Parabolic Cylinder Coordinates If you were on a desert island, how would you derive parabolic cylinder from scratch - no memorizing formulas or memorizing extraneous terms in constant surfaces. I just can't see how these things naturally fall out!
 A: I looked at this picture from the wikipage

and figured out that confocal parabolas means they all share the same focus. So first thing to work out is where does the focus of a parabola lie, since I don't remember that formula.
So take the following parabola with coordinates $y=ax^2$. A focal point for this parabola means that when a ray comes in from infinity, parallel with its symmetry axis, then the reflected ray goes through the focal point. That's what I can remember about focal points.
A ray coming from infinity has cartesian equation $x=x_0$. It hits the parabola in the point $(x_0,ax_0^2)$ where the tangent has a directional vector $(1,2ax_0)$. A normal vector to this tangent is $(-2ax_0,1)$. The angle between the incoming ray and this normal and the outgoing ray and this normal has to be the same, thus the scalar products of the vectors has to be the same,
$$(0,1)\cdot(-2ax_0,1) = \sqrt{1+4a^2x_0^2}\cos\alpha = (p,q)\cdot(-2ax_0,1)$$
provided that $(p,q)$ is normalized, i.e. $p^2+q^2=1$.
With a bit of algebra, you can then work out that a directional vector of the outgoing ray is
$$(4ax_0,4a^2x_0^2-1)$$
From this, it is easy to find a parametric equation for the outgoing ray
$$\begin{cases}x=x_0+\lambda (4ax_0) \\ y=ax_0^2+\lambda (4a^2x_0^2-1)\end{cases}$$
The focal point is the intersection of that ray with the symmetry axis of the parabola and has coordinates $(0,f)$ with $f$ being after some work
$$f=\frac{1}{4a}$$
Thus, the parabola $y=ax^2$ has this focal point. I want now to construct parabolas that all have the same focal point, and I'll choose that point to be $(0,0)$. All I have to do is shift my parabola accordingly so that I have.
$$y=ax^2-\frac{1}{4a}$$
Let's say that $a>0$, then I'll keep track of the parabolas pointing down with another parameter $b>0$ and they have equation
$$y=-bx^2+\frac{1}{4b}$$
Allow me now a slight abuse of notation, because I want to replace $a\to a/2$ and $b\to b/2$ to have nicer equations
$$y=\frac{1}{2}\left(ax^2-\frac{1}{a}\right)$$
and
$$y=\frac{1}{2}\left(-bx^2+\frac{1}{b}\right)$$
Our work is almost done. A point that lies on both parabolas for certain parameter values $(a,b)$ must have
$$\frac{1}{2}\left(ax^2-\frac{1}{a}\right)=\frac{1}{2}\left(-bx^2+\frac{1}{b}\right)$$
or after some work
$$x^2=\frac{1}{ab}$$
and plugging this in in one of the formulas for the parabolas we also have
$$y=\frac{1}{2}\left(\frac{1}{b}-\frac{1}{a}\right)$$
Introducing a final change of parameters $\tau^2=1/b$ and $\sigma^2=1/a$ we obtain
$$x=\sigma\tau \; \text{ and } \; y=\frac{1}{2}\left(\tau^2-\sigma^2\right) $$
which are indeed the same equations as on the wikipage. The corresponding equations for the parabolas are
$$y=\frac{1}{2}\left(\frac{1}{\sigma^2}x^2-\sigma^2\right)$$
and
$$y=\frac{1}{2}\left(-\frac{1}{\tau^2}x^2+\tau^2\right)$$
