Curved Mirror producing Parallel Rays is Paraboloid

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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:

\(\displaystyle \tan \theta\) \(=\) \(\displaystyle \tan 2 \beta\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {2 \tan \beta} {1 - \tan^2 \beta}\) Double Angle Formula for Tangent
\(\displaystyle \implies \ \ \) \(\displaystyle \frac y x\) \(=\) \(\displaystyle \dfrac {2 \dfrac {\mathrm d y} {\mathrm d x} } {1 - \left({\dfrac {\mathrm d y} {\mathrm d x} }\right)^2}\) Double Angle Formula for Tangent

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.

$\blacksquare$


Sources