Ellipse Locus, Distance from Two Lines

elldist.html

Exploration

A point moves so that the sum of the squares of its distances from two intersecting straight lines is a constant. Prove that its locus is an ellipse.

Approach

Compute the distances from a generic point (x,y) to the lines and show that the equation must be an ellipse.

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 the two lines and a generic point.

Clear[A1,B1,C1,A2,B2,C2,x,y];
l1=Line2D[A1,B1,C1];
l2=Line2D[A2,B2,C2];
pt=Point2D[x,y];

Sum of distances squared is a constant, K.

Clear[K];
eq1=Distance2D[pt,l1]^2+Distance2D[pt,l2]^2-K

Form the quadratic equation (without loss of generality, assume the lines are normalized).

Q1=Quadratic2D[eq1,{x,y}] /.
   {A1^2+B1^2->1, A2^2+B2^2->1}

Compute the discriminant of the quadratic, .

disc=Q1[[2]]^2-4*Q1[[1]]*Q1[[3]] //Simplify

The discriminant of the quadratic, , is negative; therefore, the curve is an ellipse.  Note that the expression cannot be zero if the lines intersect.

Discussion

This is a plot of a numerical example using three different values of K.

Sketch2D[{l1,l2,
   Map[(Q1 /. #)&, {K->2, K->3, K->6}]} /. {
   A1->1, B1->1.5, C1->-1,
   A2->-0.5, B2->2.5, C2->-1},
   CurveLength2D->5]

Graphics saved as "elldis01.eps".