take two points $(-2, cosh(2))$ and $(2, cosh(2))$. Now we take a parabola $p(x)=kx^2+d$ and a catenary, both intersecting at the two given points.
What is $k$, so that the length of both curves is equal?
Length of catenary = Length of parabola ->
$2sinh(2) = \frac{1}{2} \cdot (4k\cdot \sqrt{1+(4k)^2} + arsinh(4k))$.
It should be correct so far. How can I solve the equation for k? Do you have any hints for me?
I have found a different result
The parabola's equation is
$p(x)=k x^2-4 k+\cosh 2$
$p'(x)=2kx$
$$L=\int_{-2}^2 \sqrt{1+4 k^2 x^2} \, dx=2 \sqrt{16 k^2+1}+\frac{\text{arcsinh}(4 k)}{2 k}$$
and from catenary $$L=\int_{-2}^2 \sqrt{\sinh ^2(x)+1} \, dx=2\sinh 2$$
so the equation to solve is $$f(k)=2 \sqrt{16 k^2+1}+\frac{\text{arcsinh}(4 k)}{2 k}-2\sinh 2=0$$ we can easily see that $f(0.1)\approx -3$
while $f(1)\approx 2$
Therefore a zero is in the interval $(0.1,1)$.
Let's guess $k_0=0.5$
and define
$k_{n+1}=k_n-\dfrac{f(k_n)}{f'(k_n)}$
we get the following table
$ \begin{array}{r|l} k_0 & 0.5 \\ k_1 & 0.720893 \\ k_2 & 0.708215 \\ k_3 & 0.708188 \\ k_4 & 0.708188 \\ \end{array} $
Therefore we have $k\approx 0.708188$
as the function is even we have also the solution $k=-0.708188$
Hope this helps
Edit
I verified: $L=2\sinh 2\approx 7.25372$ for catenary and
$L \approx 7.25372$ for parabola $p(x)=0.708188 x^2+0.929444$