Note

This Mathematica notebook is from the WWW project A Visual Dictionary of Special Plane Curves. Copyright © 2004 by Xah Lee. (http://xahlee.org/)

Needs["ParaPlot`"]
Needs["PlaneCurveFormulas`"]
Needs["PlaneCurveGenerator`"]
Needs["Graphics`ImplicitPlot`"]
$HistoryLength = 5;

the Cross Curve

Packages and Initializations

Needs["ParaPlot`"]
Needs["PlaneCurveFormulas`"]
Needs["PlaneCurveGenerator`"]

Needs["Graphics`ImplicitPlot`"]
(*MathSource Item:0200-529*)

plotting the curve

?CrossCurve

ParaPlot[Evaluate@CrossCurve[][t], {t, 0, 2  Pi}, 
      AspectRatio -> Automatic, 
      PlotRange -> {{-1, 1}, {-1, 1}} 7]

The cross curve is given by the equation: x^2+y^2-x^2*y^2==0

crossCurve = Function[{x, y}, x^2 + y^2 - x^2*y^2];

crossCurve[x, y] == 0

Needs["Graphics`ImplicitPlot`"]

a = 4;
ImplicitPlot[
        crossCurve[x, y] == 0, {x, -a, a}, {y, -a, a}, 
        PlotPoints -> {50, 50}];
Clear@a;

a = 1.6;
Plot3D[crossCurve[x, y], {x, -a, a}, {y, -a, a}, 
        BoxRatios -> Automatic, 
        PlotPoints -> {30, 30}];
Clear@a;

a = 4;
Plot3D[
        crossCurve[x, y], {x, -a, a}, {y, -a, 
          a}];
Clear@a;

a = 4;
ContourPlot[crossCurve[x, y], {x, -a, a}, {y, -a, a}, 
        PlotPoints -> {50, 50}, 
        Contours -> 14];
Clear@a;

a = 4;
DensityPlot[crossCurve[x, y], {x, -a, a}, {y, -a, a}, 
        PlotPoints -> {50, 50}, 
        Mesh -> False];
Clear@a;

Formula derivation

The circle centered on the origin with radius 1 is
{Cos[t],Sin[t]}
Its tangent vector at parameter t1 is {-Sin[t1],Cos[t1]}.Using the line formula,a line passing the point at parameter t1 and parallel to the vector is then:


the equation of a line passing {a,b} and {c,d} is
y==(b*c-a*d-b*x+d*x)/(-a+c)

now our two points are {Cos[t1],Sin[t1]} and
 {-Sin[t1],Cos[t1]}+{Cos[t1],Sin[t1]}
 substitute we have

{-Sin[t1], Cos[t1]} + {Cos[t1], Sin[t1]}

y == (b*c - a*d - b*x + d*x)/(-a + c) /. {a -> 
          Cos@t1, b -> Sin@t1, c -> Cos[t1] - Sin[t1], 
        d -> Cos[t1] + Sin[t1]}

Simplify@%

To find their intersection with the x and y axis, set x==0 or y==0.
When y==0, we have x==Sec[t1]
When x==0, we have y==Csc[t1]
So we see that the cross curve is {Sec[t1],Csc[t1]}

The Cartesian equation is easily found to be x^2+y^2=x^2*y^2. We eliminate the trig functions in the parametric formula, using Cos[t]^2+Sin[t]^2==1.