Q:

Why is the Lagrangian for space-like geodesics equal to 1?

In Schwarzschild spacetime, the Lagrangian can be defined as 
$$ L = -\left( 1 - \frac{2M}{r} \right) \dot{t}^2 + \left( 1- \frac{2M}{r} \right)^{-1} \dot{r}^2 + r^2 \dot{\theta}^2 + r^2 \sin^2\theta \dot{\phi}^2 $$
where the dot denotes a derivative $\frac{\mathrm d}{\mathrm d\lambda}$ for an affine parameter $\lambda$. 
From Noether's theorem, one can easily prove that the Lagrangian itself if an integral of motion. One can also prove that $\theta = \text{const} =  \pi/2$ without loss of generality. Hence, the quantity $$Q = -\left( 1 - \frac{2M}{r} \right) \dot{t}^2 + \left( 1- \frac{2M}{r} \right)^{-1} \dot{r}^2 + r^2 \dot{\phi}^2$$ is a constant of motion. 
In my lecture notes, it is claimed without proof that $Q = 1$ for space-like geodesics. It seems like this is rather obvious from the way it is stated but I can't seem to find any proof of this anywhere in my lecture notes nor from Google (after a little bit of searching). Hence, my question is why is this the case and how can one prove this? 

A:

The solutions to the Euler-Lagrange (EL) equations for OP's non-square root Lagrangian $L$ are affinely parametrized geodesics.
Noether's theorem yields that $L$ is a constant (of motion) along a geodesic, but it is not necessarily $1$. For more information, see e.g. my Phys.SE answer here.
The constant is $1$ if one picks the parameter $\lambda$ to be arclength $s$.

