The mirror point of an ellipse's focus about the tangent line through a point is collinear to said point and the other focus

Question

I was reading a bit about ellipses, specifically why light sent out from one focus will reflect off the ellipse and converge back to the other focus. The proof I was reading involves a construction as follows:

We choose some point $P$ on the ellipse and draw the tangent line through that point. Then from the focus $F'$ we draw a segment perpendicular to the tangent line, in such a way that the tangent line bisects it (i.e. if the tangent line were a mirror, $G$ is where $F'$ would see itself).

Now the book tacitly assumes $G, P$ and $F$ are actually collinear. This seems to be of fundamental importance in the argument. But why is it true?

Edit: I have updated the diagram since the previous one was a bit misleading.

Answer 1

Here is a funny little animation showing how the reflection of the blue ellipse about its tangent line (purple) makes an orange ellipse whose foci trace the locus of two circles (green and orange). The lines joining opposite pairs of foci intersect at the point of tangency.

But why? Specifically, why should the reflection of the foci $F$, $F'$ about a tangent line at $P$ to points $G'$ and $G$, respectively, result in $FG$ and $F'G'$ intersecting at $P$?

The reason has to do with the constant distance property of ellipses; i.e., $$FP + F'P = 2a = G'P + GP$$ where $a$ is the semimajor axis. Consequently, any point $P^* \ne P$ on the tangent line will be strictly outside either ellipse, hence the sum of distances will be $$FP^* + F'P^* = FP^* + GP^* > 2a.$$ It follows that $P$ is the point that minimizes the sum of distances $FP + GP = 2a$, hence $P$ is collinear with $F$ and $G$.

Answer 2

It directly follows from the reflection property of ellipses:

When a ray leaves one of the foci and meets a point on that ellipse, it will reflect off of the ellipse and pass through the other focus.

"Reflect off the ellipse" means that it will reflect off the tangent; the ellipse is no longer an important construction.

Let's say you were in a large playground with a giant mirror nearby. The mirror is basically the tangent at $P$. You scale the above diagram appropriately; your friend stands in the place of $F'$, while you stand at $F$. Your friends image in the mirror is at $G$.

When you look at the image of your friend in the mirror, what you're really looking at are the rays of light that start from $F'$, bounce off mirror at $P$ and reach your eyes. The image you see at $G$ is the point these rays of light seem to be coming from, when traced backwards.

Since both $P$ and $G$ lie on a line formed by extending the reflected rays of light, they are collinear.

When I say "rays of light," I actually mean an extremely narrow and divergent beam of light rays, that originates at $F'$, and after all of them reflect they seem to be diverging from $G$. So in this scenario, all of $F,P,G$ and $F$ are really not points, but very small regions.

If both your friend and you were point bodies, the width of this beam tends to zero and $P$ also approaches a point.

PS:- This can very easily be proved with congruence, too, but I find reflection very intuitive.

Answer 3

The following proof is cited from the book What Is Mathematics?

Consider any line $l$ in the plane $\mathbb{R}^2$. For any two points $F,F'\notin l$ on the same side of $l$, and for any given length $L=|QF|+|QF'|$, one may find the locus of $Q$ to be an ellipse by definition.

One can find exactly one point $P\in l$ satisfying $$|PF|+|PF'|=\underset{Q\in l}{\min}(|QF|+|QF'|). $$ Now let $L=\underset{Q\in l}{\min}(|QF|+|QF'|)$, and the ellipse is tangent to $l$.

Notice how one may find the point $P$. You just find the reflection point $G$ of $F'$, and the intersection of $FG$ and $l$ is exactly $P$, which can be easily proven by triangle inequality $$|QF|+|QF'|\geqslant|PF|+|PF'|=|FF'|. $$ And this is why the reflection property of ellipse holds.