The one-sheeted hyperboloid is a ruled surface, which is seen from its equation written in the form

or

In fact, (2) may be thought to be formed by multiplying the equations in the pair

which represents a line in the space; $h$ is an arbitrary parameter. For any  $h\neq 0$,  each point  $(x,\,y,\,z)$  on the line (3) satisfies also (2). This means that the line (3) lies on the hyperboloid, i.e. it’s a question of a generatrix (= ruling) of the one-sheeted hyperboloid.

Giving distinct real values to the parameter $h$ we get an infinite family of the generatrices (3). Further, one of these lines passes through every point of the hyperboloid. Actually, if the point  $P_{1}=(x_{1},\,y_{1},\,z_{1})$  satisfies the equation (2) of the surface, we have the proportion equation

and if we assign in (3) to $h$ the value of the left hand side of the proportion, then $P_{1}$ satisfies also the equations (3).

But since the equation (2) may be splitted also as

the hyperboloid has as well the other family (4) of generatrices, containing similarly one generatrix through every point of the surface. The one-sheeted hyperboloid is doubly ruled  —  having two distinct generatrices through every point. And the families (3) and (4) have really no common members, since otherwise we had an equation

for all $x$’s; this would imply, by substituting  $x=0$,  that  $h=k$  and then the impossibility  $\displaystyle{1-\frac{x}{a}\equiv 1+\frac{x}{a}}$.

Note 1. One can solve from the equations (3) and (4) the coordinates for points of the one-sheeted hyperboloid:

This is a parametric presentation of the surface.

Note 2. Furthermore one may prove, that two lines of the same family (3) or (4) cannot lie in a same plane, but two lines of distinct families (3) and (4) lie always in a same plane.