I want to find the geodesics of the hyperbolic plane, which has the metric $$ds^2=\frac{1}{y^2} (dx^2+dy^2)$$, as functions of $s$. I am given the result which is $$\dot{x}^2+\dot{y}^2=y^2$$ $$\dot{x}=ky^2$$. Now I can understand how the first equation is derived but I have no idea how $\dot{x}=ky^2$ is derived. Can you explain how to derive it?
Also if I wanted to find the geodesics with respect to $t$, here is what I would do. $$s=\int \frac{1}{y} \sqrt{\dot{x}^2+\dot{y}^2} dt$$ And use the Lagrangian $$L=\frac{1}{y} \sqrt{\dot{x}^2+ \dot{y}^2}$$ to solve the Euler-Lagrange equations, but it seems that I would get some complicated differential equations like this. I want to ask if this approach is correct and if not what should I do instead?
$ L(t, x,y, \frac {dx}{dt}, \frac{dy}{dt}) =\frac{1}{y} \sqrt{(\frac {dx}{dt})^2 + (\frac {dy}{dt})^2} $
Noting that $ \frac{\partial L}{\partial x} = 0 $ , use $ \frac{d}{dt} \frac{\partial L}{\partial \frac{dx}{dt} } = 0 $ to get
$ \frac{\partial L}{\partial \frac{dx}{dt} } = $ constant
Now,
$ \frac {dx}{dt} = \dot x \dot s $ , $ \frac {dy}{dt} = \dot y \dot s $ ,
$ \frac {ds}{dt} = \dot s $ ,
$ \frac {dx}{ds} = \dot x $ , $ \frac {dy}{ds} = \dot y $ ,
$ {\frac{dx}{dt}}^2 + {\frac{dy}{dt}}^2 = ({\dot x}^2 + {\dot y}^2){\dot s}^2 = y^2 {\dot s}^2 $
So,
$ \frac{\partial L}{\partial \frac{dx}{dt} } = \frac {dx}{dt} /y({\frac{dx}{dt}}^2 + {\frac{dy}{dt}}^2)^{\frac {1}{2}} = c $
Then,
$ \dot x = c y^2 $