TLDR:
You want to solve for $v$ in this equation:
$$v\sqrt{v^2+1} + \sinh^{-1} v = 2a_1x_0\sqrt{u^2+1} + \frac{\sinh^{-1} u}{2a_0}$$
And then $x_1=\frac{v}{2a_1}$ is your solution.
I’ve been banging away at this for hours, not because I think I can help you—much of this is new to me—but because I found it interesting. I haven’t yet worked out (an approximation to a) solution for $v$, though. And it looks like Claude’s answer has beaten me to it, anyway.
But here is my working, just because I can’t bear to hit Discard on this whole thing.
I don’t think I’ve ever done the length of a parabolic curve before. So while this would probably go more smoothly if done from the integral definition of arc length, I’m going to take the easy(?) way out: Wikipedia has a parabolic arc length formula. Let’s give it a go!
So we have a parabola $f\left(x\right)=a_0 x^2$. Wikipedia tells us that we can find the arc length, from the vertex at $\left(0,0\right)$ to any point $\left(x,f\left(x\right)\right)$ on the parabola, using these values:
- The focal length $l$ of the parabola; in this case, $l=\frac{1}{4a_0}$
- The perpendicular distance $p$ between the point and the axis of symmetry; in this case it’s simply $p=x$
Then, given $h=\frac{p}{2}$ and $q=\sqrt{l^2+h^2}$, the arc length is:
$$s=\frac{hq}{l}+l\ln\frac{h+q}{l}$$
Let’s simplify. Given that $h=\frac{x}{2}$…
$$\begin{align}
q &= \sqrt{\frac{1}{16a_0^2}+\frac{x^2}{4}} \\
&= \sqrt{\frac{4a_0^2x^2+1}{16a_0^2}} \\
&= l\sqrt{4a_0^2x^2+1}
\end{align}$$
Thus:
$$s=\frac{x}{2}\sqrt{4a_0^2x^2+1}+\frac{1}{4a_0}\ln\left(2a_0x+\sqrt{4a_0^2x^2+1}\right)$$
Now, we have another parabola $g(x)=a_1 x^2$ such that the arc lengths of $f\left(x_0\right)$ and $g\left(x_1\right)$ are equal, i.e.:
$$\begin{align}
\frac{x_0}{2}\sqrt{4a_0^2x_0^2+1}+\frac{1}{4a_0}\ln\left(2a_0x_0+\sqrt{4a_0^2x_0^2+1}\right) &= \frac{x_1}{2}\sqrt{4a_1^2x_1^2+1}+\frac{1}{4a_1}\ln\left(2a_1x_1+\sqrt{4a_1^2x_1^2+1}\right) \\
\therefore x_0\sqrt{4a_0^2x_0^2+1}+\frac{1}{2a_0}\ln\left(2a_0x_0+\sqrt{4a_0^2x_0^2+1}\right) &= x_1\sqrt{4a_1^2x_1^2+1}+\frac{1}{2a_1}\ln\left(2a_1x_1+\sqrt{4a_1^2x_1^2+1}\right)
\end{align}$$
And we want to solve for $x_1$ in terms of $a_0$, $a_1$, and $x_0$. Nothing simpler! </sarc>
Let’s define $u=2a_0x_0$ (thus $u^2=4a_0^2x_0^2$) and $v=2a_1x_1$ (thus $v^2=4a_1^2x_1^2$). That shortens things to:
$$x_0\sqrt{u^2+1} + \frac{1}{2a_0}\ln\left(u+\sqrt{u^2+1}\right) = x_1\sqrt{v^2+1} + \frac{1}{2a_1}\ln\left(v+\sqrt{v^2+1}\right)$$
Now I see where $\sinh$ comes into the other answers! It’s because $\sinh^{-1} x = \ln\left(x+\sqrt{x^2+1}\right)$, so we get:
$$\begin{align}
x_0\sqrt{u^2+1} + \frac{\sinh^{-1} u}{2a_0} &= x_1\sqrt{v^2+1} + \frac{\sinh^{-1} v}{2a_1} \\
\therefore 2a_1x_0\sqrt{u^2+1} + \frac{\sinh^{-1} u}{2a_0} &= 2a_1x_1\sqrt{v^2+1} + \sinh^{-1} v \\
&= v\sqrt{v^2+1} + \sinh^{-1} v
\end{align}$$
That right-hand side does not look easy to invert. I admit, I copped out and asked Wolfram Alpha to do it. And of course it tells me, “no result found in terms of standard mathematical functions”. [sigh…]
Based on Tyma Gaidash’s answer, I went looking into the Lagrange inversion theorem. My engineering-oriented education never covered this, but I think I’ve grasped the basics of it. As I understand it, to solve $y=f(x)$ for $x$, we choose some $z$, such that $f(z)$ is defined and $f'(z)\ne 0$.
Let’s shorten the entire left-hand side of the equation to $w$, and define $w=g\left(v\right)=v\sqrt{v^2+1} + \sinh^{-1} v$. First, let’s find the derivative… by cheating and using Wolfram Alpha: $g'(v)=2\sqrt{v^2+1}$.
We need a value $z$ where $g'(z)\ne 0$. Conveniently, this derivative is nowhere zero on the reals, so that’s trivial. I think (but am not sure) that $z$ should approximate $v$, so let’s randomly assume that $x_1\approx 1$ and so $v\approx 2a_1 = z$.
Now, by the inversion theorem, the inverse function $v=g^{-1}\left(w\right)$ is:
$$\begin{align}
v &= z+\sum_{n=1}^\infty \left[\frac{\left(w-g\left(z\right)\right)^n}{n!} \lim_{t\to z} \frac{d^{n-1}}{dt^{n-1}}\left(\frac{t-z}{g(t)-g(z)}\right)^n\right] \\
&= z+\left(w-g\left(z\right)\right)\lim_{t\to z}\frac{t-z}{g(t)-g(z)} + \frac{\left(w - g\left(z\right)\right)^2}{2} \lim_{t\to z}\frac{d}{dt} \left(\frac{t-z}{g(t)-g(z)}\right)^2 + \cdots
\end{align}$$
Doesn’t that first limit look just like the reciprocal of the derivative? So the second term of the series becomes $\frac{w-g\left(z\right)}{2\sqrt{z^2+1}}$.
And I’ve been working on this for way too long now, so that’s where I’ll stop for the night.
