[This is the 4th and last installment in the current series on mini-golf and ellipse geometry. See the previous ones here: #1, #2, #3.]

We must settle one more question to round out our elliptical arc: Why does light, when shot from one focus of an ellipse-shaped mirrored room, reflect back to the other focus? To answer this question, we’ll need a Fact, a Formalism, and a Fairy Tale.

### The Fact

Recall that, in the previous post, we saw that ellipses can be described by distances: Any ellipse has two focus points *F _{1}* and

*F*so that the total length of broken path

_{2}*F*is the same for every point

_{1}XF_{2}*X*on the ellipse; let

*d*be this common total distance. In fact, more is true: the length of

*F*is smaller than

_{1}XF_{2}*d*if

*X*is inside the ellipse, and larger than

*d*if

*X*is outside.

To prove this fairly intuitive fact, we’ll use the “straight line principle”: the shortest distance between two points is a straight line. Indeed, when *X* is outside the ellipse (see right diagram above), straight-line path *YF _{2}* is shorter than the path

*YXF*that detours through

_{2}*X*, and so \(\)d = F_1 Y F_2 < F_1 Y X F_2[/latex]. See if you can fill in the case where

*X*is inside the ellipse.

### The Formalism

Recall that when light bounces off a straight mirror, the angle of incidence equals the angle of reflection. But here we’re discussing light bouncing off an *ellipse*, which is decidedly not straight. So we need to formally describe how light reflects off *curved* mirrors.

If we zoom into where the incoming light ray strikes a curved mirror (illustrated above), the mirror closely resembles a straight line, specifically its tangent line. This suggests that the light should behave as if it is reflecting off of this line, with equal angles as marked. This is indeed the rule governing ideal reflections on curved mirrors: the angles of incidence and reflection, as measured from the tangent line, should be equal.

### The Fairy Tale

The last ingredient involves Little Red Riding Hood and her thirsty grandmother. Red is delivering cake and wine from her mother (point *M*) to her grandmother (point *G*), but she must first fill a bucket of water at the nearby stream *S*, which is conveniently shaped like a straight line. She was warned by her Brothers to watch out for a big bad wolf, so she must minimize her total walking distance. Where on the stream should she fill her bucket to minimize this distance?

To answer this, imagine reflecting the first leg of Red’s journey across line *S*, so her path from *M* to *S* to *G* gets reflected to a path from *M’* to *G*. The reverse may be done as well: any path from *M’* to *G* turns into a path from *M* to *G* that stops somewhere along *S*. So we just need to find the shortest path from *M’* to *G*. But this is easy: it’s just the straight path *M’G*. So Red’s shortest path from *M* to *S* to *G* is the one that stops at *Z*.

Notice that this shortest path is the one with equal angles as marked. This means Red’s best strategy is to pretend the stream is a mirror and to follow the light ray that bounces directly to grandma’s house. This neatly exemplifies Fermat’s principle, which says that light tends to follow the fastest routes.

### The Proof

With these pieces in place, we can finish today’s question in a flash. Let’s say light from focus *F _{1}* hits an ellipse at point

*X*, as illustrated below. Why does this ray bounce off the ellipse toward

*F*? If we draw the tangent line

_{2}*L*at point

*X*, by The Formalism above, this question is equivalent to: why does the light ray bounce off of line

*L*toward

*F*?

_{2}Let’s reimagine this Grimm scenario by thinking of *F _{1}* as Red’s mother’s house,

*F*as grandma’s house, and

_{2}*L*as the stream. I claim

*F*is the shortest path for Red to take. Why? If

_{1}XF_{2}*Y*is any other point on line

*L*, then

*Y*is outside the ellipse, so by The Fact above,

*F*has distance longer than

_{1}YF_{2}*d*. So

*F*is indeed the shortest. But by The Fairy Tale, we know that this shortest route behaves like light bouncing off of line

_{1}XF_{2}*L*, i.e., the marked angles are indeed equal. So we’re done!