OrthopticSurfaceOneLeafHyperboloid::usage="The orthoptic surface of a surface is the geometric locus of the points through which pass 3 planes orthogonal to each other and tangents to the original surface. References: AI Assistant; Personal.";
Development:
Obtaining the orthoptic surface of the one leaf hyperboloid
2
x
4
+
2
y
9
-
2
z
1
=1
:
Let the tangent plane (where the normal vector is an unitary vector):
Ax+By+Cz==D;
Condition of tangent between a plane and an one leaf hyperboloid centered at the origin:
Tangent plane:
x
0
x
2
a
+
y
0
y
2
b
-
z
0
z
2
c
==1;​​
A
D
x+
B
D
y+
C
D
z==1;​​
x
0
2
a
==
A
D
;
x
0
==
A
2
a
D
;​​
y
0
2
b
==
B
D
;
y
0
==
B
2
b
D
;​​-
z
0
2
c
==
C
D
;
z
0
==-
C
2
c
D
;​​
Sustituting the point {
x
0
,
y
0
,
z
0
}
in the one leaf hyperboloid’s equation and simplifying:
1
2
a
2
A
2
a
D
+
1
2
b
2
B
2
b
D
-
1
2
c
2
C
2
c
D
==1;​​
2
D
==
2
A
2
a
+
2
B
2
b
-
2
C
2
c
;
So for the one leaf hyperboloid that concerns us:
2
D
==4
2
A
+9
2
B
-
2
C
;
So for the three planes we have:
2
D
1
==4
2
A
1
+9
2
B
1
-
2
C
1
;​​
2
D
2
==4
2
A
2
+9
2
B
2
-
2
C
2
;​​
2
D
3
==4
2
A
3
+9
2
B
3
-
2
C
3
;
Adding member to member the three equations and considering that the rows of an orthogonal matrix are also unit vectors:
2
D
1
+
2
D
2
+
2
D
3
==4
2
A
1
+
2
A
2
+
2
A
3
+9
2
B
1
+
2
B
2
+
2
B
3
-1
2
C
1
+
2
C
2
+
2
C
3
;​​
2
D
1
+
2
D
2
+
2
D
3
==4(1)+9(1)-1(1)==12;
Since the three normal vectors are mutually orthogonal and unitary, (and by properties of orthogonal matrices) it is simplified to:
In[]:=
2
D
1
+
2
D
2
+
2
D
3
==Expand[
2
(
A
1
x+
B
1
y+
C
1
z)
+
2
(
A
2
x+
B
2
y+
C
2
z)
+
2
(
A
3
x+
B
3
y+
C
3
z)
]
2
D
1
+
2
D
2
+
2
D
3

2
x
2
A
1
+
2
x
2
A
2
+
2
x
2
A
3
+2xy
A
1
B
1
+
2
y
2
B
1
+2xy
A
2
B
2
+
2
y
2
B
2
+2xy
A
3
B
3
+
2
y
2
B
3
+2xz
A
1
C
1
+2yz
B
1
C
1
+
2
z
2
C
1
+2xz
A
2
C
2
+2yz
B
2
C
2
+
2
z
2
C
2
+2xz
A
3
C
3
+2yz
B
3
C
3
+
2
z
2
C
3
;​​
2
D
1
+
2
D
2
+
2
D
3
==
2
A
1
+
2
A
2
+
2
A
3

2
x
+
2
B
1
+
2
B
2
+
2
B
3

2
y
+
2
C
1
+
2
C
2
+
2
C
3

2
z
+(
A
1
B
1
+
A
2
B
2
+
A
3
B
3
)2xy+(
A
1
C
1
+
A
2
C
2
+
A
3
C
3
)2xz+(
B
1
C
1
+
B
2
C
2
+
B
3
C
3
)2yz;​​
2
D
1
+
2
D
2
+
2
D
3
==
2
x
+
2
y
+
2
z
;
So it turns out how we expected concentric sphere:
2
x
+
2
y
+
2
z
==
2
a
+
2
b
-
2
c
which is the orthoptic surface of the one leaf hyperboloid. There are many similarities in 2D with the hyperbola and its orthoptic curve, the circumference
2
x
+
2
y
=
2
a
-
2
b
. Finally, the equation sought is:
2
x
+
2
y
+
2
z
==12;
Animation:
In[]:=
Manipulate​​Module​​{​​a2=4,b2=9,c2=1,​​u1,u2,u3,r1,r2,r3,d1,d2,d3,q​​},​​​​u1[α_,β_]:={Cos[α]Cos[β],Sin[α]Cos[β],Sin[β]};​​u2[α_]:={-Sin[α],Cos[α],0};​​u3[α_,β_]:={-Cos[α]Sin[β],-Sin[α]Sin[β],Cos[β]};​​​​r1[α_,β_]:=a2u1[α,β][[1]]^2+b2u1[α,β][[2]]^2-c2u1[α,β][[3]]^2;​​r2[α_]:=a2u2[α][[1]]^2+b2u2[α][[2]]^2-c2u2[α][[3]]^2;​​r3[α_,β_]:=a2u3[α,β][[1]]^2+b2u3[α,β][[2]]^2-c2u3[α,β][[3]]^2;​​​​d1[α_,β_]:=
r1[α,β]
;​​d2[α_]:=
r2[α]
;​​d3[α_,β_]:=
r3[α,β]
;​​​​q[α_,β_]:=d1[α,β]u1[α,β]+d2[α]u2[α]+d3[α,β]u3[α,β];​​​​Column​​​​Show​​​​ParametricPlot3D[​​{2Cosh[v]Cos[u],3Cosh[v]Sin[u],Sinh[v]},​​{u,0,2π},{v,-1.2,1.2},​​ColorFunction->"SolarColors",​​Mesh->None,​​MaxRecursion->2,​​PlotPoints->50​​],​​​​ParametricPlot3D​​​​
a2+b2-c2
Cos[u]Sin[v],​​
a2+b2-c2
Sin[u]Sin[v],​​
a2+b2-c2
Cos[v]​​,​​{u,0,2π},{v,0,π},​​ColorFunction->"BlueGreenYellow",​​PlotStyle->Opacity[0.6],​​Mesh->None,​​MaxRecursion->2​​,​​​​If[​​r1[α,β]>0,​​Graphics3D[{Gray,Opacity[0.6],Hyperplane[u1[α,β],d1[α,β]]}],Nothing],​​​​If[​​r2[α]>0,​​Graphics3D[{Gray,Opacity[0.6],Hyperplane[u2[α],d2[α]]}],Nothing],​​​​​​If[​​r3[α,β]>0,​​Graphics3D[{Gray,Opacity[0.6],Hyperplane[u3[α,β],d3[α,β]]}],Nothing],​​​​If[​​r1[α,β]>0&&r2[α]>0&&r3[α,β]>0,​​Graphics3D[​​{​​Black,Thick,​​Sphere[q[α,β],0.1],​​Line[{q[α,β],q[α,β]+u1[α,β]}],​​Line[{q[α,β],q[α,β]+u2[α]}],​​Line[{q[α,β],q[α,β]+u3[α,β]}]​​}​​],​​Nothing​​]​​,​​BoxRatios->Automatic,​​Axes->True,​​Boxed->False,​​AxesOrigin->{0,0,0},​​AxesLabel->{x,y,z},​​PlotRange->6,​​ImageSize->900,​​ViewPoint->{1.3,-2.4,2.}​​,​​​​Style["Orthoptic surface:
2
x
+
2
y
+
2
z
= 12",Bold,Large],​​​​Style[​​Row[{​​"Tangency condition: ",​​TraditionalForm[4n
2
x
+9n
2
y
-n
2
z
>0]​​}],​​14​​]​​​​​​,​​{{α,0.3,"α angle (Turn H)"},0.000001,2π,0.01},​​Delimiter,​​{{β,π/6,"β angle (Turn V)"},-π/2,π/2,0.01},​​ControlPlacement->Top,​​AutorunSequencing->{1,2}​​
Note: It should be remembered that the tangent plane at any point of a hyperboloid of one leaf intersects two generatrix lines (it is a double-ruled surface) that intersect at that point, unlike an ellipsoid or a sphere where the tangent point is an isolated point.
These points on a hyperboloid of one leaf are called hyperbolic points (negative Gauss curvature).