Asymptotic geodesic on hyperboloid. Consider a geodesic which starts at a point $P$ in the upper part $(z>0)$ of a hyperboloid of revolution $x^2+y^2−z^2=1$ and makes an angle $\theta$ with the parallel passing through $p$ in such a way that $cos(\theta)=1/r$, where $r$ is the distance from $P$ to the $z$-axis.
I have to prove that by following the geodesic asymptote in the direction of decreasing parallels, it approaches the parallel circle $x^2+y^2=1$, $z=0$. Thank you!
 A: It is a revolution surface That is we can parametrize $$ X(u,v)= 
(f(v)\cos\ u,f(v)\sin\ u,g(v) ) \ \ast$$
Especially since it is a hyperboloid, then we have $$f(x)=\cosh\ x,\ g(x)=\sinh\ x$$ Generally, geodesic equation on the surface with the parametrization $\ast$ is $$ u''  + \frac{2f'}{f}u'v'=0 $$ $$ v''-\frac{ff'}{(f')^2+(g')^2}(u')^2+ \frac{f'f'' + g'g''}{(f')^2+(g')^2}(v')^2 =0 $$
(cf. do Carmo's book) Assume that $c(t)=X(u(t),v(t))$ is a curve s.t. $$|c'|=1\ -\ (1),\  
\langle c'(t), -W\rangle = \frac{1}{f} \ -\ (2)$$ where $V = (\cos\ u,\sin\ u,0),\ W=(-\sin\ u,\cos\ u,0),\ e=(0,0,1)$
Hence we have a claim that $c$ satisfies the geodesic equation : From (1) we have $$
 c'=f'v'V + fu'W +g'v'e$$ $$ 1= g^2(v')^2+f^2(u')^2 +f^2(v')^2 $$
If we differentiate this we have $$ v'' + \frac{2fg (v')^2 +fg(u')^2}{f^2+g^2 } + \frac{f^2 u'u''}{(f^2+g^2) v' } =0 \ - \ (3)$$
From (2) we have $$ f^2u'=-1,\ 2ff'v'u'+f^2u''=0 \ -\ (4)$$
Note that from $(3) $ and $(4)$ we have a geodesic equation. Hence we complete the proof.
