# generatrices of one-sheeted hyperboloid

## Primary tabs

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\begin{document}
The one-sheeted hyperboloid is a ruled surface, which is seen from its equation written in the form
\begin{align}
\frac{y^2}{b^2}-\frac{z^2}{c^2} = 1-\frac{x^2}{a^2},
\end{align}
or
\begin{align}
\left(\frac{y}{b}+\frac{z}{c}\right)\left(\frac{y}{b}-\frac{z}{c}\right) =
\left(1+\frac{x}{a}\right)\left(1-\frac{x}{a}\right).
\end{align}
In fact, (2) may be thought to be formed by multiplying the equations in the pair
\begin{align}
\begin{cases}
\displaystyle{\frac{y}{b}+\frac{z}{c} = h\left(1-\frac{x}{a}\right)} \vspace{15pt} \\
\displaystyle{\frac{y}{b}-\frac{z}{c} = \frac{1}{h}\left(1+\frac{x}{a}\right),}
\end{cases}
\end{align}
which \PMlinkescapetext{represents} a \PMlinkname{line in the space}{LineInSpace}; $h$ 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
$$\frac{\frac{y_1}{b}+\frac{z_1}{c}}{1-\frac{x_1}{a}} = \frac{1+\frac{x_1}{a}}{\frac{y_1}{b}-\frac{z_1}{c}},$$
and if we assign in (3) to $h$ the value of the left hand \PMlinkescapetext{side} of the \PMlinkescapetext{proportion}, then $P_1$ satisfies also the equations (3).\\

But since the equation (2) may be splitted also as
\begin{align}
\begin{cases}
\displaystyle{\frac{y}{b}+\frac{z}{c} = k\left(1+\frac{x}{a}\right)} \vspace{15pt} \\
\displaystyle{\frac{y}{b}-\frac{z}{c} = \frac{1}{k}\left(1-\frac{x}{a}\right),}
\end{cases}
\end{align}
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 {\em 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
$$h\left(1-\frac{x}{a}\right) = k\left(1+\frac{x}{a}\right)$$
for all $x$'s; this would imply, by substituting\, $x = 0$,\, that\, $h = k$\, and then the impossibility\,
$\displaystyle{1-\frac{x}{a} \equiv 1+\frac{x}{a}}$.\\

\textbf{Note 1.}  One can solve from the equations (3) and (4) the coordinates for points of the one-sheeted hyperboloid:
$$x = a\frac{h-k}{h+k},\quad y = b\frac{hk+1}{h+k},\quad z = c\frac{hk-1}{h+k}$$
This is a parametric presentation of the surface.\\

\textbf{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.

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\begin{thebibliography}{8}
\bibitem{LL}{\sc L. Lindel\"of}: {\em Analyyttisen geometrian oppikirja}.\, Kolmas painos.\, Suomalaisen Kirjallisuuden Seura, Helsinki (1924).
\bibitem{LP}{\sc Lauri Pimi\"a}: {\em Analyyttinen geometria}.\, Werner S\"oderstr\"om Osakeyhti\"o, Porvoo and Helsinki (1958).
\end{thebibliography}

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