Curved Mirror producing Parallel Rays is Paraboloid

Theorem
Let $M$ be a curved mirror embedded in a real cartesian $3$- space.

Let there be a source of light at the origin.

Let $M$ reflect the light in a beam parallel to the $x$-axis.

Then $M$ is the solid of revolution produced by rotating about the $x$-axis the parabola whose equation is:
 * $y^2 = 2 c x + c^2$

Proof
The mirror will have the shape of a surface of revolution generated by revolving a curve $APB$ in the cartesian plane around the $x$-axis.

Let $P = \left({x, y}\right)$ be an arbitrary point on $APB$.


 * ParabolicMirror.png

From the Law of Reflection:
 * $\alpha = \beta$

By the geometry of the situation:


 * $\phi = \beta$
 * $\theta = \alpha + \phi = 2 \beta$

By definition of tangent:
 * $\tan \theta = \dfrac y x$

and so:

Using the Quadratic Formula:


 * $\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {-x \pm \sqrt {x^2 + y^2} \, \mathrm d x} y$

which can be expressed as:
 * $x \, \mathrm d x + y \, \mathrm d y = \pm \sqrt {x^2 + y^2} \, \mathrm d x$

Using Differential of Sum of Squares:
 * $\pm \dfrac {\mathrm d \left({x^2 + y^2}\right)} {2 \sqrt {x^2 + y^2} } = \mathrm d x$

and so:
 * $\pm \sqrt {x^2 + y^2} = x + c$

Hence the result.