I'm trying to find that the equation of a particle under the influence of a gravitationel field is $z(x)=Ax^2+Bx+C$.
I'm using the Maupertuis principle (because why not ...): $\delta\displaystyle{S(x)=\delta\int \sqrt{2m(H-mgz)}\sqrt{1+\left(\frac{\mathrm{d}x}{\mathrm{d}z}\right)^2}~\mathrm{d}z}=\delta\displaystyle{\int L(z,x') \mathrm{d}z}=0$
Which leads (using the Euler-Lagrange equation) to $\displaystyle{\frac{\partial L}{\partial x}-\frac{d}{\mathrm{d}z}\left(\dfrac{\partial L}{\partial {x}'}\right)=0\Rightarrow \dfrac{\partial L}{\partial x'}=a}$ and finally: $\displaystyle{x'(z)\sqrt{\frac{2m(H-mgz)}{1+x'(z)^2}}}=a$.
I know that $x(z)$ must lead to an equation: $z(x)$ being a parabola but I've no idea how to find $x(z)$...
Maybe there is an easiest way to find $z(x)$ using the Euler lagrange equation (with the Maupertuis's principle not the Hamiton's principle) but I can't find it.
For starters, rearrange to express $x^\prime$ as a function of $z$. Then you can integrate to get $x$ as a function of $z$.