## You are here

Homegeneratrices of one-sheeted hyperboloid

## Primary tabs

# generatrices of one-sheeted hyperboloid

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

$\displaystyle\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1-\frac{x^{2}}{a^{2}},$ | (1) |

or

$\displaystyle\left(\frac{y}{b}+\frac{z}{c}\right)\left(\frac{y}{b}-\frac{z}{c}% \right)=\left(1+\frac{x}{a}\right)\left(1-\frac{x}{a}\right).$ | (2) |

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

$\displaystyle\begin{cases}\displaystyle{\frac{y}{b}+\frac{z}{c}=h\left(1-\frac% {x}{a}\right)}\\ \displaystyle{\frac{y}{b}-\frac{z}{c}=\frac{1}{h}\left(1+\frac{x}{a}\right),}% \end{cases}$ | (3) |

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

$\frac{\frac{y_{1}}{b}+\frac{z_{1}}{c}}{1-\frac{x_{1}}{a}}=\frac{1+\frac{x_{1}}% {a}}{\frac{y_{1}}{b}-\frac{z_{1}}{c}},$ |

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

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

$\displaystyle\begin{cases}\displaystyle{\frac{y}{b}+\frac{z}{c}=k\left(1+\frac% {x}{a}\right)}\\ \displaystyle{\frac{y}{b}-\frac{z}{c}=\frac{1}{k}\left(1-\frac{x}{a}\right),}% \end{cases}$ | (4) |

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

$h\left(1-\frac{x}{a}\right)=k\left(1+\frac{x}{a}\right)$ |

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

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.

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.

# References

- 1 L. Lindelöf: Analyyttisen geometrian oppikirja. Kolmas painos. Suomalaisen Kirjallisuuden Seura, Helsinki (1924).
- 2 Lauri Pimiä: Analyyttinen geometria. Werner Söderström Osakeyhtiö, Porvoo and Helsinki (1958).

## Mathematics Subject Classification

51N20*no label found*51M04

*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections

## Comments

## A picture

Hi, I think it were nice in "generatrices of one-sheeted hyperboloid" to be a picture containing maybe 8 or 10 pairs of generatrices between two horizontal intersection ellipses. Perhaps someone kind picture-expert could easily make such a picture.

Jussi