Exploring Analyic Geometry with Mathematica®

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Equilateral Hyperbolas

hyp4pts.html

Exploration

Describe a method for finding the equilateral hyperbola(s) passing through four points. Show that the technique produces the correct results for the points (2,1), (-1,1), (-2,-1) and (4,-3) by plotting the hyperbola(s) and the four points.

Approach

Form a quadratic, parameterized by the variable k, representing the pencil of quadratics passing through the four points. The first and third coefficients of the quadratic, a and c, must satisfy the relationship a=-c, if the quadratic represents an equilateral hyperbola. Solve the equation for k.

Initialize

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<<Descarta2D`

Solution

This is a function that implements the approach.

Quadratic2D[p1:Point2D[{x1_,y1_}],p2:Point2D[{x2_,y2_}],
            p3:Point2D[{x3_,y3_}],p4:Point2D[{x4_,y4_}],
            Hyperbola2D] :=
Module[{Q1,k,a,b,c},
      Q1=Quadratic2D[p1,p2,p3,p4,k,Pencil2D];
      {a,b,c}=List @@ Take[Q1,3];
      ans=Solve[a==-c,k];
      Map[(Q1 /. #)&, ans] ];

Discussion

Here's the plot of the solution for the four points specified.

pts={p1=Point2D[{2,1}],p2=Point2D[{-1,1}],
     p3=Point2D[{-2,-1}],p4=Point2D[{4,-3}]};
q1=Quadratic2D[p1,p2,p3,p4,Hyperbola2D] //N

"hyp4pts_1.gif"

hyp1=Map[Loci2D,q1]

"hyp4pts_2.gif"

Sketch2D[{pts,hyp1},CurveLength2D->20]

"hyp4pts_3.gif"

Graphics saved as "hyp4pt01.eps".


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