s\) and the unit normal is \(n = (u_1 + u_2)/\sqrt {2s}\). Then \(d \cdot n = s / \sqrt {2s}\) and
which is the direction from \(P\) towards \(F_2\).
5 Second proof: Heron’s shortest path
This proof needs no calculus, only the triangle inequality.
Lemma 5.1 (Supporting line). Let \(\ell \) be the tangent line to \(E\) at \(P\). Then \(g(X) > 2a\) for every \(X \in \ell \) with \(X \neq P\).
Proof. The function \(g\) is convex, as a sum of two convex functions. Restricted to \(\ell \), it is a convex function of one variable whose derivative at \(P\) is \(\nabla g(P) \cdot v = 0\) for a direction \(v\) of \(\ell \). So \(P\) minimises \(g\) on \(\ell \), and \(g \geq 2a\) there. If \(g(X) = 2a\) for some \(X \neq P\) on \(\ell \), then by convexity \(g = 2a\) on the whole segment \(PX\). Each \(\abs {Y - F_j}\) is convex, so each must then be affine along \(PX\). A distance \(\abs {Y - F_j}\) is affine along a segment only if the line through it passes through \(F_j\), so \(\ell \) would contain both foci. But \(g(F_1) = 2c < 2a\), contradicting \(g \geq 2a\) on \(\ell \).
Second proof of Theorem 1.1. Let \(F_2'\) be the reflection of \(F_2\) in \(\ell \), so that \(\abs {X - F_2} = \abs {X - F_2'}\) for all \(X \in \ell \). Both foci satisfy \(g < 2a\), and the set \(\set {g < 2a}\) is convex and does not meet \(\ell \) by Lemma 5.1, so \(F_1\) and \(F_2\) lie on the same side of \(\ell \). Hence \(F_1\) and \(F_2'\) lie on opposite sides, and the segment \(F_1F_2'\) meets \(\ell \) at a point \(X_0\).
For any \(X \in \ell \), the triangle inequality gives
So \(X_0\) minimises \(g\) on \(\ell \), and by Lemma 5.1 the only minimiser is \(P\). Therefore \(P\) lies on the segment \(F_1F_2'\). The line \(F_1P\) continues straight to \(F_2'\), and reflecting that continuation in \(\ell \) sends it to \(PF_2\). This is exactly the law of reflection at \(P\).
6 Remarks
Remark 6.1. The same computation proves the hyperbola’s property. With \(h(X) = \abs {X - F_1} - \abs {X - F_2}\) we have \(\nabla h = u_1 - u_2\), and
So the tangent bisects the angle between the focal lines. A ray aimed at one focus is reflected towards the other.