Length of cable parabola with height $\propto$ (horizontal distance from midpoint)² Question:
The cable to a certain suspension bridge has the shape of a parabola. The height at a certain point is proportional to the square of the horizontal distance from the midpoint. The dimensions are given in the image.
Determine the length of the cable.

Attempted solution:
I add a few things to it.

The height at a certain point is something that I interpret as the $y$-value of a currently unknown function. Once we have this function, we will use the arc length integration to get the length of the cable, either by integrating from one end to the other, or integrate from the middle to the end and double it:
$$L = 2 \cdot \int_0^{25} \sqrt{1+ f'(x)^2}$$
The height is proportional to the square of the horizontal distance from the midpoint:
$h(x) = kx^2$
How do we determine the $k$ value? Well, there is an initial condition in the image. When the distance is 25 meters from center, the height is $10 + h_0$. Putting this in:
$10 + h_0 = 25^2 k \Rightarrow 625k - 10 = h_0$
But this leaves me with one equation and two unknowns and cannot really progress. I think I have not understood the basic setup for the question. What are some productive ways to proceed and finish this question off?
The expected answer is:
$$5\sqrt{41} + \frac{125}{4}\ln \frac{4 + \sqrt{41}}{5} m$$

Your formula, since $h(0)=h_0$, is $h(x)=kx^2+h_0$. And the $h_0$ plays no role in the length formula, because the derivative will kill it. Then your integral gives the value you want. 
