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Reflecting a point through a line (or hyperplane) is a fairly trivial bit of linear algebra, but I'm wondering if it is meaningful to reflect a point through some other line. Say, $y=x^2$ or something wacky like $y^3=|\sin(x)|+\sqrt{x}$.

To expand: I could define inversion of a point $P$ with respect to an arbitrary curve $C$ as, say, being reflection about the line which is tangent to $C$ at the point nearest to $P$, but one would naturally want nice properties such as uniqueness and invertibility. I expect this would be equivalent to something like finding a smooth bijection between points and their reflections for a given curve.

Is this well-defined? Is it a field of study?

maximusavant25
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spraff
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    https://en.wikipedia.org/wiki/Inversive_geometry#Inversion_in_a_circle has something for you, I think. – Arthur May 08 '24 at 12:34
  • Did you mean reflected in the title? – J. W. Tanner May 08 '24 at 12:41
  • Not quite a duplicate https://math.stackexchange.com/questions/1967501/is-square-inversion-possible but raises similar issues; especially, we first have to agree what reflect through a curve means. – Andrew D. Hwang May 08 '24 at 16:33
  • @spraff, one meaning of reflection can be the symmetry of the point about the curve, is this what you mean? – sirous May 08 '24 at 18:04
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    One approach is to consider your curve as a subset of the complex plane, find a conformal mapping that maps your curve (or a portion of it including the point you want to reflect) to the real line, then apply the complex conjugate, and then apply the inverse conformal mapping to get the "reflection". I think this generalizes the notion of reflection across a line and inversion of a circle. I'm not sure about under what conditions this is possible and gives a unique result, though. See this question. – Jair Taylor May 09 '24 at 16:25
  • From a physics perspective, what you describe as reflection could be interpreted as constructing the mirror image of a point light source, when the light it emits is reflected across an arbitrary curve. I believe that not every possible curve will produce a well-defined mirror image. Lines and circles have this property but I doubt that there are any other surfaces that can focus light on a single point, irrespective to where the point source is (which then would allow a definition of reflection/inversion). – K. Grammatikos Feb 14 '25 at 19:34
  • @K.Grammatikos Circle does not have the property, except for the circle center. – user Mar 13 '25 at 20:42

1 Answers1

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Alright, I got my answer using the properties of reflection in Euclidean geometry. The properties of reflection, with $\ell$ being the line of reflection, are:

  • $\overline{AA'}$ bisects $\ell$
  • $\overline{AA'}\perp\ell$

I have found a new class of function relationships I called perpendicular functions and looked foward to applying this to my answer. Moving on from here I can sketch a rough overview of my answer:

  1. Define a curve $C(x)$ by a given rule, for example, $C(x)=x^2$, and a pre-image point $A$ such that $A(x_A,y_A)$
  2. Find such a perpendicular function (the definition of perpendicular functions is in this Desmos graph, notated by $C_\perp(x)$) that includes the point $A$. This can be achieved by solving the equation $y=\int^{x-h}_0[-\frac{1}{C'(t+h)}]dt+f(h)$ for $h$, substituting in the values of $x_A$ and $y_A$ for $x$ and $y$, respectively. Your result $h$ will determine the intersection point, $P(h,C(h))$.
  3. Finally, determine the coordinates of $A'(x_{A'},C_\perp(x_{A'}))$ by solving for $x_{A'}$ in the following equation: $$\int^h_{x_A}[\sqrt{1+f'(t)^2}]dt=\int^{x_{A'}}_h[\sqrt{1+f'(t)^2}]dt$$ Or $$\int^h_{x_A'}[\sqrt{1+f'(t)^2}]dt=\int^{x_{A}}_h[\sqrt{1+f'(t)^2}]dt$$ Edit: A new answer is underway that instead synthesizes the answer to these two questions: https://math.stackexchange.com/questions/5048474/how-do-i-drop-a-perpendicular-from-a-point-onto-a-parametrized-curve\ How do I find the other endpoint of this restricted parametrized curve?
maximusavant25
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  • This approach seems to be limited to curves expressed in the form $y=f(x),$ which is a small subset of all curves. To try to apply it to a circle, I replaced your $f$ by $f(x)=\sqrt{16-x^2},$ which gave half the circle, but the "perpendicular function" (which I was hoping would just be a radial line) was a strange curved thing. – David K Feb 14 '25 at 21:51
  • I have read your comment and plugged in the formula $f(x)=\sqrt{16-x^2}$ and yes, it was a curved line except for when $P=(0,4)$, when it can be concluded as a radial line, but it does not show as anything on the graph. The advantage of this curve is that, when you sketch out a set of these curves at different intersection points, despite the fact that $\lim\limits_{x \to 0} [C_\perp(x)]=-\infty$ (All the perpendicular functions his limit does not exist, but this just shows their asymptote), they intersect nowhere in the plane. This is just half a circle, so it wouldn't fill in the full circle – maximusavant25 Feb 17 '25 at 12:21
  • I think you're correct that the perpendicular functions never intersect. I think you could also start at any point on one of these curves, move in a direction perpendicular to the perpendicular function, and continuing to cross perpendicular functions at a right angle trace a new function. Do this everywhere and you have a new grid that is a transformation of the Cartesian coordinate grid. – David K Feb 17 '25 at 17:13
  • @DavidK You are correct on the point that "curves expressed in the form $y=f(x)$ are a small subset of all curves." But then this leads me whether or not the curve should be in parametric form $C(t)=(x(t), y(t))$, but that will raise two questions that I have yet to ask, about the "dropping a perpendicular from point $P$ to a curve $C$" and "the midpoint of a parametrized curve between two points." (which might already have an answer here: https://math.stackexchange.com/questions/3365903/how-to-determine-the-midpoint-of-the-parametric-curve ) – maximusavant25 Mar 22 '25 at 12:13