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In this other question the user asks for a parametric curve and "imposing" one curve on another. You can find a demonstration here.

I have been meaning to use the tangent to answer a similar question on how to find a reflected curve across another and have such a desmos demonstration. However I am not sure what the equations should turn out to be in that case.

By reflecting curve $F$ across $G$, I mean that for every point of $G$, we calculate the normal vector, and copying the intersection point between the normal line and curve $F$ at the same distance but on the opposite side of the curve $G$ along the normal direction. The resulting curve $F'$ of all those points is the "reflection" of $F$ along $G$.

This can also be thought of as a generalization of the function inverse. When we let the curve we are reflecting over be the $y=x$ line, the reflection $F'$ is precisely the inverse of the curve $F$, in the sense that the $x$ and $y$ coordinates of the points have been swapped.

vallev
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  • I suggest that you call this "reflecting $F$ across $G$" rather than onto, as the points generally don't land on $G$ at all, unless they began there, in which case they're fixed points. – Sammy Black Feb 27 '24 at 19:14
  • That is a good suggestion and I took it. Hopefully we get some answers. – vallev Mar 06 '24 at 15:36
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    I have discussed a similar problem here in the comments; perhaps this may be helpful? https://math.stackexchange.com/questions/4849213/reflection-of-a-continuous-differentiable-curve-about-a-line – Cognoscenti Mar 12 '24 at 07:39
  • I think that might be worth a try. How did you find this method? – vallev Mar 12 '24 at 19:12

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