A Question on Ellipses, Part 3

In the last two posts, we’ve looked at the question

What is the set of points of intersection of a normal line of an ellipse with its major axis, barring the normal lines at the endpoints of the major axis?

We found the answer is the segment stretching from $\left(-\dfrac{a^2-b^2}{a},0\right)$ to $\left(\dfrac{a^2-b^2}{a},0\right)$, if we assume (for simplicity, mostly) that the equation of the ellipse is $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b>0$.

So, we see what the answer is, but why? Why shouldn’t the segment stretch from focus-to-focus, as I would have initially thought?

We can find an answer by considering the optical property of ellipses: a beam of light from one focus will bounce off of the ellipse toward the other focus. If we call our foci $F_1$ and $F_2$, let $T$ be a point on the ellipse, and define $I$ to be where the normal line at $T$ intersects the major axis, then the physical law of reflection tells us that $$\text{(angle of incidence)}=\text{(angle of reflection)}$$ $$\angle F_1TI=\angle F_2TI.$$

In other words, segment $\overline{TI}$ bisects $\angle F_1TF_2$. By the Angle Bisector Theorem, we know that $$\dfrac{IF_1}{TF_1}=\dfrac{IF_2}{TF_2}.$$ It is this equation, really, that shows us why the intersection segment cannot stretch from focus to focus. If it did, there would be intersection points $I$ with $IF_2$ arbitrarily close to $0$. However, that would also force $IF_1$ to be close to $0$ to satisfy the other side of the equation, which cannot happen simultaneously with $I$ being near $F_2$.

In fact, we can extend this argument to see just how close $I$ can get to $F_2$. Let us use $c$ to stand for $\sqrt{a^2-b^2}$, so that the foci are at $(\pm c,0)$. Then, as $T$ moves to the rightmost point on the ellipse, $(a,0)$, we have $TF_2\to a-c$ and $TF_1\to a+c$. Thus, $$\dfrac{IF_1}{IF_2}=\dfrac{TF_1}{TF_2}\to\dfrac{a+c}{a-c}.$$ $IF_1$ is equal to $F_1F_2-IF_2=2c-IF_2$, so we have $$\dfrac{2c-IF_2}{IF_2}\to\dfrac{a+c}{a-c}=1+\dfrac{2c}{a-c},$$ or $$\dfrac{2c}{IF_2}\to 2+\dfrac{2c}{a-c}.$$ Solving for $IF_2$, we conclude that $IF_2\to\dfrac{c}{a}(a-c)$.

So, if the intersection segment we found in the last post was initially unexpected, it was because of the unexpected appearance of the ellipse’s optical property.