"If we write the equation of the hyperboloid in the form \frac{x^2}{ a^2} - \frac{z^2}{c^2} = 1 - \frac{y^2}{b^2}, \qquad (1)It is evident that (1) is the product of two equations\begin{array}{lcl} \frac{x}{a} - \frac{z}{c} = k_1(1 - \frac{y}{b}), \\ \frac{x}{a} + \frac{z}{c} = \frac{1}{k_1}(1 + \frac{y}{b}), \qquad (2) \end{array}for any value of k_1. But (2) are the equations of a straight line... Moreover this straight line lies entirely on the surface, since the coordinates of every point of it satisfy (2) and hence (1). As different values are assigned to k_1, we obtain a series of straight lines lying entirely on the surface. Conversely if P_1( x_1, y_1, z_1) is any point of (1), \frac{\frac{x_1}{a} - \frac{z_1}{c}}{1 - \frac{y_1}{b}} = \frac{1 + \frac{y_1}{b}}{\frac{x_1}{a} + \frac{z_1}{c}}Therefore P_1 determines the same value of k_1 from both equations (2). Hence every point of (1) lies in one and only one line (2). We may also regard (1) as the product of the two equations\begin{array}{lcl} \frac{x}{a} - \frac{z}{c} = k_2(1 + \frac{y}{b}), \\ \frac{x}{a} + \frac{z}{c} = \frac{1}{k_2}(1 - \frac{y}{b}), \qquad (3) \end{array}whence it is evident that there is a second set of straight lines lying entirely on the surface, one and only one of which may be drawn through any point of the surface. Equations (2) and (3) are the equations of the rectilinear generators, and every point of the surface may be regarded as the point of intersection of one line from each set."