OrthopticSurfaceEllipsoid::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; Personal.";
Development:
Obtaining the orthoptic surface of the ellipsoid
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 ellipsoid 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 ellipsoid’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 ellipsoid 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)==14;
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 Monge’s concentric sphere:
2
x
+
2
y
+
2
z
==
2
a
+
2
b
+
2
c
which is the orthoptic surface of the ellipsoid. There are many similarities in 2D with the ellipse and its orthoptic curve, the circumference of Monge. Finally, the equation sought is:
2
x
+
2
y
+
2
z
==14;
Animation:
In[]:=
Manipulatea2=4;b2=9;c2=1;​​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]};​​d1[α_,β_]:=
a2
2
u1[α,β][[1]]
+b2
2
u1[α,β][[2]]
+c2
2
u1[α,β][[3]]
;​​d2[α_]:=
a2
2
u2[α][[1]]
+b2
2
u2[α][[2]]
+c2
2
u2[α][[3]]
;​​d3[α_,β_]:=
a2
2
u3[α,β][[1]]
+b2
2
u3[α,β][[2]]
+c2
2
u3[α,β][[3]]
;​​Q[α_,β_]:=d1[α,β]u1[α,β]+d2[α]u2[α]+d3[α,β]u3[α,β];​​​​​​Show​​​​ParametricPlot3D[{2Cos[u]Sin[v],3Sin[u]Sin[v],Cos[v]},{u,0,2π},{v,0,π},ColorFunction->"SolarColors",MeshNone,MaxRecursion->3],​​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[0.2],MeshNone,MaxRecursion->3,​​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]}],{Gray,Opacity[0.4],Hyperplane[u1[α,β],d1[α,β]],Hyperplane[u2[α],d2[α]],Hyperplane[u3[α,β],d3[α,β]]}}]​​,​​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: Monge's sphere",Bold,Large]​​,​​{{α,0.00002,"α angle (Turn H)"},0.000001,2π,0.00001},Delimiter,β,
π
4
,"β angle (Turn V)",-
π
2
,
π
2
,0.00001,ControlPlacement->Top,AutorunSequencing->{1,2}​​
Out[]=
​
Orthoptic surface: Monge's sphere
α angle (Turn H)
β angle (Turn V)