Deriving the equation of a parabola? Imagine you wanted to derive the equation of a parabola by imposing that all parallel rays that bounce on the parabola, end up in the foci of the curve.

By using geometry and vectors, I arrived to something like:
$$
F = (a, b) \\
-\sqrt{(a - x)^2 + (b - y(x))^2} = (a - x)y'(x) + y(x) + b \Rightarrow \\
(a - x)^2 y'(x)^2 + 2(a - x)y'(x)y(x) + 2(a - x)y'(x)b = a^2 + x^2 +2ax
$$
I've just imposing that the $\cos{\theta}$ has to be equal when the ray bounces (using dot products):

This is far from what I was expecting:
$$
y''(x) = k
$$
For $a = b = 0$ we have the following:
$$
xy'(x)^2 - 2y'(x)y(x) = x
$$
This is a pretty intimidating equation. My question is: Is this line of reasoning correct? Is this strange looking equation right? Is there any other way of doing this more efficiently?
 A: I worked this out and was able to recover the parabola. Here's how I did it:
Consider a curve defined by $y=f(x)$, $f$ differentiable, and rays coming from the positive $y$-direction. The reflected rays will be a line with some slope, completely determined by two separate conditions.

*

*The ray goes through the focus $(a,b)$ and the point on the curve $(x,y)$

*The angle of incidence is equal to the angle of reflection

In order for both of these to be true, we must have that the value of the slope the two methods determines coincide.
The first method is easy, we have
$$ m = \frac{y-b}{x-a} $$
The second method is a bit more complicated, but one may determine that the slope of the reflected line is
$$ m = \frac{y'^2-1}{2y'} $$
So, the differential equation that determines the curve is the equality of these two values
$$ \frac{y-b}{x-a} = \frac{y'^2-1}{2y'} $$
Letting $(a,b) = (0,0)$ and rearranging a bit, this becomes
$$ xy'^2 - 2yy' = x $$
(You should have this equation, but you dropped an $x$ in the $yy'$ term when plugging in $(0,0)$ which gave you the wrong equation at this point)
This equation has solutions of the form
$$ y = \pm x\sinh(k+\log x) $$
which can be rewritten as
$$ y = \frac{c}{2}x^2 - \frac{1}{2c} $$
which is indeed the family of parabolas with focus at $(0,0)$.
A: One can use a different treatment by connecting the reflection property to the focus-directrix property.
See the following figure adapted from this excellent site:

where the directrix is depicted in red.
Let us recall and establish the focus-directrix characterization of a parabola:
"The locus of points $A$ situated at the same distance from a point $F$ and a line  (the directrix $(D)$) is a parabola".
Indeed with axes as depicted (dotted lines) in the figure, with $F(0,f)$ and $(D)$ with equation $y=-f$ we have, with current point $A(x,y)$:
$$AF^2=AA'^2 \ \ \iff \ \ x^2+(y-f)^2=(y+f)^2 \ \ \iff \ \ y=\tfrac{1}{4f}x^2$$
Now, how is the reflecting property of the parabola equivalent to the focus-directrix property ? Plainly, through the consideration of "magic" triangle $AFA'$ which is easily established as isosceles with the tangent to the parabola in point $A$ as its altitude $AH$ with $H$ situated on the $x$ axis (see the site referenced above for detailed explanations). Corresponding angles in blue explain the connection with the reflection property.
It is of interest to note (see figure below) that this equilateral triangle can be found in a similar configuration but for an ellipse where the directrix has been replaced by the so-called director circle of the ellipse ; please note that if focus $F'$ is kept fixed and focus $F$ is sent to infinity on the left, the ellipse becomes a parabola and the director circle becomes its directrix.

