Given a quadratic curve $y = ax^2+bx+c$ and a point $(n,y(n))$ on the curve, I need to find the coordinate of a new point $(m,y(m))$ on the curve and at certain arc distance $L$ from the original point $(n,y(n))$. Please note, $L$ is not straight euclidean distance, it is, however, the curve segment length between two points.
I first tried to solve this using the arc length formula: The length $L$ of the quadratic arc segment between points $(n,y(n))$ and $(m,y(m))$ can be calculated as: $$L = \int_n^m\sqrt{1+(b+2ax)^2}dx$$ Assuming that $n, a, b$ are known parameters, now the question is how to find a closed form equation for $m$ given the arc segment length $L$.
I tried to solve the integral and expand $L$: $$L = \frac{-arsinh(2an+b)-(2an+b)\sqrt{4a^2n^2+4abn+b^2+1}}{4a}+\frac{arsinh(2am+b)+(2am+b)\sqrt{4a^2m^2+4abm+b^2+1}}{4a}$$
I am, however, not able to find an explicit solution to this equation for $m$.
I appreciate it if anyone could come up with a solution.
-- assumptions: $x,y$ can be positive or negative.
So, you want to solve for $m$ the equation $$4a L+\sinh ^{-1}(2 a n+b)+(2 a n+b) \sqrt{1+(2 a n+b)^2}=$$ $$\sinh ^{-1}(2 a m+b)+\sqrt{1+(2 a m+b)^2} (2 a m+b)$$
Let $$k=4a L+\sinh ^{-1}(2 a n+b)+(2 a n+b) \sqrt{1+(2 a n+b)^2}$$ and $x=2am+b$ to make $$k=\sinh ^{-1}(x)+x\sqrt{1+x^2} $$ This is an highly transcendental equation and,as @Gerry Myerson commented, a numerical method would be required (Newton method being the simplest) and, as usual, you need a starting guess (as good as possible if you want to reduce the number of iterations).
A reasonable approximation is $$\sinh ^{-1}(x)+x\sqrt{1+x^2} \sim x^2\implies x_0=\sqrt k$$
Let us try with $a=3$, $b=5$, $n=7$ and $L=1234$ which give $k=14808+47 \sqrt{2210}+\sinh ^{-1}(47)$ which is $\approx 17022.$.
This would generate the following Newton iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 130.4685531 \\ 1 & 130.4453134 \\ 2 & 130.4453113 \end{array} \right)$$ corresponding to $m=20.9075518$.
With this result, repeating the calculations, we get $L=1233.9999889$