How to straighten a parabola?

Question

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.

Answer 1

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.

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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

Answer 2

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.

Answer 3

$\large{\text{Method 1:}}$

Here is a recursive answer for $g(a_0,a_1,x_1)$. Please see this graph for result verification. You can plug in some value for $x_1$ at the RHS of $x_1=g(x_1)=g(a_0,a_1,x_0)$:

$$\mathrm{\text{Arclength from 0 to }x_1\,\big(a_0x^2\big)=Arclength\ from \ 0\ to\ x_1\ \big(a_1x^2\big),x_0+c=x_1\implies\frac{2a_0x_0\sqrt{4a_0^2x_0^2+1}+sinh^{-1}(2a_0x_0)}{4a_0}=\frac{2a_1x_1\sqrt{4a_1^2x_1^2+1}+sinh^{-1}(2a_1x_1)}{4a_1}\implies A_0(x_1)=x_1=A_1(x_1)=\frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1x_1\sqrt{4a_1^2x_1^2+1} \right)}$$

Then we define and create the following recursive relation converging to $x=x_1$:

$$\mathrm{A_{n+1}(x_1)= \frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1A_n(x_1)\sqrt{4a_1^2A_n^2(x_1)+1} \right)\implies x_1=g(a_0,a_1,x_0)=\lim_{n\to\infty}A_n(x_1)=A_\infty(x_1)= \frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1(…)\sqrt{4a_1^2(…)+1} \right)\implies A_2(x_1)= \frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1{\frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1x_1\sqrt{4a_1^2x_1^2+1} \right)}\sqrt{4a_1^2{\frac1{2a_1}sinh \left(\sinh^{-1}(2a_0x_0)-2a_1x_0\sqrt{4a_0^2x_0^2+1}-2a_1x_1\sqrt{4a_1^2x_1^2+1} \right)}^2+1} \right)}$$ $\large{\text{Method 2:}}$

There is also another recursive method as seen in this other graph. This forms a horizontal line at y=$x_1$. There also may be another $\pm$ branch where the sign is chosen as needed, all + or all -. Notice the main square root argument is also a difference of squares:

$$\mathrm{\text{Arclength from 0 to }x_1\,\big(a_0x^2\big)=Arclength\ from \ 0\ to\ x_1\ \big(a_1x^2\big),x_0+c=x_1 \frac{2a_0x_0\sqrt{4a_0^2x_0^2+1}+sinh^{-1}(2a_0x_0)}{4a_0}=\frac{2a_1x_1\sqrt{4a_1^2x_1^2+1}+sinh^{-1}(2a_1x_1)}{4a_1}\implies B_0(x_1) =x_1= B_1(x_1)=\pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1x_1^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1x_1\right)\right)^2-1}}$$ Recursive solution: $$\mathrm{ x_1=g(a_0,a_1,x_0)=\lim_{n\to\infty}B_n(x_1)=B_\infty(x_1)= \pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1(…)^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1(…)\right)\right)^2-1},B_{n+1}(x_1)= \pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1B^2_n(x_1)}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1B_n(x_1)\right)\right)^2-1}\implies B_2(x_1)= \pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1{\left(\pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1x_1^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1x_1\right)\right)^2-1}\right)}^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1{\pm\frac1{2|a_1|}\sqrt{\frac1{4a^2_1x_1^2}\left(2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1x_1\right)\right)^2-1}}\right)\right)^2-1}}$$

$\large{\text{Method 3:}}$

Here is graphical proof of the solution. Now that we have eliminated the rest of the ways of solving for $x_1$, the last and simpler method is as follows.:

$$\mathrm{\text{Arclength from 0 to }x_1\,\big(a_0x^2\big)=Arclength\ from \ 0\ to\ x_1\ \big(a_1x^2\big),x_0+c=x_1 \frac{2a_0x_0\sqrt{4a_0^2x_0^2+1}+sinh^{-1}(2a_0x_0)}{4a_0}=\frac{2a_1x_1\sqrt{4a_1^2x_1^2+1}+sinh^{-1}(2a_1x_1)}{4a_1}\implies C_0(x_1)=x_1=C_1(x_1)=\frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1x_1)} {2a_1\sqrt{4a_1^2x_1^2+1}}}$$

Recursive solution for third method which converges to $x=x_1$ $$\mathrm{x_1=g(a_0,a_1,x_0)=\lim_{n\to\infty}C_n(x_1)=C_\infty(x_1), C_{n+1}(x_1)= \frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1C_n(x_1))} {2a_1\sqrt{4a_1^2C^2_n(x_1)+1}} = \frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1(…))} {2a_1\sqrt{4a_1^2(…)^2+1}}\implies C_2(x_1)= \frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}\left(2a_0x_0\right)-sinh^{-1}\left(2a_1{\frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1x_1)} {2a_1\sqrt{4a_1^2x_1^2+1}}}\right)} {2a_1\sqrt{4a_1^2\left(\frac{2a_1x_0\sqrt{4a_0^2x_0^2+1}+\frac{a_1}{a_0}sinh^{-1}(2a_0x_0)-sinh^{-1}(2a_1x_1)} {2a_1\sqrt{4a_1^2x_1^2+1}}\right)^2+1}}}$$

As you can see, $\mathrm{g(a_0,a_1,x_0)=x_1}$ does not have an easy form. This is like the W-Lambert/ Product Logarithm function where $\mathrm{ \ if \ xe^x=y,\ then \ x=W(y)=ln(y)-ln(ln(y)-ln(x))=…\,}$. We need a recurvise definition for W(x) seen in this graph. Therefore, one may need to derive a new function to solve recursively for $g(a_0,a_1,x_0)$ and these definitions definitely work.

Note that the MathJax may render differently based on your computer. See @Tim Pederick’s solution for an Inverse Lagrange theorem solution. I may work more on this. Please correct me and give me feedback!

$\large{\text{Method 4:}}$

The problem reduces to finding the inverse of $x-\sinh(x)$. You can get better results for the inverse of the arc length of $ix^2$ from

here

but the inverse of the arc length of $x^2$ can only solve

ArcLength$(x^2)=ia,a\in\Bbb R$

because of the domain of Mathematica’s Inverse Beta Regularized $\text I^{-1}_s(a,b)$. Choose the appropriate signs:

$$\text{ArcLength}(x^2)=\frac12\sqrt{4x^2+1}+\frac14\sinh^{-1}(2x)=z\in\Bbb I\text{ pure imaginary}\implies x\mathop=^{|\text{Im}(z)|\le \frac{\pi}8}\mp\frac i2\sqrt{\text I^{-1}_\frac{\pm8i z}\pi\left(\frac12,\frac32\right)}$$

Plot of $k\in\Bbb R$ vs $\mp\frac i2\sqrt{\text I^{-1}_\frac{\pm8i z}\pi\left(\frac12,\frac32\right)},z=ik$ where opposite signs are taken:

Test the function here by plugging in a real $k$

Answer 4

$L$ 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.

Answer 5

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.

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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!

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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:

$$s=\frac{x}{2}\sqrt{4a_0^2x^2+1}+\frac{1}{4a_0}\ln\left(2a_0x+\sqrt{4a_0^2x^2+1}\right)$$

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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>

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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…]

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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.

Answer 6

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.

Answer 7

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). $