Parabolic antenna I'm currently attempting the following question:
Consider the parabolic antenna below. Suppose, the dish is a surface patch parametrised
by
\begin{align}
& r:\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1\} \rightarrow \mathbb{R}^3; \\[10pt]
& (x,y) \mapsto \left(x,y,\frac{1}{4} (x^2 + y^2)\right) \in \mathbb{R}^3
\end{align}
In which point do you have to mount the actual antenna in order to guarantee an
ideal signal?
I started off by trying to show that all incoming rays parallel to the central axis of symmetry are reflected to a common point, with some basic geomtry but this isn't rigorous enough. I think I should find the normal vector field and then show that they focus on one point, but am not sure how to go about doing this. 
Any help is much appreciated, thanks.
 A: The antenna prime focus is the focal point.
As for geometric optics, let focal length $f$ be =1.
Let the incident rays be parallel to x-axis.The reflected rays all pass through focal point F.
Then the focal point F of paraboloid $ z = r^2/(4 f)= (x^2+y^2)/(4 f) $
is 
$$ (x_F,z_F)= (1,2)*f =(1,2) $$ for the central section $y=0$
This follows directly from consideration of bisection by focal ray to the incident and reflected rays as a law of reflection ... and also by definition of a conic eccentricity $e=1$ with Fermat's principle.
To check reflection, differentiate relation w.r.t  $z$ to find direction of $ x^{'}= 2f/x $ is equally inclined to horizontal ray of zero slope and reflected ray of slope $\dfrac{x-0}{z-f}$.  
A: By symmetry, we can work in the plane $y=0$. Let the focus be at $(0,f)$ (in $2D$).
A ray from $\left(x,\dfrac{x^2}2\right)$ to $(0,f)$ has the slope
$$\frac{\dfrac{x^2}4-f}{x}$$ while the tangent to the parabola has the slope
$$\frac x2.$$
If this is the tangent of an angle $\theta$, the tangent of $2\theta$ is
$$\frac{2\dfrac x2}{1-\dfrac{x^2}4}.$$
You immediately see the connection with $f=1$.
