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Using @PaulSinclair's answer for how to drop a perpendicular onto a curve, this provides some information to get an answer to this next question:
Point $A=(x_A,y_A)$ is an endpoint of the parametrized curve $Q_t$ by definition, the parametric curve the point will be reflected in is: $$P_t=(x(t),y(t))$$ $s$ and $t$ can be solved for here: $$x_A=x(t)+sy'(t)$$ $$y_A=y(t)-sx'(t)$$ The perpendicular that is dropped from $P$ onto the parametric curve $P_t$ has the formula: $$Q_t=(x(t)-sy'(t),y(t)+sx'(t))$$ Define a new point $M$ that is the intersection of the parametric curves $P_t$ and $Q_t$. How do I find the coordinates of new point $B$ such that $$AM=BM$$ along the parametric curve $Q_t$?
Edit: Answer reference here: How do I drop a perpendicular from a point onto a parametrized curve?

maximusavant25
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    Your question here is quite confused. You expect people to read through the other post to even figure out what you are asking in this one (don't do that - make the question itself self-contained, though referencing the other post is acceptable). In addition to the lack of clarity in the wording, you also denote the second curve with a prime, even though the post also uses the prime to indicate derivatives of the first curve's coordinates, Please try to read your post as if you were someone seeing it for the very first time, having no idea what anything is, Does it seem clear to you? – Paul Sinclair May 05 '25 at 20:00

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To find lengths along curves, you need to use an arclength parametrization of the curve instead of an arbitrary parametrization. That is, in

$$P(r) = (x(r), y(r))$$ $r$ needs to measure arclength along the curve from some point $(x_0, y_0)$ on the curve to $(x(r), y(r))$.

For an arbitrary parametrization $P_t = (x(t), y(t))$, the arclength between points $t=t_0$ and $t$ is found by $$r(t) = \int_{t_0}^t \sqrt{x'^2(t) +y'^2(t)}\,dt$$ This defines $r$ as strictly increasing function of $t$ (assuming $(x'(t), y'(t))$ is never $(0,0)$ except at isolated points). Because it is increasing, it is one-to-one, and therefore invertible. Denote the inverse function as $t(r)$. Then $P(r) = \big(x(t(r)), y(t(r))\big)$.

Do this for your curve $Q_t$ (I originally said "both" curves, but later realized that you are only measuring distances on the $Q$ curve, so there is no reason to reparametrize the $P$ curve).

Now your question becomes simple. Because $M$ is on $Q$, there is some $m_q$ such that under the arclength parametrizations $M = Q(m_q)$. $A$ is also expressed as $Q(r_q) = Q(m_q + (r_q - m_q))$ for some $r_q$. The distance along $Q$ from $A$ to $M$ is $|m_q - r_q|$. And the reflection along $Q$ of $A$ through $M$ is obtained by moving that same distance in the other direction. I.e., $B = Q(m_p - (r_q - m_q)) = Q(2m_q - r_q)$.

Paul Sinclair
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