Asymptotic curves of one-sheet hyperboloid I want to prove that the asymptotic curves of $x^2+y^2-z^2=1$ are straight lines. There are two parametrization of this surface that I used:
$$x(u,v)=(\cosh(v)\cos(u),\cosh(v)\sin(u),\sinh(v))\;\;\;\;\;(u,v)\in(0,2\pi)\times\mathbb{R},$$
$$x(u,v)=(\cos(u)+v\sin(u),\sin(u)-v\cos(u),v)\;\;\;\;\;(u,v)\in(0,2\pi)\times\mathbb{R}.$$
A curve $\alpha(t)=x(u(t),v(t))$ is asymptotic if and only if
$$e(u')^2+2fu'v'+g(v')^2=0.$$
Where $e,f,g$ are the coeficients of the second fundamental form. After several calculations, it follows that $\alpha(t)$ is asymtotic if and only if 
$$-\cosh(v)(u')^2+(v')^2=0,$$
$$-(1+v^2)(u')^2+2u'v'=0$$
depending on the parametrizations. So,
$$\sinh(v)=\tan(u+C_1)\;\;\;\text{ or }\;\;\;\sinh(v)=\tan(-u+C_2),$$
$$u=C_3\;\;\;\text{ or }\;\;\;v=\tan\left(\frac{u}{2}+C_4\right).$$
With $C_1,C_2,C_3$ and $C_4$ arbitrary constants. Is clear that if $u=C_3$, $\alpha(t)$ is a striaght line because of the parametrization. My question is why the others curves are also straight lines? 
Thank you.
 A: Your question is a little scrambled because you're using the same letters for different parametrizations. I assume you're asking about the first parametrization $x(u,v)=(\cosh(v)\cos(u),\cosh(v)\sin(u),\sinh(v))$. After calculating $e,f,g$ and solving the differential equation for asymptotic curves, you should get the solution $u=\pm \arctan(\sinh(v)) + \phi$. To proceed from here is straightforward: just substitute that value of $u$ into the parametrization and use a couple of trig and hyperbolic trig identities. For example, $\cos(u) = \cos(\pm\arctan(\sinh(v)) + \phi) = \cos(\pm\arctan(\sinh(v))) \cos(\phi) -\sin(\pm\arctan(\sinh(v)))\sin(\phi)$.
Using formulas for $\sin(\arctan)$ and $\cos(\arctan)$, $\cos(u) = \frac{1}{\sqrt{1+\sinh^2(v)}} \cos(\phi) - \frac{\pm\sinh(v)}{\sqrt{1+\sinh^2(v)}} \sin(\phi)$. Using the Pythagorean hyperbolic trig identity, $\cos(u) = \frac{1}{\cosh(v)} \cos(\phi) - \frac{\pm\sinh(v)}{\cosh(v)}\sin(\phi)$.
Then $\cosh(v)\cos(u) = \cos(\phi) \mp \sinh(v)\sin(\phi)$.
Similarly, $\cosh(v)\sin(u) = \pm\sinh(v)\cos(\phi) + \sin(\phi)$. 
Finally, $x(u(v),v) = (\cos(\phi)\mp\sinh(v)\sin(\phi),\pm\sinh(v)\cos(\phi)+\sin(\phi),\sinh(v)) =(\cos(\phi),\sin(\phi),0) + \sinh(v)(\mp\sin(\phi),\pm\cos(\phi),1)$ which is definitely a nice representation of the two families of asymptotic lines. You can even substitute $z=\sinh(v)$ to transform the families back to Cartesian coordinates.
In retrospect I think I got a sign or two wrong, so do check my calculation carefully.
