In[]:=
JerabekHyperbola::usage="The Jerabek hyperbola passes through the three vertices of the triangle and passes through the circumcenter, the orthocenter and the symmedian point K. The Jerabek hyperbola has the center of the point
X
125
.References: Personal. ";​​​​JerabekHyperbola[tri={{x1_,y1_},{x2_,y2_},{x3_,y3_}},{X0_,X1_},{Y0_,Y1_}]:=​​Module{},DEP=Det[{{x1,y1,1},{x2,y2,1},{x3,y3,1}}];​​​​IfDEP==0||X0≥X1||Y0≥Y1,Print[PasteButton["Incorrect data or points do not form a triangle"]],​​ρ=TriangleCenter[tri,"Orthocenter"];​​xo=ρ[[1]];​​yo=ρ[[2]];​​ξ=TriangleCenter[tri,"Circumcenter"];​​xc=ξ[[1]];​​yc=ξ[[2]];​​xm1=
x1+x2
2
;ym1=
y1+y2
2
;xm2=
x2+x3
2
;ym2=
y2+y3
2
;xm3=
x3+x1
2
;ym3=
y3+y1
2
;​​tri1={{xm1,ym1},{xm2,ym2},{xm3,ym3}};​​ϕ=TriangleCenter[tri1,"Circumcenter"];​​xce=ϕ[[1]];​​yce=ϕ[[2]];​​ϕ1=TriangleMeasurement[tri1,"Circumradius"];​​rce=ϕ1;​​aa=
2
(x3-x2)
+
2
(y3-y2)
;bb=
2
(x3-x1)
+
2
(y3-y1)
;cc1=
2
(x2-x1)
+
2
(y2-y1)
;​​​​α=ArcCos
2
bb
+
2
cc1
-
2
aa
2bbcc1
;β=ArcCos
2
aa
+
2
cc1
-
2
bb
2aacc1
;γ=ArcCos
2
aa
+
2
bb
-
2
cc1
2aabb
;​​​​xx=Sin[2α]
2
(Sin[β-γ])
;yy=Sin[2β]
2
(Sin[γ-α])
;zz=Sin[2γ]
2
(Sin[α-β])
;​​
λ
1
=
xx
xx+yy+zz
;
λ
2
=
yy
xx+yy+zz
;
λ
3
=
zz
xx+yy+zz
;​​PX125=
λ
1
{x1,y1}+
λ
2
{x2,y2}+
λ
3
{x3,y3};​​xX125=Take[PX125,{1}].{1};yY125=Take[PX125,{2}].{1};​​Print[PasteButton["Coordinates of the center of the Jerabek hyperbola (​
X
125
​) "]];​​Print[" "];​​Print["
X
125
","(",N[xX125]," , ",N[yY125],")"];​​Print[" "];​​​​s5=NSolve[aex1^2+x1y1+cey1^2+dex1+eey1+fe==0,​​ aex2^2+x2y2+cey2^2+dex2+eey2+fe==0,​​ aex3^2+x3y3+cey3^2+dex3+eey3+fe==0,​​ aexc^2+xcyc+ceyc^2+dexc+eeyc+fe==0,​​ aexo^2+xoyo+ceyo^2+dexo+eeyo+fe==0,{ae,ce,de,ee,fe}];​​ a11=ae/.s5;c11=ce/.s5;d11=de/.s5;e11=ee/.s5;f11=fe/.s5;​​ at=Dot[a11,{1}];​​ bt=1;​​ ct=Dot[c11,{1}];​​ dt=Dot[d11,{1}];​​ et=Dot[e11,{1}];​​ ft2=Dot[f11,{1}];​​ Print[PasteButton["The Jerabek hyperbola equation:"]];Print[" "];Print[{at,bt,ct,dt,et,ft2}.{
2
x
,xy,
2
y
,x,y,1},"= 0"];Print[""];​​Print[PasteButton["Information about the Jerabek hyperbola:"]];​​μ=ResourceFunction["ConicProperties"][{at,bt,ct,dt,et,ft2}.{
2
x
,xy,
2
y
,x,y,1}==0,{x,y}];​​Print[μ];​​p=μ[[6,1]];​​p1=μ[[6,2]];​​g1=Line[{{x1,y1},{x2,y2},{x3,y3},{x1,y1}}];​​g2=Point[{xo,yo}];g3=Point[{xc,yc}];g4=Point[{x1,y1}];​​g5=Point[{x2,y2}];g6=Point[{x3,y3}];g7=Point[{xX125,yY125}];​​g8=ContourPlot[atx^2+btxy+cty^2+dtx+ety+ft2==0,{x,X0,X1},{y,Y0,Y1}];​​g9=Circle[{xce,yce},rce];​​g10=ContourPlot[p,{x,X0,X1},{y,Y0,Y1},ContourStyle->{Gray,Dashed}];​​g11=ContourPlot[p1,{x,X0,X1},{y,Y0,Y1},ContourStyle->{Gray,Dashed}];​​SwatchLegend[{Red,Green,Orange,Purple,Gray},{"
X
125
point","Orthocenter","Circumcenter","Nine points circumference","Asymptotes"}]​​Show[{Graphics[{Green,PointSize[0.01],g2}],Graphics[{Blue,g1}],Graphics[{Orange,PointSize[0.01],g3}],Graphics[{Black,PointSize[0.01],g4}],Graphics[{Black,PointSize[0.01],g5}],Graphics[{Black,PointSize[0.01],g6}],Graphics[{Red,PointSize[0.01],g7}],g8,Graphics[{Purple,g9}],g10,g11},AxesTrue,AspectRatioAutomatic,AxesOrigin{0,0},ImageSize900,PlotLabelStyle[Framed["The hyperbola of Jerabek and detail of the triangle"],16,Blue,BackgroundYellow]]​​​​​​
In the command below, type the new coordinates of the points. You can also change the extension to X and Y. Evaluate the notebook.
In[]:=
JerabekHyperbola[tri={{-2,5},{7,12},{9,-4}},{-20,20},{-20,20}]
Coordinates of the center of the Jerabek hyperbola (​
X
125
​)
X
125
(6.75885 , 7.61153)
The Jerabek hyperbola equation:
48.6858-4.30171x-0.244851
2
x
-10.4862y+xy+0.244851
2
y
= 0
Information about the Jerabek hyperbola:
TypeHyperbola,Equation48.6858-4.30171x-0.244851
2
x
-10.4862y+xy+0.244851
2
y
0,Foci{{9.16622,11.4709},{4.35147,3.75211}},Vertices{{8.46112,10.3405},{5.05658,4.8825}},Eccentricity1.41421,Asymptotes{y36.7815-4.31583x,y6.04547+0.231705x},Center{6.75885,7.61153},SemimajorAxisLength3.21641,SemiminorAxisLength3.21641,Directrices{y14.4278-0.623765x,y14.486-0.623765x},AxesOfSymmetry{y-3.22404+1.60317x,y11.8275-0.623765x},SemiLatusRectum3.21641,RotationAngle1.01309,ConjugateHyperbola1.(-668.965+17.5687x+1.
2
x
+42.827y-4.08412xy-1.
2
y
)0,BaseConicEquation0.0966624
2
x
-0.0966624
2
y
1,RotationAngle1.01309,DisplacementFromOrigin{6.75885,7.61153}
Out[]=
X
125
point
Orthocenter
Circumcenter
Nine points circumference
Asymptotes