Exploring Analyic Geometry with Mathematica® |
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Rectangular Hyperbola Distances
hypinv.html
Exploration
Show that the distance of any point on a rectangular hyperbola from its center varies inversely as the perpendicular distance from its polar to the center.
Approach
Construct a generic point on a rectangular hyperbola and compare the appropriate distances.
Initialize
To initialize Descarta2D, select the input cell bracket and press SHIFT-Enter.
This initialization assumes that the Descarta2D software has been copied into one of the standard directories for AddOns which are on the Mathematica search path, $Path.
<<Descarta2D`
Solution
Create a generic point on a rectangular hyperbola.
Clear[a,t];
h1=Hyperbola2D[{0,0},a,a,0];
p1=Point2D[a*Cosh[t],a*Sinh[t]]
Compute the distances.
p0=Point2D[0,0];
{D1,D2}=
{Distance2D[p0,p1],
Distance2D[p0,l1=Line2D[p1,h1]]} //Simplify
Use a trigonometric identity.
1/Sech[2t] //Simplify
Therefore, since
is a constant,
varies inversely as
.
Clear[E1];
D1*D2 //. {
Sqrt[a^2*E1_]->a*Sqrt[E1],
Sqrt[Cosh[E1_]]*Sqrt[Sech[E1_]]->1}
Discussion
This is a plot of a numerical example of the geometric objects.
Sketch2D[{h1,p1,p0,l1} /. {a->1,t->0.5}]
Graphics saved as "hypinv01.eps".