Tangent line to an ellipse I know that if a line $r$ is tangent to an ellipse with foci $F$ and $F'$ at some point $T$, then $r$ is perpendicular to the bisector of the angle $FTF'$.
Is there a simple proof of that? By 'simple' I mean a proof suitable for high school students.
 A: The unique point at which the tangent line hits the ellipse must be the point on that line at which the sum of the distances to the foci is minimized. Given two points on one side of a line, to minimize the sum of the distances in this way to any point on the line the best thing is to reflect one point over the line, and then the shortest path between them is a straight line. This shows the reflection property of ellipses, and what you are looking for is immediate from that.
Hope that helps.
Greg 
A: Disclaimer: I'm assuming your students are familiar with (partial) derivatives...

Let the ellipse be in standard position with foci at $F=(a, 0)$ and $F'=(-a, 0)$, and let $T=(x,y)$ be a general point on the ellipse. Then we have the invariance
$\|FT\| + \|F'T\| = \text{constant}$, that is
$$
f(x,y) = \text{constant}.
$$
where $f(x,y) := \|FT\| + \|F'T\| = \sqrt{(x-a)^2 + y^2} + \sqrt{(x+a)^2+y^2}$.
Let $(x,y) = (x(t), y(t))$ by any smooth paremetrization of the ellipse. For a tangent to the ellipse, we must have $\frac{d f(x,y)}{d t}=0$. For convenience, we will be using the following notation


*

*$\mathbf{\widehat{u}}:=\frac{x-a}{\|FT\|}\mathbf{\widehat{i}} + \frac{y}{\|FT\|}\mathbf{\widehat{j}}$ is the unit vector along $\overset{\longrightarrow}{FT}$

*$\mathbf{\widehat{v}}:=\frac{x+a}{\|FT\|}\mathbf{\widehat{i}} + \frac{y}{\|FT\|}\mathbf{\widehat{j}}$ is the unit vector along $\overset{\longrightarrow}{F'T}$

*$\mathbf{T}=\frac{\partial x(t)}{\partial t}\mathbf{\hat{i}} + \frac{\partial y(t)}{\partial t}\mathbf{\hat{j}}$ is the tangent at $T$.



Note that the perpendicular bisector of the angle $\measuredangle FTF'$ is the vector $\mathbf{\widehat{u}} + \mathbf{\widehat{v}}$.

Now, one computes $\frac{\partial f(x,y)}{\partial x} = \frac{x-a}{\sqrt{(x-a)^2 + y^2}}+\frac{x+a}{\sqrt{(x+a)^2 + y^2}}=\frac{x-a}{\|FM\|}+\frac{x+a}{\|F'M\|}$ and 
$\frac{\partial f(x,y)}{\partial xy} = \frac{y}{\sqrt{(x-a)^2 + y^2}}+\frac{y}{\sqrt{(x+a)^2 + y^2}}=\frac{y}{\|FM\|}+\frac{y}{\|F'M\|}$. Thus
$$
\begin{split}
0 = \frac{d f(x,y)}{d t} &= \left(\frac{x-a}{\|FM\|} + \frac{x+a}{\|F'M\|}\right)\frac{\partial x}{\partial t} + \left(\frac{y}{\|FM\|} + \frac{y}{\|F'M\|}\right)\frac{\partial y}{\partial t}\\
&= (\mathbf{\widehat{u}} + \mathbf{\widehat{v}}).\textbf{T}
\end{split}
\tag{1}
$$
(1) says that the perpendicular bisector $\mathbf{\widehat{u}} + \mathbf{\widehat{v}}$ of the angle $\measuredangle FTF'$ is perpendicular to the tangent at $T$, and we are done. $\quad\quad\Box$
