Here is a rather nice differential argument in a general setting that works for all conics as we are going to show it.
An ellipse can be defined in different equivalent ways. Among the most important are these two ones:
- as the set of points $M(x,y)$ such that
$$\tag{1}MF+MF'= k.$$
where $k$ is a constant, and $F$ and $F'$ are the foci (have a look at the first figure below). We are going to assume that the axes are chosen in such a way that the coordinates of $F$ and $F'$ are resp. $(f,0)$ and $(-f,0)$.
- as a parameterized curve, the most simple being, with respect to the axes just described:
$$\tag{2}\begin{cases}x&=&a \cos(t)\\y&=&b \sin(t)\end{cases}$$
(with $a^2=b^2+f^2$).
Analytically, (1) becomes:
$$\tag{3}f(x,y)=\sqrt{(x-f)^2+y^2}+\sqrt{(x+f)^2+y^2}=k$$
If we consider $x$ and $y$ defined by (2), $f$ is a function of $t$ through $x$ and $y$ ; let us differentiate (3) with respect to $t$
$$\dfrac{\partial f}{\partial t}=0 \ \ \ \iff \ \ \ \dfrac{\partial f}{\partial x}\dfrac{\partial x}{\partial t}+\dfrac{\partial f}{\partial y}\dfrac{\partial y}{\partial t}=0$$
which is equivalent, using the fact that $\sqrt{u}$ is differentiated as $\tfrac{u'}{2 \sqrt{u}}$, to:
$$\left(\tfrac{(x-f)}{\sqrt{(x-f)^2+y^2}}+\tfrac{(x+f)}{\sqrt{(x+f)^2+y^2}}\right)\dfrac{\partial x}{\partial t}+\left(\tfrac{y}{\sqrt{(x-f)^2+y^2}}+\tfrac{y}{\sqrt{(x+f)^2+y^2}}\right)\dfrac{\partial y}{\partial t}=0.$$
Let us rearrange the expression above in the following way:
$$\tag{4} \dfrac{1}{\sqrt{(x-f)^2+y^2}}\left((x-f)\dfrac{\partial x}{\partial t}+y\dfrac{\partial y}{\partial t}\right)+ \dfrac{1}{\sqrt{(x+f)^2+y^2}}\left((x+f)\dfrac{\partial x}{\partial t}+y\dfrac{\partial y}{\partial t}\right)$$
In (4), we recognize dot products with $\vec{T}$, the tangent vector in $M$. This gives:
$$\tag{5}\dfrac{1}{\|\vec{FM}\|}\vec{FM} . \vec{T}+\dfrac{1}{\|\vec{F'M}\|}\vec{F'M} . \vec{T}=0 \ \ \iff$$
$$\tag{6}\underbrace{\left(\dfrac{1}{\|\vec{FM}\|}\vec{FM}+\dfrac{1}{\|\vec{F'M}\|}\vec{F'M}\right)}_{\text{directing vector of the angle biss. (*)}} . \vec{T}=0$$
expressing that the directing vector of the angle bissector of $\widehat{FMF'}$ is orthogonal to the tangent vector, thus is a directing vector of the normal line to the ellipse at $M$, ending the proof.
Remarks:
1) One can be convinced of characterization $(*)$ by thinking to the "manual" construction of an angle bissector using a compass with the same aperture on $MF$ and $MF'$.
2) This method works in the same way for parabolas considered as limit cases of ellipses for which foci $F'$ tends to infinity.
3) A different phenomena occurs for hyperbolas. One should consider one branch at a time (see fig. 2 below). For example, the branch for which (see the difference with (1)
$$\tag{7}MF\color{red}{-}MF'=k$$
It is enough to change all along the calculations above, the central "plus" sign into a minus sign, and at the end obtain:
$$\tag{8}\underbrace{\left(\dfrac{1}{\|\vec{FM}\|}\vec{FM}\color{red}{-}\dfrac{1}{\|\vec{F'M}\|}\vec{F'M}\right)}_{\text{directing vector of the angle biss. (*)}} . \vec{T}=0$$
This minus sign explains the fact that the rays issued from $F$ (red point), reflect on the hyperbola and, instead of passing through the second focus $F'$, make it a virtual focus, in the sense that, for an observer situated on the right, looking leftwads, the rays seem to emanate from $F'$.
4) This method is connected with the "first variation formula" (see p. 104 of "Problems in Geometry", Marcel Berger, P. Pansu, J.-P. Berry, X. Saint-Raymond, Springer.)
Fig. 1: Illustration of formula (6). In red, the unit vectors $\tfrac{1}{\|\vec{FM}\|}\vec{FM}$ and $\tfrac{1}{\|\vec{F'M}\|}\vec{F'M}$; the sum of these vectors is a directing vector of the normal.
Fig. 2: Illustration of formula (8).
Remark : take a look page 12 of the interesting document (https://www.math.psu.edu/tabachni/Books/billiardsgeometry.pdf)
