If we are given a surface S given by $z^2 = a(x^2 + y^2)$, $z>0$. I want to find the Riemannian metric of the cone and an explicit formula for the geodesics.
I parametrise it by $\sigma \colon U \to S$ where $U$ is an open subset of $\mathbb{R}^2$. $U = \{(u,v): 0<u<2\pi, 0<v<\infty\}$. $\sigma (u,v) = (v \sin(u), v \cos(u), \sqrt{a}v)$ is a smooth parametrisation of the cone minus a line.
We can show that the Riemannian metric is given by $E=v^2, F=0, G=1+a$.
To solve the geodesic ODEs, we find that given a curve $\gamma = (x(t), y(t)) \colon [a,b] \to U$ $$y^2\dot x = c$$ $$(1+a)\ddot y = y\dot x^2$$
I am lookin for an answer as to whether what I have done so far is correct and to complete the argument.
If your length is $$l=\int\sqrt{v^2u'^2+(1+a)v'^2}dt\qquad\qquad (1)$$ we can change it by $t=v$ to get $$\int\sqrt{1+a+v^2\dot{u}^2}dv$$ In terms of variational calculus the Lagrangian depends as $${\cal{L}}={\cal{L}}(v,u,\dot{u})={\cal{L}}(v,\dot{u}),$$ and to find extremes the integrand must comply $$\frac{d}{dv}\frac{\partial{\cal{L}}}{\partial\dot{u}}-\frac{\partial{\cal{L}}}{\partial u}=0.$$ So, since $\frac{\partial{\cal{L}}}{\partial u}=0$ and $\frac{\partial{\cal{L}}}{\partial\dot{u}}=\frac{v^2\dot{u}}{\sqrt{1+a+v^2\dot{u}^2}}$ but $\frac{d}{dv}\frac{\partial{\cal{L}}}{\partial\dot{u}}=0$ then $\frac{v^2\dot{u}}{\sqrt{1+a+v^2\dot{u}^2}}=K$ so $$\dot{u}=\frac{K\sqrt{1+a}}{v\sqrt{v^2-K^2}},$$ which upon integration gives $$u=M+\sqrt{1+a}\cos^{-1}\left(\frac{K}{v}\right).$$ So $$v=K\sec\left(\frac{u-M}{\sqrt{1+a}}\right).$$
With this your curve in the cone is $$u\mapsto K\sec\left(\frac{u-M}{\sqrt{1+a}}\right) \left( \begin{array}{c} \sin(u)\\ \\ \cos(u)\\ \\ \sqrt{a} \end{array}\right),$$ in angular terms, with $K,M$ some constants of integration.
Note: If you were to choose directly from integral $(1)$ to find extremes, where the Lagrangian is $$L=L(t,u,u',v,v')=L(u',v,v')=\sqrt{v^2u'^2+(1+a)v'^2},$$ the equations would be $\frac{d}{dt}\frac{\partial{L}}{\partial\dot{u}}-\frac{\partial{{L}}}{\partial u}=0$ and $\frac{d}{dt}\frac{\partial{L}}{\partial\dot{v}}-\frac{\partial{L}}{\partial v}=0$, which are the same (for geodesics) if you were using differential geometry slang.