How to straighten a parabola? Consider the function $f(x)=a_0x^2$ for some $a_0\in \mathbb{R}^+$. Take $x_0\in\mathbb{R}^+$ so that the arc length $L$ between $(0,0)$ and $(x_0,f(x_0))$ is fixed. Given a different arbitrary $a_1$, how does one find the point $(x_1,y_1)$ so that the arc length is the same?
Schematically,

In other words, I'm looking for a function $g:\mathbb{R}^3\to\mathbb{R}$, $g(a_0,a_1,x_0)$, that takes an initial fixed quadratic coefficient $a_0$ and point and returns the corresponding point after "straightening" via the new coefficient $a_1$, keeping the arc length with respect to $(0,0)$. Note that the $y$ coordinates are simply given by $y_0=f(x_0)$ and $y_1=a_1x_1^2$. Any ideas?
My approach: Knowing that the arc length is given by
$$
L=\int_0^{x_0}\sqrt{1+(f'(x))^2}\,dx=\int_0^{x_0}\sqrt{1+(2a_0x)^2}\,dx
$$
we can use the conservation of $L$ to write
$$
\int_0^{x_0}\sqrt{1+(2a_0x)^2}\,dx=\int_0^{x_1}\sqrt{1+(2a_1x)^2}\,dx
$$
which we solve for $x_1$. This works, but it is not very fast computationally and can only be done numerically (I think), since
$$
\int_0^{x_1}\sqrt{1+(2a_1x)^2}\,dx=\frac{1}{4a_1}\left(2a_1x_1\sqrt{1+(a_1x_1)^2}+\arcsin{(2a_1x_1)}\right)
$$
Any ideas on how to do this more efficiently? Perhaps using the tangent lines of the parabola?
More generally, for fixed arc lengths, I guess my question really is what are the expressions of the following red curves for fixed arc lengths:

Furthermore, could this be determined for any $f$?
Edit: Interestingly enough, I found this clip from 3Blue1Brown. The origin point isn't fixed as in my case, but I wonder how the animation was made (couldn't find the original video, only a clip, but here's the link)

For any Mathematica enthusiasts out there, a computational implementation of the straightening effect is also being discussed here, with some applications.
 A: Phrased differently, what we want are the level curves of the function
$$\frac{1}{2}f(x,y) = \int_0^x\sqrt{1+\frac{4y^2t^2}{x^4}}\:dt = \frac{1}{2}\int_0^2 \sqrt{x^2+y^2t^2}\:dt$$
which will always be perpendicular to the gradient at that point
$$\nabla f = \int_0^2 dt\left(\frac{x}{\sqrt{x^2+y^2t^2}},\frac{yt^2}{\sqrt{x^2+y^2t^2}}\right)$$
Now is the time to naturally reintroduce $a$ as the parameter for these curves. Therefore what we want is to solve the differential equation
$$x'(a) = \int_0^2 \frac{-axt^2}{\sqrt{1+a^2x^2t^2}}dt \hspace{20 pt} x(0) = L$$
where we substitute $y(a) = a\cdot x^2(a)$, thus solving for one component automatically gives us the other.

EDIT: Further investigation has led me to some interesting conclusions. It seems like if $y=f_a(x)$ is a family strictly monotonically increasing continuous functions and $$\lim_{a\to0^+}f_a(x) = \lim_{a\to\infty}f_a^{-1}(y) = 0$$
Then the curves of constant arclength will start and end at the points $(0,L)$ and $(L,0)$. Take for example the similar looking family of curves
$$y = \frac{\cosh(ax)-1}{a}\implies L = \frac{\sinh(ax)}{a}$$
The curves of constant arclength are of the form
$$\vec{r}(a) = \left(\frac{\sinh^{-1}(aL)}{a},\frac{\sqrt{1+a^2L^2}-1}{a}\right)$$
Below is a (sideways) plot of the curve of arclength $L=1$ (along with the family of curves evaluated at $a=\frac{1}{2},1,2,4,$ and $10$), which has an explicit equation of the form
$$x = \frac{\tanh^{-1}y}{y}\cdot(1-y^2)$$

These curves and the original family of parabolas in question both have this property, as well as the perfect circles obtained from the family $f_a(x) = ax$. The reason the original question was hard to tractably solve was because of the non analytically invertible arclength formula
A: For a parabola with parametrization
$$ x= at,y= a t^2/2 ;\;x^2= 2 a y \;; \text{slope} \;t=\tan \phi; \tag1$$
Differentiate $x^2$, primed wrt arc length
$$ 2 x \cos \phi = 2 a \sin \phi ;\; x= a \tan \phi ;\;x'= \cos \phi= a \sec^2 \phi \;\phi' \tag 2 $$
from which comes the curvature
$$  a \phi'=a \kappa=\cos^3 \phi \tag 3$$
An easier/direct way out is by direct numerical integration of ode in(3). A fraction k of max arc length  can be set as a parameter  for required integrands  ( k= 2/3 in this particular case), making the subset parabolas flatter or deeper by adjusting $a$.

Total length on one side is a given constant $L$
$$ L = \int _0^{\phi_m} \frac{ d \phi}{\kappa} ; \text{ now plug in curvature from(3) and integrate }$$
$$ \frac{L}{a}=\int _0^{\phi_m} \sec^3 \phi\;  d\phi = \frac12\bigg[\log\bigg(\tan\bigg(\frac{\pi}{4} + \frac{\phi_m}{2} \bigg)\bigg) +\sec \phi_m \tan \phi_m \bigg]\tag 4 $$
Plug in from (2) $x_{m}=a \tan \phi_m  $
$$ \frac{2L}{a}= \log\left(\tan\bigg(\frac{\pi}{4} + \frac{\tan^{-1}(x_m/a)}{2}\right)\bigg)+ (x_m/a) \sqrt{(x_m/a)^2-1} \tag 5 $$
which is a neat implicit function $ f(a,x_m,L) $ , plotted assuming an arm of parabola has given arc length $=1.8,$ on  Mathematica, enabling   flatter or deeper parabola plots.

It has become clear that there are two criteria for equal parabolic arc lengths  connecting $x_{max} $ to $ \text { a = 2* focal-length }$.
EDIT 1/2:
To come out of the apparent dilemma I have carefully calculated/plotted special cases for $( x_{max},a)$ combinations:
$$(0.4,0.0458),(0.8,0.1948), (1.2,0.4704),(1.6,0.8874)(2.0, 1.42264),(2.4,2.0101),(2.8,2.5825)$$
All arcs are all of same length but for $( x_{max} =2.0,2.4,2.8) $ the choice of $a$ seems to have got switched over to the second critirion. Relation $(x_{max},a) $ is not unique by the first plot...it is now examined further.

A: $L$ being the known arc length, let $x_1=\frac t{2a}$ and $k=4a_1L$; then you need to solve for $t$ the equation
$$k=t\sqrt{t^2+1} +\sinh ^{-1}(t)$$A good approximation is given by $t_0=\sqrt k$.
Now, using a Taylor series around $t=t_0$ and then series reversion gives
$$t_1=\sqrt{k}+z-\frac{\sqrt{k} }{2 (k+1)}z^2+\frac{(3 k-1) }{6 (k+1)^2}z^3+\frac{(13-15
   k) \sqrt{k} }{24 (k+1)^3}z^4+\cdots$$ where $$z=-\frac{\sqrt{k(k+1)} +\sinh ^{-1}\left(\sqrt{k}\right)-k}{2 \sqrt{k+1}}$$
Let us try for $k=10^n$
$$\left(
\begin{array}{ccc}
n & \text{estimate} & \text{solution} \\
 0 & 0.4810185 &  0.4819447 \\
 1 & 2.7868504 &  2.7868171 \\
 2 & 9.8244940 &  9.8244940 \\
 3 & 31.549250 &  31.549250 \\
 4 & 99.971006 &  99.971006 \\
 5 & 316.21678 &  316.21678 \\
 6 & 999.99595 &  999.99595
\end{array}
\right)$$  This seems to be quite decent.
A: Here's about the most efficient thing I can see is this:
Take your antiderivative (replacing the sin with a sinh) and define $a_1 x \equiv y$ so that
$$f(y) = a_1 * (\textrm{arc length}),$$
$$f(y) \equiv \frac{1}{2}(y \sqrt{1+y^2}+ \sinh^{-1} y).$$
$f(y)$ is monotonic and has some nice approximations when $y \ll 1, y \gg 1$. In those cases it might be possible to obtain it analytically. In general, invert it numerically. Then,
$$x = a_1^{-1} f^{-1}(a_1 * (\textrm{arc length}).$$
The utility of doing it this way is that you don't have to keep inverting a new function for each new $a_1$, you only have to do it once to be able to flatten any parabola you want.
A: Adding another short ODE method that finds a direct relation between maximum slope at parabola tip and its focal length for invariant arc length while bending.
$$ x = 2 f \tan \phi \;\text {is differentiated w.r.t. arc, } 2f \sec^2\phi \frac{d\phi}{ds}=\cos \phi$$
Integrate back but w.r.t. slope  $ \phi, s= 2 f \int \sec^3\phi\; d\phi$
$$ s= 2 f \left(\frac12 \log \frac{\cos (\phi/2)+ \sin (\phi/2) }{\cos (\phi/2)- \sin (\phi/2) } +\sec \phi +\tan \phi \right)= \text{constant}$$
Computed focal length $ f=F(\phi_{max})$ as the relation between bending variables on Mathematicafor maximum slopes $(0.2,0.5,0.8,1.1)$ radians with fixed arc length of parabola 3 units assumed and plotted as shown; $ (x,y)= (2ft,ft^2). $

