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 $P = \tuple {x, y}$ be an arbitrary point on $APB$.
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:
\(\ds \tan \theta\) | \(=\) | \(\ds \tan 2 \beta\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 \tan \beta} {1 - \tan^2 \beta}\) | Double Angle Formula for Tangent | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac y x\) | \(=\) | \(\ds \dfrac {2 \dfrac {\d y} {\d x} } {1 - \paren {\dfrac {\d y} {\d x} }^2}\) | Double Angle Formula for Tangent |
Using the Quadratic Formula:
- $\dfrac {\d y} {\d x} = \dfrac {-x \pm \sqrt {x^2 + y^2} \rd x} y$
which can be expressed as:
- $x \rd x + y \rd y = \pm \sqrt {x^2 + y^2} \rd x$
Using Differential of Sum of Squares:
- $\pm \dfrac {\map \d {x^2 + y^2} } {2 \sqrt {x^2 + y^2} } = \d x$
and so:
- $\pm \sqrt {x^2 + y^2} = x + c$
Hence the result.
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.9$: Integrating Factors: Example $2$