Riemannian metric and geodesics on a cone 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.
 A: 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.
