Actually small question, but i need also small hints
So I have the canonical equation: $$y^2=2px \space (0 \le x \le x_0)$$
As far as I know there's the following integral:
$$\int^b_a\sqrt{1+f'(y)^2}dy$$
but should I simplify the function like this?
$$y = \sqrt{2px}$$
And only then find $y'$ ?
Either way same result. We can take
$$ x=f(y )$$
or
$$ y= g(x) $$
Prime denotes differentiation wrt other independent variable
$$x'=f'(y )$$
or
$$y'= g'(x) $$
$$s = \int \sqrt{1+ f'(y)^2} dy = \int \sqrt{1+ g'(x)^2} dx$$