I have this question; Consider the parabolic antenna below. Suppose, the dish is a surface patch parametrised by
$$r:\{(x,y)\in\mathbb{R}^2 :x^2 +y^2 =1\}\to \mathbb{R}^3;\\ (x,y)\mapsto(x,y,1-(x^2 +y^2))\in\mathbb{R}^3.$$
Prove that all incoming rays parallel to the $z$ axis are reflected into a common point.
Find the coordinates of this point (the point at which to mount the actual antenna).
[Hint: use and prove the fact that an incoming ray with direction $v$ is reflected in $r(x, y)$ into an outgoing ray of direction $v′ = v − 2(N(x, y)\cdot v)N(x, y)$.]
I'm really stuck and don't know what to do. I've found the normal vector but that's as far as I've got. Any help would be much appreciated! This is for a Maths course, not a Physics course so I don't have any knowledge of general physics properties.
$z = 1 - (x^2+y^2)$ This is a parabola that points down?
Our normal:
$N(x,y,z) = (\frac {dz}{dx}, \frac{dz}{dy}, -1) = (-2x, -2y, -1)$ points the right way.
The unit normal $n = \frac {N}{\|N\|} = (\frac {-2x}{\sqrt{ 1+4x^2 + 4y^2}}, \frac {-2y}{\sqrt{ 1+4x^2 + 4y^2}}, \frac {-1}{\sqrt{ 1+4x^2 + 4y^2}})$
Angle of incidence:
$\cos \theta = (0,0,-1) \cdot n = \frac {1}{\sqrt{ 1+4x^2 + 4y^2}}$
Equals the angle of reflection.
$\cos 2\theta = 2\cos^2\theta - 1 = \frac {1-4x^2 - 4y^2}{1+4x^2 + 4y^2}$
The direction of our reflected beam.
$(\frac {-4x}{1+4x^2 + 4y^2}, \frac {-4y}{1+4x^2 + 4y^2}, \frac {1-4x^2-4y^2}{1+4x^2 + 4y^2})$
If take the point of incidence and add this vector, scaled such that we get back to the axis of symmetry.
$(x,y,1-x^2 - y^2) + \frac {1+4x^2 + 4y^2}{4}(\frac {-4x}{1+4x^2 + 4y^2}, \frac {-4y}{1+4x^2 + 4y^2}, \frac {1-4x^2-4y^2}{1+4x^2 + 4y^2}) = (0,0,\frac 14)$