Reflections in Ellipse

Choose two points T₁, T₂ on an ellipse. Imagine a light source at T₁ sending a ray of light in the direction of T₂. Assuming the inner surface of the ellipse fully reflective, the ray will bounce off at T₂ and reach the ellipse again at T₃ where it will bounce again and then again at point T₄ and so on.

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There are three possibilities:

T₁T₂ passes through one of the foci. Then T₂T₃, T₃T₄, ... will pass alternately through the two foci of the ellipse. (Due to a limited accuracy and also because of the difficulty in positioning the points accurately, this case is only partially validated by the applet.)
All the pieces T_kT_k+1 remain outside the segment F₁F₂ of the two foci. In which case there is an ellipse confocal with the given one for which T_kT_k+1's serve as tangents.
All the pieces T_kT_k+1 cross the segment F₁F₂ of the two foci. In which case there is a hyperbola confocal with the given ellipse for which T_kT_k+1's serve as tangents.

In the latter two cases, the straight lines T_kT_k+1 are said to form an envelope of the ellipse (case 2) or of the hyperbola (case 3.)

Proof

Reflections in Ellipse

The question is, What happens to a ray of light that hops off an elliptic mirror with a light source inside the ellipse?

One case, i.e. where the light source is located at a focus of the ellipse has been discussed from the first principles elsewhere. In this case, after every reflection the ray will proceed from one focus to the other indefinitely. This answer clearly applies to the case where the first the light source is not at a focus but emits a ray that passes through one.

Only other two possibilities are possible: either every piece of the light trajectory crosses between the foci of the ellipse or none does. In the latter case, the pieces envelope an ellipse, in the former a hyperbola. Bellow, we shall look into the case of ellipse.

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Let the foci of the ellipse be F₁ and F₂, and assume that the light source located at T₁ on the ellipse emits a ray that reflects of the ellipse at point T₂ and reaches it again at T₃. If T₁T₂ does not cross F₁F₂, there is an ellipse confocal with the given one that touches T₁T₂. This is because the whole plane is split into the level curves of the function f(M) = |F₁M| + |F₂M|, each of these curves being an ellipse.

We are going to prove that T₂T₃ is tangent to the same ellipse.

Find F'₁, the reflection of F₁ in T₁T₂, and F'₂, the reflection of F₂ in T₂T₃. By the optical property of ellipse, the angles between the tangent at T₂ and F₁T₂ and F₂T₂ are equal. By the construction, so are the angles between the tangent and T₁T₂ and T₃T₂. It follows that the angles F₁T₂T₁ and F₂T₂T₃ are equal and so are their doubles:

F₁T₂F'₁ =

F₂T₂F'₂,

implying that isosceles triangles F₁T₂F'₁ and F₂T₂F'₂ are similar. It then follows that triangles F₁T₂F'₂ and F'₁T₂F₂ are equal. In particular,

F₂F'₁ = F₁F'₂.

With respect to the constructed ellipse, the locus of points F'₁ is a circle with center F₂ and radius, say, R = |F₁T| + |F₂T|, where T is the point of tangency if the ellipse with T₁T₂. But then also F₁F'₂ = R; so that F'₂ lies on the circle with center F₁ and radius |F₁T| + |F₂T|, which is the locus of reflections of F₂ in all the tangents to the constructed circle. This proves that T₂T₃ is one of these tangents.

References

V. Gutenmacher, N. Vasilyev, Lines and Curves: A Practical Geometry Handbook , Birkhauser, 1 edition (July 23, 2004)

Ellipse

Conic Sections > Ellipse

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