Find the curve with the shortest path on a surface(geodesic) 1. The problem statement, all variables and given/known data
Let  $U$ be a plane given by $\frac{x^2}{2}-z=0$
Find the curve with the shortest path on $U$ between the points $A(-1,0,\frac{1}{2})$ and $B(1,1,\frac{1}{2})$
I have a question regarding the answer we got in class.
2. Relevant equations
Euler-Lagrange
$$I(y)=\int L(x,y,y')dx$$ has extremes when $$L_y-\frac{d}{dx}L_{y'}=0$$
3. The attempt at a solution
So how what we did in class was.
Let $\gamma (x)=(x,y(x),\frac{x^2}{2})$ then the shortest path is going to be the minimum of the functional
$$I(\gamma)=\displaystyle\int_{-1}^{1}\sqrt{dx^2+dy^2+dz^2}=\displaystyle\int_{-1}^{1}\sqrt{1+(y')^2+x^2}dx$$
Now using the Euler-Lagrange equation for the extremes of a functional we get that:
$L_y=0$
$L_{y'}=\frac{y'}{\sqrt{1+(y')^2+x^2}}$
Therefore we are going to have extremes when
$\frac{d}{dx}\frac{y'}{\sqrt{1+(y')^2+x^2}}=0$ which means that $\frac{y'}{\sqrt{1+(y')^2+x^2}}=C$
Now solving this DE we get
$y'=D\sqrt{x^2+1}\implies  y=\frac{D}{2}( (\sqrt{x^2 + 1} x + \sinh^{-1}(x)) +E$
Which would mean that the shortes path on the curve would be
$\gamma(x)=(x,y(x),\frac{x^2}{2})$ where we could get $E,D$ from the initial conditions $y(-1)=0,y(1)=1$
All seems good. However last week in another class we said the a curve $\gamma\,\text{is a geodesic}\iff \gamma''|| N$. Where $N$ is the normal of the surface $U$
I decided to check if this holds for the curve we got and got
$N=\nabla( \frac{x^2}{2}-z)=(x,0,-1)$
$\gamma ''(x)=(0, \frac{D x}{\sqrt{x^2 + 1}}, 1)$
And here we see that $N\nparallel \gamma '' $. Can anyone explain me why this is the case? does that mean that the shortest path is not a geodesic?
 A: It depends on how one parametrize the line. A curve $\gamma(t)$ is geodesic if $\gamma''(t)=0$ or $\gamma''$ is parallel to the normal. So a line $\gamma(x)=ax+b$ is geodesic. However, you can parametrize the same line as $\gamma=a\tan \alpha+b$, with $\alpha \in(-\pi/2,\pi/2)$. Then $$\gamma''(u)=\frac{2a\sin \alpha}{\cos^3\alpha}$$ 
This expression is zero only when $\alpha=0$. In principle is possible to parametrize every curve such that it's geodesic. Not sure how easy it is for this particular case.
A: A minimizing curve is geodesic only up to reparameterization.
Assume that exists a reparameterization $\phi$ such that $\delta = \gamma \circ \phi$ is a geodesic.
We have
$$\dot\delta(t)= \dot\gamma(\phi(t))\dot\phi(t) = \left(1, D\sqrt{\phi(t)^2+1}, \phi(t)\right)\dot\phi(t)$$
In particular, a geodesic $\delta$ has to be of unit speed so $$1 = \|\dot\delta\|^2 = (D^2+1)(\phi^2+1)\dot\phi^2 \implies \dot\phi = \frac{1}{\sqrt{(D^2+1)(\phi^2+1)}} \implies \phi(t) = \frac1{\sqrt{D^2+1}} \operatorname{Arsinh}(t)$$
Then $$\dot\delta(t) = \left(\frac{1}{\sqrt{D^2+1} \sqrt{\frac{\operatorname{Arsinh}^2(t)}{D^2+1}+1}},\frac{D}{\sqrt{D^2+1}},\frac{\operatorname{Arsinh}(t)}{\left(D^2+1\right) \sqrt{\frac{\operatorname{Arsinh}^2(t)}{D^2+1}+1}}\right)$$
and
$$\ddot\delta(t) = \left(-\frac{\operatorname{Arsinh}(t)}{(D^2+1)^{3/2}\sqrt{t^2+1} \left(\frac{\operatorname{Arsinh}^2(t)}{D^2+1}+1\right)^{3/2}},0,\frac{1}{\sqrt{D^2+1}\sqrt{t^2+1} \left(\frac{\operatorname{Arsinh}^2(t)}{D^2+1}+1\right)^{3/2}}\right)$$
Evidently $\ddot\delta \parallel N\circ \delta$ since $N(\delta(t)) = (\phi(t),0,-1) = \left(\frac1{\sqrt{D^2+1}} \operatorname{Arsinh}(t),0,-1\right)$.
