Given a parabola and a fixed arc length, how do I find the matching point on a different parabola?

The arc length of a parabola has a closed form, and inverting it does not — which is the whole difficulty.

For y = a x², the arc length from the origin to x is an integral of the square root of (1 + 4a²x²). That integral evaluates in terms of a square-root term plus an inverse hyperbolic sine. Writing t = 2ax makes it tidy: the length is proportional to

t·√(t² + 1) + sinh⁻¹(t)

divided by 4a.

So given a and x you get L directly. What you want is the reverse — given L and a new a, solve for t. That equation mixes a polynomial term with an inverse hyperbolic one, and there is no closed-form inverse.

So the honest answer is: the forward direction is exact, the backward direction is numerical.

The practical method, which converges very fast:

1. Compute L once from the original curve using the closed form.
2. For the new coefficient, define k = 4·a_new·L and solve k = t·√(t²+1) + sinh⁻¹(t) for t.
3. Recover the point from x = t / (2·a_new).

For step 2, a good starting guess is t₀ = √k, which comes from noticing that for large t the expression is dominated by the t² term. From there, Newton's method converges in a handful of iterations because the function is smooth and strictly increasing.

Strict monotonicity is worth noting: the right-hand side increases without bound as t increases, so there is exactly one solution and no ambiguity about which root you found. That makes even a crude bisection perfectly reliable if you would rather not differentiate anything.

There is a nice way to think about what you are computing, which also tells you the answer's qualitative shape.

You are asking for the level curves of the arc-length function on the plane of (x, a) pairs. Each curve joins all the points that are the same distance along their respective parabolas.

Those level curves are always perpendicular to the gradient of the arc-length function, which gives you a differential equation describing them. Solving that numerically traces the whole family at once, which is more efficient than solving separately for each new coefficient if you want many of them.

Qualitatively: as the parabola gets tighter, the matching point moves inward in x and the curve climbs faster, exactly as a bent wire would. That the answer behaves like a wire is not a coincidence — arc length is precisely the quantity a wire conserves.

If you need this inside something interactive rather than as a one-off, precompute.

Tabulate the forward function on a grid of t, then invert by interpolation. Since it is smooth and monotonic, a modest table plus cubic interpolation gives you accuracy far beyond what any drawing needs, at a fraction of the cost of running Newton's method every frame.

That is the standard engineering answer to "this function has no closed-form inverse" and it is almost always the right one.