Parabola arc length minimization

Question

Consider a parabola: $y = a x^2 + b x + c$. I want that parabola to have the vertex on the $x=0$ axis (so $b = 0$), to be downward concave (so $a<0$) and to pass through the point $(r,h)$ with $r$ and $h$ both positive. If $x=s$ is the point where the parabola intersect the $y=0$ axis, is it possible to find the value of $s$ (as a closed form function of $r$ and $h$) that minimize the length of the parabolic arc in $0<x<s$?

Answer 1

Assuming $(r,h)$ is fixed, you'll find that the parabola satisfying those conditions has the formula

$$ y = h\frac{s^2-x^2}{s^2-r^2} $$

The arc length in $(0,s)$ is given by

$$ L(s) = \int_0^s \sqrt{(y')^2+1}\ dx = \int_0^s \sqrt{\left(\frac{-2hx}{s^2-r^2}\right)^2+1}\ dx $$

Let $t = \frac{s^2-r^2}{2h}$, this becomes a common integral

\begin{align} L(s) &= \frac{1}{t} \int_0^s \sqrt{x^2+t^2}\ dx \\ &= \frac{1}{2t}\left( x\sqrt{x^2+t^2} + t^2\ln \Big(x + \sqrt{x^2+t^2}\Big) \right)\bigg|_0^s \\ &= \cdots \end{align}

Once you have $L(s)$, it remains to solve $L'(s)=0$. There might not be a closed form solution.