Calculus of Variations help I've been studying Chapter 6 in Taylor's Mechanics book. And am working through the odd-numbered problems. I am struggling with 6.13, which reads:

In relativity theory, velocities can be represented by points in a certain "rapidity space" in which the distance between two neighboring points is $$ds = \frac{2}{1 - r^2}\sqrt{dr^2 + r^2d\phi^2},$$ where $r$ and $\phi$ are polar coordinates, and we consider just a two-dimensional space.(An expression like this for the distance in a non-Euclidean space is often called the metric of the space.) Use the Euler-Lagrange equation to show that the shortest path from the origin to any other point is a straight line.

I'd appreciate some assistance please.
Distance $= S = \int\frac{2}{1 - r^2}\sqrt{dr^2 + r^2d\phi^2}$
I'm trying to show $\phi(r) = c$. So I shall write the above as:
$\int\frac{2}{1 - r^2}\sqrt{1 + r^2\phi'^2} dr$, where $\phi' = \frac{d\phi}{dr}$ 
The Euler-Lagrange equation in this case is:
$\frac{\partial f}{\partial \phi} - \frac{d}{dr} \frac{\partial f}{\partial \phi'} = 0$, where $f = \frac{2}{1 - r^2}\sqrt{1 + r^2\phi'^2)}$
This gives: $0 = \frac{d}{dr} (\frac{2}{1 - r^2} \frac{r^2\phi'}{\sqrt{1 + r^2\phi'^2}})$
So $(\frac{2}{1 - r^2} \frac{r^2\phi'}{\sqrt{1 + r^2\phi'^2}}) = k$
And at this point I am not sure what to do. I tried solving for $\phi'$, but that doesn't seem to help. I know seeing as I want to show $\phi = c$, that I must find that $\phi' = 0$. But I am not sure how to get there.
The best I came up with, which I think is most likely wrong, is to argue that
$(\frac{2}{1 - r^2} \frac{r^2\phi'}{\sqrt(1 + r^2\phi'^2)}) = k$ is of the form $g(r) = 0$. And so the LHS must equal $0$ for any choice of $r$. And so, plugging in $0$ to the LHS, gives that $k = 0$. And, then using that fact, it is clear that $\phi' = 0$, and so $\phi = c$ as desired.
I'd appreciate some pointers on how to approach this problem. And why my argument above is wrong. Thanks. 
 A: I will offer a partial answer. I have worked out the solution mostly. I will then do an analysis of the limiting behavior to try to solve the differential equation.
$$ L= \frac{2}{1 - r^2}\sqrt{dr^2 + r^2d\phi^2} $$
Just to be careful with the differential, define the coordinate $d\xi(r) = rd\phi$. Since $\phi(r)$ we can just write it as a pure function of $r$. Then, we have that 
$$ L= \frac{2}{1 - r^2}\sqrt{1 + \xi'^2} $$
where the prime is the derivative wrt $r$. Then, we have that 
$$\frac{\partial L}{d \xi} = 0  $$
and hence that 
$$\frac{d}{dr}\frac{d L}{d\xi'} = \frac{d}{dr}\frac{2}{1 - r^2} \frac{\xi'}{\sqrt{1 + \xi'^2}} =0 $$
So that 
$$\ \frac{\xi'}{\sqrt{1 + \xi'^2}} = \frac{k}{2}(1-r^2) $$
implies that 
$$\ \frac{\xi'^2}{1 + \xi'^2} = \frac{k^2}{4}(1-r^2)^2 $$
Or rather that 
$$\xi'^2 - \xi'^2\frac{k^2}{4}(1-r^2)^2 = \frac{k^2}{4}(1-r^2)^2 $$
so that 
$$ \xi'^2 = \frac{\frac{k^2}{4}(1-r^2)^2}{1 - \frac{k^2}{4}(1-r^2)^2}  .$$
This is gross, but we are able to gain some insight. Namely, in the special case where $r\to 1$ we have that 
$$ \xi'^2 = const \implies \xi \propto r \implies  r\phi \propto r$$
and hence only in the case of $r\to 1$ do we have that $\phi = const$. Similarly for $r\to 0$.
Edit: This paper derives the fact that $\phi = const$ on page 18. It looks like your original intuition was close! 
