Function Repository Resource:

FivePointConic

Find a conic equation that passes through five given points

Contributed by: Ed Pegg Jr

ResourceFunction["FivePointConic"][pts,{x,y}]

returns the implicit Cartesian equation in the variables x and y of the conic section that goes through the points pts.

ResourceFunction["FivePointConic"][pts]

uses the formal variables x and y.

Details

For random, uniformly-distributed points in a rectangle, the probability of a hyperbola is

and the probability of an ellipse is

Examples

Basic Examples (2)

Find a conic section through five points:

In[1]:=

Out[2]=

Show the conic and points:

In[3]:=

Out[3]=

Scope (7)

Use formal variables:

In[4]:=

Out[4]=

The results of a five point conic are usually a hyperbola:

In[5]:=

points = {{-2, 1}, {-1, -2}, {1, -2}, {2, 1}, {0, 0}};
conic = ResourceFunction["FivePointConic"][points, {x, y}];
curve = ContourPlot[conic == 0, {x, -3, 3}, {y, -4, 4}][[1]];
Labeled[Graphics[{curve, {Directive[
AbsolutePointSize[5],
RGBColor[0.922526, 0.385626, 0.209179]], Point[points]}}, ImageSize -> Small], conic]

Out[8]=

A random five point conic is also frequently an ellipse:

In[9]:=

points = {{-2, 1}, {-1, -2}, {1, -2}, {2, 1}, {0, 2}};
conic = ResourceFunction["FivePointConic"][points, {x, y}];
curve = ContourPlot[conic == 0, {x, -3, 3}, {y, -4, 4}][[1]];
Labeled[Graphics[{curve, {Directive[
AbsolutePointSize[5],
RGBColor[0.922526, 0.385626, 0.209179]], Point[points]}}, ImageSize -> Small], conic]

Out[12]=

A circle can also be the result of a five point conic:

In[13]:=

points = {{-2, 1}, {-1, -2}, {1, -2}, {2, 1}, {1, 2}};
conic = ResourceFunction["FivePointConic"][points, {x, y}];
curve = ContourPlot[conic == 0, {x, -3, 3}, {y, -4, 4}][[1]];
Labeled[Graphics[{curve, {Directive[
AbsolutePointSize[5],
RGBColor[0.922526, 0.385626, 0.209179]], Point[points]}}, ImageSize -> Small], conic]

Out[16]=

A parabola may also appear:

In[17]:=

points = {{-2, 1}, {-1, -2}, {1, -2}, {2, 1}, {0, -3}};
conic = ResourceFunction["FivePointConic"][points, {x, y}];
curve = ContourPlot[conic == 0, {x, -3, 3}, {y, -4, 4}][[1]];
Labeled[Graphics[{curve, {Directive[
AbsolutePointSize[5],
RGBColor[0.922526, 0.385626, 0.209179]], Point[points]}}, ImageSize -> Small], conic]

Out[20]=

Degenerate conics are also possible:

In[21]:=

points = {{-2, 1}, {-1, -2}, {1, -2}, {2, 1}, {0, -1}};
conic = ResourceFunction["FivePointConic"][points, {x, y}];
curve = ContourPlot[conic == 0, {x, -3, 3}, {y, -4, 4}][[1]];
Labeled[Graphics[{curve, {Directive[
AbsolutePointSize[5],
RGBColor[0.922526, 0.385626, 0.209179]], Point[points]}}, ImageSize -> Small], conic]

Out[24]=

See the properties of the non-degenerate conics with the resource function ConicProperties:

In[25]:=

$base = {{-2, 1}, {-1, -2}, {1, -2}, {2, 1}}; last = {{0, 0}, {0, 2}, {0, -3}, {1, 2}}; ResourceFunction["ConicProperties"][ ResourceFunction["FivePointConic"][Append[base, #]] == 0, {\[FormalX], \[FormalY]}] & /@ last$

Out[26]=

Neat Examples (5)

A function for characterizing a conic section:

In[27]:=

$CharacterizeConic[poly_] := Module[{\[FormalCapitalA], \[FormalCapitalB], \[FormalCapitalC], \[FormalCapitalD], \[FormalCapitalE], \[FormalCapitalF], discriminant, coeff}, If[Length[FactorList[poly]] > 2 || NumericQ[poly], Return["D" (*degenerate*)]]; coeff = Flatten[CoefficientList[poly, {\[FormalX], \[FormalY]}, {3, 3}]]; {\[FormalCapitalA], \[FormalCapitalB], \[FormalCapitalC], \[FormalCapitalD], \[FormalCapitalE], \[FormalCapitalF]} = coeff[[{7, 5, 3, 4, 2, 1}]]; discriminant = \[FormalCapitalB]^2 - 4 \[FormalCapitalA] \[FormalCapitalC]; Which[ \[FormalCapitalB] == 0 && \[FormalCapitalA] == \[FormalCapitalC], "C" (*circle*), discriminant == 0, "P" (*parabola*), discriminant < 0, "E" (*ellipse*), discriminant > 0, "H" (*hyperbola*), True, "D" (*degenerate*) ]]$