MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

2 1

2

2

2

− =

b

z

a

x

; c^2 =a^2 +b^2 (1)

hence, the equation of a hyperboloid of revolution (Fig. 8) is:

2 1

2

2

2 2

− =

+

b

z

a

x y

(2)

To get a and b we should have two points (x 1 y 1 z 1 ) (x 2 y 2 z 2 ) as given on the hyperboloid,

then to substitute their co-ordinates in (2) and finally solve the resultant system of two equa-

tions with two unknowns (a and b).

If we make

2

1

2

A=x 1 +y ;

2

B=z 1 ;

2

2

2

C=x 2 +y ;

2

D=z 2

we get:

()()

B D

AB D BA C

a

−

− − −

= ;

()()

A C

AB D B A C

b

−

− − −

= ( 3 )

¿Which pair of points could we chose? To answer we should recall that the hyperboloid

is a ruled surface and therefore all its generatrices (mountain folds) are straight lines resting

on it; in consequence, any point of these lines belong to the hyperboloid.

Off hand we have points B and A in ∆ABC, but they induce division by zero in (3) be-

cause of the symmetry with respect to the co-ordinate plane Z = 0 that obviously contains the

center of the hyperboloid.

Then we may keep A and look for another point in AB: the intersection point of AB

with a horizontal plane distinct from the hyperboloid bases. The operation is easy when the

program INTERPR.BAS that yields the intersection point of a plane and a straight line, helps

CAD. By so doing we get

a = 24,2384 ; b = 26,5738

Substituting these values in (1) we obtain c which in turn determines the position of fo-

cus F in the hyperbola of Fig. 9 c= a^2 +b^2 = 35 , 9676

Only to add that to draw the hyperbolas of Figs. 8 and 9 we have to give values to x,

take them to (1), obtain the corresponding z and then carry both (x,z) to the drawing.

asymptote

a

c

F ́ O

9

Z

b

X

F