Finding the Asymptotic Curves of a Given Surface I have to find the asymptotic curves of the surface given by $$z = a \left( \frac{x}{y} + \frac{y}{x} \right),$$ for constant $a \neq 0$.
I guess that what was meant by that statement is that surface $S$ can be locally parametrized by $$X(u,v) = \left( u, v, a \left( \frac{u}{v} + \frac{u}{v} \right) \right).$$  Do you think that my parametrization is correct (meaning that I read the description of the surface correctly), and do you know of a more convenient parametrization?
Assuming that parametrization, I derived the following ($E$, $F$, $G$, are the coefficients of the first fundamental form; $e$, $f$, $g$ are coefficients of the second fundamental form; $N$ is the normal vector to surface $S$ at a point; these quantities are all functions of local coordinates $(u,v)$):
$$E = 1 + a^2 \left( \frac{1}{v} - \frac{v}{u^2} \right)^2,$$ 
$$F = -\frac{a^2 (u^2 - v^2)^2}{u^3 v^3},$$
$$G = 1 + a^2 \left( \frac{1}{u} - \frac{u}{v^2} \right)^2.$$
$$N = \frac{1}{\sqrt{E G - F^2}} \left( a \left( \frac{v}{u^2}-\frac{1}{v} \right), a \left( \frac{u}{v^2}-\frac{1}{u} \right), 1 \right).$$
$$X_{u,u} = \left( 0,0, \frac{2 a v}{u^3} \right), X_{u,v} = \left( 0,0, -a \left( \frac{1}{u^2} + \frac{1}{v^2} \right) \right), X_{v,v} = \left( 0, 0, \frac{2 a u}{v^3} \right).$$
$$e = \frac{2 a v}{u^3 \sqrt{E G - F^2}},$$
$$f = - \frac{a (\frac{1}{u^2} + \frac{1}{v^2})}{\sqrt{E G - F^2}},$$
$$g = \frac{2 a u}{v3 \sqrt{E G - F^2}}.$$
Thus, the Gaussian curvature (from these calculations) is:
$$K = -\frac{a^2 u^4 v^4 (u^2 - v^2)^2}{(u^4 v^4 + 
  a^2 (u^2 - v^2)^2 (u^2 + v^2))^2}.$$
And the mean curvature would be:
$$H = \frac{a u^3 v^3 (u^4 + v^4)}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2
 }}.$$
So, the principal curvatures are:
$$k_{\pm} = H \pm \sqrt{H^2 - K} = a u^2 v^2 \frac{u v (u^4 + v^4) \pm \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}}{(u^4 v^4 + a^2 (u^2 - v^2)^2 (u^2 + v^2))^{3/2}}.$$
In order to find the asymptotic curves, but trying to avoid the differential equation, I was hoping to find the angles $\theta (u,v)$ such that the normal curvature would always be $0$.  In other words I was trying:
$0 = k_n = k_{+} \cos{(\theta)}^2 + k_{-} \sin{(\theta)}^2$, and solving for $\theta$.
Assuming sufficient niceness, this calculate would result in: $$(u v (u^4 + v^4) + \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \cos{(\theta)}^2 + (u v (u^4 + v^4) - \sqrt{(u^2 + v^2) (a^2 (u^2 - v^2)^4 + u^2 v^2 (u^6 + v^6))}) \sin{(\theta)}^2$$
First of all, is this approach (solving for $\theta$ rather than solving the differential equation) valid? If it is, after I find that angle $\theta$, determined by location $(u,v)$ on $S$, what more work do I have to do? How do I find the equations for the asymptotic curves based on this angle?
If this whole method was for naught, how does one solve the differential equation. in this case, of: $$e (u')^2 + 2f u' v' + g (v')^2 = 2a v^4 (u')^2 - 2a u^3 v^3 \left( \frac{1}{u^2} + \frac{1}{v^2} \right)u' v' + 2a u^4 (v')^2 = 0?$$ (Again, assuming sufficient niceness.)
(See: https://math.stackexchange.com/questions/762195/differential-equation-for-the-asymptotic-directions-of-a-given-surface)
Thank you!


I believe this surface can be described by two networks of lines. ( Shown in Red and Cyan). Make the substitution $nt$ for $u$ , and separately $nt$ for $t$ ,
so that the new parametrization might look like : $<t,nt,a(1/n + n/1)> $ and/or $<nt,t,a(1/n+ n)>$ these are the asymptotic lines. Then switch the sign of n to pick up the surface below the z axis. I've used $n$ as an integer , but every line in the surface chosen in this way is an asymptotic curve, and a geodesic.
(Added a guess for the solution of the equations based on a study of the surface.)  
