So I wrote a program a while back that does reflections along mirrors defined by arbitrary functions and I noticed an interesting property of mirrors defined by functions that were power greater than 2. For a cubic mirror, you get something that looks kind of like this:

Now, to me, those two symmetric spaces (filled with the beams going down before they were reflected) look kind of like hyperbolas. I'll give images of some other higher degree mirrors and their patterns if you want, but the gist of my question is this: Is there any way you could prove that those lines, after reflected by a mirror defined by a function, say m(x), are tangent to another function, r(x)?

## Reflection from functions

**Moderators:** gmalivuk, Moderators General, Prelates

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### Re: Reflection from functions

Sure you can, just compute the trajectories. You can even find the function this way. It's not actually a hyperbola, however - this is clear since a hyperbola won't actually fit between the positive y-axis and the positive part of the graph y=x

When you say cubic mirror, what do you mean? I'll assume the mirror is given by |x|

If we see a vertical ray coming in from (t,+∞) and intersecting the mirror at the point (t,|t|

^{3}.When you say cubic mirror, what do you mean? I'll assume the mirror is given by |x|

^{3}.If we see a vertical ray coming in from (t,+∞) and intersecting the mirror at the point (t,|t|

^{3}), it's new trajectory is the vertical line of the original trajectory, reflected across the normal to the curve at that point. The unit tangent vector is parallel to (1,3t|t|), so the unit normal vector is parallel to (-3t|t|,1), which makes an angle of tan^{-1}-3t|t| with the vertical. Thus the new trajectory makes an angle of 2tan^{-1}3t|t|, which points in the direction of (-6t|t|, 1-9t^{4}). Thus the trajectory is given by the line [math]y=f_t(x)=\frac{9t^4-1}{6t|t|}(x-t)+t^3.[/math] All of these trajectories together determine a curve, which can be found by maximizing over t: let g(x)=max{f_{t}(x) : t>0}, and y=g(x) should give you the curve you see in the picture. This is a one variable maximization problem, which can in principle be solved by one variable techniques. However, an analytic solution requires finding the zero of a 5th degree polynomial, so a purely analytic solution may be impossible.I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

"With math, all things are possible." —Rebecca Watson

"With math, all things are possible." —Rebecca Watson

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