Doubly Ruled Surfaces I get stuck in the following exercice:
Let $S$ be a doubly ruled surfaces by orthogonal lines.


*

*Use the Gauss equations to prove that $K\equiv 0$;

*Conclude that $S$ is a plane.


I tried to use a parametrization $X(u,v)=\alpha(u)+v\beta(u)$, usinge the fact that the lines are orthogonal  and calculate the Christoffel symbols, but it did't worked.
 A: We don't even need Christoffel symbols or curvature:
We may assume that $S$ is given in the form
$${\bf f}: \quad (x,y)\mapsto \bigl(x,y, f(x,y)\bigl)\qquad(*)$$
with $f(x,0)\equiv0$, $f(0,y)\equiv0$, i.e., the $x$-axis and the $y$-axis are lines on $S$. 
By assumption through each point $(x,0,0)$ there is a line $g_x$ on $S$ which is orthogonal to the $x$-axis; therefore $g_x$ can be written in the form
$${\bf g}_x:\quad s\mapsto\bigl(x, s, p(x) s\bigr)\ ,$$
where $p(x)$ denotes the slope of $g_x$ with respect to the $(x,y)$-plane.
Similarly, through each point $(0,y,0)$ there is a line $h_y$ on $S$ which is orthogonal to the $y$-axis; therefore $h_y$ can be written in the form
$${\bf h}_y:\quad t\mapsto\bigl(t, y, q(y) t\bigr)\ ,$$
where again $q(y)$ denotes the slope of $h_y$ with respect to the $(x,y)$-plane.
The two lines $g_x$ and $h_y$ intersect at the point 
${\bf r}_{xy}\in S$ characterized by  ${\bf g}_x(y)={\bf h}_y(x)$. Looking at the third coordinate of ${\bf r}_{xy}$ we see that
$$p(x)y=q(y) x\ .$$
As this is true for all $x$ and $y$ near $0$ there has to be a constant $c\in{\mathbb R}$ with $p(x)=cx$, $q(y)=cy$. This implies that the $f$ in $(*)$ is given by $f(x,y)=c\>xy$, or that $S$ is a parabolic hyperboloid, when $c\ne0$.
Since $g_x$ and $h_y$ should intersect orthogonally at ${\bf r}_{xy}$ we must have
$$0={\bf g}_x'(y)\cdot {\bf h}_y'(x)=(0,1, cx)\cdot(1,0, cy)=c^2 x y\ ,$$
which implies $c=0$, so that in fact $f(x,y)\equiv0$, and $S$ is a plane.
