Is the reflective property of an ellipse unique?

Question

Let F1 and F2 be the foci. It is known that in every ellipse with F1 and F2 being its foci, verifies the following property:a ray of light which starts from F1 and is reflected in that curve, will then pass through F2. Proof that only the ellipses verify that property using differential equations.

I've been having a hard time trying to prove this question,if someone can give me at least a hint, it'll be much appreciated. There is a similar question here:Is the reflective property of a parabola unique?. In this case is a parabola.

Answer 1

One can assume WLOG that $F_1(−1,0),F_2(1,0)$.

Let $y=f(x)$ be the unknown cartesian equation (assumed differentiable) of a curve having the reflecting property.

Let us consider a point of the curve $T(x_0,y_0)$ with $y_0:=f(x_0)$.

The tangent to the curve in $T$ has equation

$$y−y_0=f′(x_0)(x−x_0)\tag{1}$$

Let $S(x_1,y_1)$ be the symmetrical point of $F_2$ with respect to this tangent and

$$I\left(\frac{1+x_1}{2},\frac{y_1}{2}\right)$$

the midpoint of line segment $[F_2S]$.

Due to orthogonality, we have the following zero dot-product : $\vec{F_2S}.\vec{TI}=0$, giving the constraint

$$x_1-1=-y_1f'(x_0)\tag{2}$$

As point $I$ belongs to the tangent, we have :

$$\frac{y_1}{2}−y_0=f′(x_0)\left(\frac{1+x_1}{2}−x_0\right)\tag{3}$$

A last constraint, the most important because it is equivalent to the reflection property, is that $S(x_1,y_1)$ is aligned with $F_1$ and $T$, a relationship that can be expressed in the following way (consider it as an equality between slopes):

$$\frac{y_1}{x_1+1}=\frac{y_0}{x_0+1}\tag{4}$$

Using (2), (3) and (4), one obtains the following relationship :

$$1+x_1=2-y_1f'(x_0)=\frac{y_1-2y_0+2x_0f'(x_0)}{f'(x_0)}=\frac{y_1(x_0+1)}{y_0}\tag{5}$$

Eliminating $y_1$ between the two last equations, we get the following condition:

$$f'(x_0)^2x_0 f(x_0)+f'(x_0)(x_0^2-f(x_0)^2-1)- x_0 f(x_0) = 0\tag{6}$$

(without shame, I have done it with the assistance of a Computer Algebra System... see details in remark 3 below),

(6) can be written, taking a neutral notation :

$$f'(x)^2 (x f(x))-f'(x)(f(x)^2-x^2+1) - (x f(x)) = 0\tag{6'}$$

Considering (6') as a quadratic expression in $f'(x)$, we get $f'(x)$ by using on it the roots' formula, giving the following differential equation :

$$f'(x)=\frac{1}{2 x f(x)}\left((f(x)^2-x^2+1)\color{red}{-} \sqrt{(f(x)^2-x^2+1)^2+(2 x f(x))^2}\right)\tag{7}$$

One can check graphically, using for example Geogebra (see Fig. below) that this differential equation and its twin differential equation (with a $\color{red}{+}$ in front of the square root) give rise to elliptical arcs, knowing that it can be done in a rigorous manner (see Remark 2 below).

The big blue dot illustrates an initial point with, attached to it, its trajectory. Please note that these trajectories stay either into quadrants (II) and (III) or into quadrants (I) and (IV). The small blue dots are the foci.

Remarks :

1. There is a lot to say about the (maximum) domains of validity, but this is not the objective here...

2. One can check that the solutions of differential equation (6) and its "twin differential equation" are the curves with cartesian equations :

$$y=b\sqrt{1-\frac{x^2}{b^2+1}}$$

i.e., ellipses intersecting the axes in $(0,\pm b)$ and $(\pm \sqrt{b^2+1},0)$.

3. The symbolic Matlab program I have written for the obtention of eq. (6) :

    syms x0 y0 x1 y1 fp
    E1=2-y1*fp;E2=(y1-2*y0+2*x0*fp)/fp;E3=y1*(x0+1)/y0;
    Y11=solve(E1==E2,y1); % first way to express y1
    Y12=solve(E2==E3,y1); % second way
    [N,D]=numden(Y11-Y12); % reduction to a same denominator
    factor(N)
    % gives -2*fp*(fp^2*x0*y0 + fp*x0^2 - fp*y0^2 - fp - x0*y0)
   

Answer 2

An arbitrary curve in 2D $f(x,y)=0$. E.G. $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0, $the equation of an ellipse .

Arbitrary foci: $\vec{r_1}=<a_1,b_1>, \vec{r_2}=<a_2,b_2>$.

Law of reflection says angle of incidence with respect to the normal is angle of reflection with respect to the normal.The angles have different orientations though, so some sign complications can arise.

The normal to a curve in 2D is $\nabla f$. Let $\vec{r}$ be a vector pointing to the point on the curve.

We want to find an $f$ so that $\frac{\vec{r}-\vec{r_1}}{|\vec{r}-\vec{r_1}|}\cdot \nabla f =- \frac{\vec{r}-\vec{r_2}}{|\vec{r}-\vec{r_2}|}\cdot \nabla f$ where we use the minus sign because of the different orientation of the relevant angle mentioned before.

This suggests $\vec{q}=\frac{\vec{r}-\vec{r_1}}{|\vec{r}-\vec{r_1}|}+\frac{\vec{r}-\vec{r_2}}{|\vec{r}-\vec{r_2}|}$ is $\vec{0}$ or perpendicular to $\nabla f$.

Consider the derivative with respect to a parametrization of the curve $\vec{r}(\lambda): \frac{d\vec{r}}{d\lambda}$. That's a tangent vector at $\vec{r}$. It's in the same plane as $\nabla f$ and perpendicular, so up to a sign, $d\vec{r}/d\lambda = \hat{k}\times \nabla f$.

Using some vector identities:

$\vec{q}\cdot (\hat{k}\times d\vec{r})=\hat{k}(\vec{q}\cdot d\vec{r})-d\vec{r}(\vec{q}\cdot \hat{k})=0\implies \vec{q} \cdot d\vec{r}=0$

Now suppose $|\vec{r}-\vec{r_1}|+|\vec{r}-\vec{r_2}|=c$, the definition of an ellipse.

$(d/d\lambda)(|\vec{r}-\vec{r_1}|+|\vec{r}-\vec{r_2}|)=\frac{d\vec{r}}{d\lambda}\cdot (\frac{\vec{r}-\vec{r_1}}{|\vec{r}-\vec{r_1}|}+\frac{\vec{r}-\vec{r_2}}{|\vec{r}-\vec{r_2}|})=0$

So $ \int \frac{d\vec{r}}{d\lambda} \cdot \vec{q} d\lambda=$ sum of distances to foci= constant. We have an ellipse.