A rope is hanging between two poles find this Two poles of 50 m height are connected with 80m rope hanging between them. The rope is hanging 20m above the ground. What is the distance between the two poles?
My attempt: I tried finding the equation as $f(x)=kx^2$ with the length of half of the rope 40 m. So:
$$\int_0^a\sqrt{1+f'(x)^2}dx=40$$
I made some attempts but was unable to solve it.
One more thing: I thought about the catenary equation but didn't get it after searching a lot.
 A: 
First of all, the shape of rope is not a parabola but a catenary, and all catenaries are similar, defined by:
$$y=a\cosh\frac xa$$
You just have to figure out where the origin is (see picture). The hight of the lowest point on the rope is 20 and the pole is 50 meters high. So the end point must be $a+(50-20)$ above the $x$-axis. In other words $(d/2, a+30)$ must be a point on the catenary:
$$
a+30=a\cosh\frac{d}{2a}\tag{1}
$$
The lenght of the catenary is given by the following formula (which can be proved easily):
$$s=a\sinh\frac{x_2}{a}-a\sinh\frac{x_1}{a}$$
where $x_1,x_2$ are $x$-cooridanates of ending points. In our case:
$$80=2a\sinh\frac{d}{2a}$$
$$40=a\sinh\frac{d}{2a}\tag{2}$$
You have to solve the system of two equations, (1) and (2), with two unknowns ($a,d$). It's fairly straightforward.
Square (1) and (2) and subtract. You will get:
$$(a+30)^2-40^2=a^2$$
Calculate $a$ from this equation, replace that value into (1) or (2) to evaluate $d$.
My calculation:
$$a=\frac{35}{3}\approx 11.67$$
$$d=\frac{70}{3}\text{arccosh}\frac{25}{7}\approx 45.40$$
