In[]:=
ParaboloidCircumscribedCone::usage="Find the equation of a cone that has the vertex at the point M{-1, 0, 0} and is circumscribed around the paraboloid 2
2
y
+
2
z
= 4x.References: Fedenko, A.S. Problemas de Geometría Diferencial. Editorial Mir: p.74, Moscú, 1981. ";
Development:
The equation of the cone generatrix is:
In[]:=
x+1
a
==
y
b
==
z
c
==t;
The coordinates of any point of the generatrix are:
In[]:=
{-1+at,bt,ct};
It is very easy to see that the focus of the paraboloid is at the point {1, 0, 0} , then the plane x = 1 cuts the paraboloid in the ellipse:
In[]:=
2
2
y
+
2
z
==4;
The one that will be the directrix of the cone sought.
Then we have:
In[]:=
2
2
(bt)
+
2
(ct)
==4;-1+at==1;t==
2
a
;
Then:
In[]:=
2
2
2b
a
+
2
2c
a
==4;​​8
2
y
x+1
+4
2
z
x+1
==4;
The equation of cone is:
In[]:=
2
2
y
+
2
z
==
2
(x+1)
;​​
2
(x+1)
==2
2
y
+
2
z
;
In[]:=
​​Manipulate[ClearAll;​​Show[ContourPlot3D[2
2
y
+
2
z
-
2
(x+1)
==0,{x,-4,2},{y,-4,4},{z,-4,4},ContourStyle->{Opacity[0.2],Yellow},MeshFalse,PerformanceGoal"Quality",BoundaryStyleDirective[Black,Thick],PlotPoints->60],ContourPlot3D[2
2
y
+
2
z
-4x==0,{x,-4,t1},{y,-4,4},{z,-4,4},ContourStyle->{Opacity[0.8],Red},MeshFalse,PerformanceGoal"Quality"],Graphics3D[{Black,Ball[{-1,0,0},0.05]}],ParametricPlot3D[1,
2
Cos[u],2Sin[u],{u,0,π(t1+2)},PlotStyle{Blue,Thick}],​​AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRangeAll,ViewPoint{-3,3,-10},ImageSize500],Style["Cone:",Bold,Large],{{t1,-1.9,"Value (t1)"},-2+0.01,2,0.01},ControlPlacementTop]
Out[]=
​
Cone:
Value (t1)