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I attempted to answer a question (Can a point be reflected through a line which is not straight?) which, probably because of my perpendicular functions theory, got a downvote and needed improvement. I then reread the comment posted by @DavidK on my answer:

This approach seems to be limited to curves expressed in the form $y=f(x)$, which is a small subset of all curves.

and realized that he was right. But that would motivate two questions, the first one of which shall be presented in this post:

The Question: How do I drop a perpendicular from point $P$ to a parametrized curve $C$ in the form $C(t)=(x(t), y(t))$?

I was hoping this could be extended from functions to curves that are not functions, but that part was what I was struggling with. It could be in a form such as (everyone had heard this one!) $(\cos t, \sin t)$ or something more contrived and complex such as $(\sqrt{6-3\tan (2t)},2+5\sin(\frac{2}{t})\cos(\frac{3}{t}))$. But I am looking for a general rule for parametric curves.

maximusavant25
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  • Convert the curve into the level curve of a function and take the gradient. The gradient will be perpendicular to the curve. That vector and the point will give a line through the point. The line's intersection with the curve will give you your point. – John Douma Mar 22 '25 at 12:49

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If you prefer to use the parametric form, then in the plane, the perpendicular direction to $(x(t), y(t))$ is $(y'(t), -x'(t))$ or its opposite. If $P=(x_p, y_p)$, then to find the perpendicular projection, solve the equations $$x_p = x(t) + sy'(t)\\y_p = y(t) - sx'(t)$$ for $t$ and $s$. There may be multiple solutions for $t$, each with its own single corresponding $s$. Each $t$ provides a point where the perpendicular passes through $P$. The reflection of $P$ through that point is obtained by changing the sign of $s$: $$P_t' = (x(t) - sy'(t), y(t) + sx'(t))$$

For example, points inside a circle other than its center will have two reflections - one through the closer side, and the other through the opposite side. The reflections of the center form another circle of twice the radius.

Paul Sinclair
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  • "Conformal" = "preserves angles", so your conformal transformation applied to the perpendicular is going to still be perpendicular to the image of the original curve. But it need not be a line anymore, and the distances are not preserved in general either, so an equidistant reflection along the image of the perpendicular need not be the image of the reflection through the original curve. – Paul Sinclair Mar 24 '25 at 17:04
  • Okay, my next question uses this post. I just put it here to get more attention to this question, for it only has 14 views and still has no answers: https://math.stackexchange.com/questions/5062118/how-do-i-find-the-other-endpoint-of-this-restricted-parametrized-curve – maximusavant25 May 05 '25 at 19:34