RuledSurfaceEllipticalHyperboloid::usage="Demonstrate that the surface
2
x
- 9
2
y
+ 9
2
z
= 9it is also generated by the family of straight lines kx - 3ky - 3z = 3, x + 3y + 3kz = 3k. Demonstrate that the previous family of straight lines are cut with x - 3y + 3kz = 3k, kx + 3ky - 3z = 3.References: Lehmann, Charles H. Geometría Analítica. Ed. Limusa: p.418, 2003; Personal.";
Obtaining the surface equation:
k(x-3y)==3z+3;k==
3z+3
x-3y
;​​x+3y==k(3-3z);k==
x+3y
3-3z
;​​
3z+3
x-3y
==
x+3y
-(3z-3)
;​​-(9
2
z
-9)==
2
x
-9
2
y
;​​
2
x
-9
2
y
+9
2
z
-9==0;
Demonstration that families of straight lines intersect:
In[]:=
{k,-3k,-3}{1,3,3k}
{9-9
2
k
,-3-3
2
k
,6k};
In[]:=
Solve[{kx-3ky-3z==3,x+3y+3kz==3k},{x,y}]
Out[]=
x-
3(-1-
2
k
-z+
2
k
z)
2k
,y-
1-
2
k
+z+
2
k
z
2k

Doing z = 0, the first family remains:
ℒ
1
:
3+3
2
k
2k
,
-1+
2
k
2k
,0+t{9-9
2
k
,-3-3
2
k
,6k};
In[]:=
{1,-3,3k}{k,3k,-3}
{9-9
2
k
,3+3
2
k
,6k};
In[]:=
Solve[{x-3y+3kz==3k,kx+3ky-3z==3},{x,y}]
Out[]=
x-
3(-1-
2
k
-z+
2
k
z)
2k
,y-
-1+
2
k
-z-
2
k
z
2k

Doing z = 0, the second family remains:
ℒ
2
:
3+3
2
k
2k
,
1-
2
k
2k
,0+t{9-9
2
k
,3+3
2
k
,6k};
It is observed that each pair of elements of the families will intersect at a point for the same value of k.
It will be seen if the first family of lines represents a developable surface:
In[]:=
ℒ:=
3+3
2
u
2u
,
-1+
2
u
2u
,0+v{9-9
2
u
,-3-3
2
u
,6u};​​γ[u_]:=
3+3
2
u
2u
,
-1+
2
u
2u
,0;​​ω[u_]:={9-9
2
u
,-3-3
2
u
,6u};​​p[u]=Det[{γ'[u],ω[u],ω'[u]}]​​
Out[]=
-108
It is not a developable surface.
You will get the line of striction that is the directrix:
In[]:=
β[u]=γ[u]-
(ω'[u]ω[u]).(γ'[u]ω[u])
2
Norm[ω'[u]ω[u]]
ω[u]
In[]:=
Simplify
3+3
2
u
2u
-
(9-9
2
u
)
18-18
2
u
6u+6
3
u

2
2
u
-
54+54
2
u
-18u+18
3
u

2
2
u
11664
2
u
+
2
(18-18
2
u
)
+
2
(54+54
2
u
)
,
-1+
2
u
2u
-
(-3-3
2
u
)
18-18
2
u
6u+6
3
u

2
2
u
-
54+54
2
u
-18u+18
3
u

2
2
u
11664
2
u
+
2
(18-18
2
u
)
+
2
(54+54
2
u
)
,-
6u
18-18
2
u
6u+6
3
u

2
2
u
-
54+54
2
u
-18u+18
3
u

2
2
u
11664
2
u
+
2
(18-18
2
u
)
+
2
(54+54
2
u
)


54u(1+
2
u
)
5+26
2
u
+5
4
u
,
8u(-1+
2
u
)
5+26
2
u
+5
4
u
,
5(-1+
4
u
)
5+26
2
u
+5
4
u
;
Animation:
In[]:=
​​SwatchLegend[{Green,Yellow,Blue},{"Ruled surface: One leaf hyperboloid","Directrix (striction line)","Straight line generatrix"}]​​Manipulateβ[u_]:=
54u(1+
2
u
)
5+26
2
u
+5
4
u
,
8u(-1+
2
u
)
5+26
2
u
+5
4
u
,
5(-1+
4
u
)
5+26
2
u
+5
4
u
;ℒ[t2_]:=
54t2(1+
2
t2
)
5+26
2
t2
+5
4
t2
,
8t2(-1+
2
t2
)
5+26
2
t2
+5
4
t2
,
5(-1+
4
t2
)
5+26
2
t2
+5
4
t2
+t2{9-9
2
t2
,-3-3
2
t2
,6t2};σ[u_,v_]:=
54u(1+
2
u
)
5+26
2
u
+5
4
u
,
8u(-1+
2
u
)
5+26
2
u
+5
4
u
,
5(-1+
4
u
)
5+26
2
u
+5
4
u
+v{9-9
2
u
,-3-3
2
u
,6u};​​Show[{Graphics3D[{Blue,Thickness[0.004],InfiniteLine[{β[t2],ℒ[t2]}]}],ParametricPlot3D[β[u],{u,-11,11},PlotStyleYellow]/.Line[pts_,rest___]:>Tube[pts,0.04,rest],Graphics3D[{Yellow,Ball[β[t2],0.1]}],ParametricPlot3D[σ[u,v],{v,-11,11},{u,-11,t2},PlotStyle->Directive[Opacity[0.6],Specularity[White,30]],MeshFalse,PerformanceGoal->"Quality",PlotPoints->60,ColorFunction->"BlueGreenYellow",BoundaryStyleDirective[Black,Thick]]},​​AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRange6,ImageSize900,ViewPoint{1.3,-2.4,2.}],Style["Ruled surface: One leaf hyperboloid",Bold,Large],{{t2,-11+0.00002,"Value (t2)"},-11+0.00001,11,0.00001},ControlPlacementTop
Out[]=
Ruled surface: One leaf hyperboloid
Directrix (striction line)
Straight line generatrix
Out[]=
​
Ruled surface: One leaf hyperboloid
Value (t2)
Note: The directrix (striction line ) presentsel the following behavior:
Therefore, for practical reasons, the directrix in the animation will not be closed, because in the graph we do use very large amounts for u. The detail of the graph will be lost.
Cheking that the directrix is correct and also the parametric equations of the surface:
View from all angles (below):