Parabolic Mirrors & Focal Points

At one point or another in your dealings with high school physics, you've probably been told that concave parabolic mirrors are the only concave mirrors that have focal points when parallel light rays are reflected. Just considering the shape of a parabola, this seems to make a lot of sense, but what about other mirrors? Would quartic functions such as $y=x^4$ or sextic functions like $y=x^6$ also generate mirrors with focal points? (This post will model mirrors using 2-dimensional functions, but keep in mind that 3-dimensional mirrors are simply the shapes generated when these two-dimensional functions are rotated around the axis of rotational symmetry.)

Well, let's first derive a formula that all mirrors with focal points must satisfy. We'll represent the mirror as an even function, $f(x)$ with the focal point located at $(0,F)$, when rays parallel to the $y$-axis strike the mirror. If we zoom in infinitely on an arbitrary point on the mirror, $(x,f(x))$, we'll see the image:

In the image, $df(x)$ and $dx$ are differentials. From this, we can say that $\tan^{-1}{f'(x)}$ gives the angle opposite to $df(x)$. Furthermore, the following can be said of the angle marked as $\theta$:

$$\frac{\pi}{2}-\tan^{-1}{f'(x)} = \theta$$

Because $2\theta$ represents the angle between the vertical and the reflected ray, and because the reflected ray passes through $(0,F)$, we can also say:

$$\tan{2\theta} = \frac{x}{f(x)-F}$$

Combining the two equations, we get the mother-of-all mirror equations defining mirrors with focal points:

\begin{equation}
\tan\left(\pi-2\tan^{-1}{f'(x)}\right) = \frac{x}{f(x)-F}
\end{equation}

Because the tangent function repeats every $\pi$ and is an odd function, the equation can be rewritten as:

$$-\tan\left(2\tan^{-1}{f'(x)}\right) = \frac{x}{f(x)-F}$$

or more simply:

\begin{equation}
\tan\left(2\tan^{-1}{f'(x)}\right)=\frac{x}{F-f(x)}
\end{equation}

Those tangent and arctangent functions don't look so good together so let's get rid of them using the tangent's double angle identity:

\begin{equation}
\frac{2f'(x)}{1-{f'(x)}^2}=\frac{x}{F-f(x)}
\end{equation}
\begin{equation}
2[F-f(x)]f'(x)=\left[1-{f'(x)}^2\right]x
\end{equation}

That looks much better. It may seem counter-intuitive, but now, we'll divide sides by ${f'(x)}^2$. Surprisingly, this gives rise to a rather similar equation:

\begin{equation}
2[F-f(x)]\frac{1}{f'(x)} = \left[\frac{1}{f'(x)^2}-1\right]x
\end{equation}
\begin{equation}
-x\frac{1}{f'(x)^2} + 2[F-f(x)]\frac{1}{f'(x)} + x = 0
\end{equation}
\begin{equation}
\frac{1}{f'(x)} = \frac{2[f(x)-F]\pm\sqrt{4[F-f(x)]^2+4x^2}}{-2x}
\end{equation}
\begin{equation}
\frac{-2x}{f'(x)}=2f(x)-2F\pm2\sqrt{[F-f(x)]^2+x^2}
\end{equation}
\begin{equation}
0 = 2f(x)f'(x) - 2Ff'(x) + 2x\pm2f'(x) \sqrt{F^2-2Ff(x) + {f(x)}^2 + x^2}
\end{equation}

Now we see why we divided by ${f'(x)}^2$ earlier on. Notice how in the above equation, the the terms before the last are the derivatives of the terms in the radicand. That means we can do some cool things. Dividing both sides of the equation by $\pm2 \sqrt{F^2-2Ff(x)+{f(x)}^2 + x^2}$ gives:

\begin{equation}
0=\pm\frac{2f(x)f'(x)-2Ff'(x)+2x}{2\sqrt{F^2-2Ff(x)+{f(x)}^2+x^2}}+f'(x)
\end{equation}
\begin{equation}
\int0dx = \int\left[\pm\frac{2f(x)f'(x) - 2Ff'(x) + 2x}{2\sqrt{F^2-2Ff(x) + {f(x)}^2+x^2}} + f'(x)\right]dx
\end{equation}
\begin{equation}
C = \pm\sqrt{F^2-2Ff(x) + {f(x)}^2+x^2} + f(x)
\end{equation}
\begin{equation}
[C-f(x)]^2 = F^2 - 2Ff(x) + {f(x)}^2 + x^2
\end{equation}
\begin{equation}
C^2-2Cf(x) = F^2-2Ff(x)+x^2
\end{equation}
\begin{equation}
2f(x)(F-C) = x^2+\left(F^2-C^2\right)
\end{equation}
\begin{equation}
f(x) = \frac{1}{2} \left[\frac{x^2}{F-C} + F+C\right]
\end{equation}

And so, any concave mirror with a focal point at $(0,F)$ must have an equation of the form $f(x) = \frac{1}{2} \left[\frac{x^2}{F-C} + F+C\right]$ where $C$ is an arbitrary constant. In other words, your high school teacher was right after all: parabolic mirrors are the only concave mirrors which have focal points. Note that if the light in the system we modeled above came from below, the mirror would be considered to be convex. In that case, simply re-visualizing the image presented above will reveal that the reflected rays would then travel in the exact opposite direction as before, all diverging away from a single focus. Therefore, convex mirrors with focal points must also satisfy the equations in this post. It would then follow that parabolic mirrors are also the only convex mirrors to have imaginary focal points.