Locus Problem.

Let ABC be a triangle constrained to move in the plane so that its vertices A and B always lie on a pair l[1] and l[2] of intersecting lines so that if A is on one line then B is on the other. Show that the locus of the point C is an ellipse.

A Solution.

Without loss of generality, assume that the lines are the x and y axes, and that the length of side AB is 1. Parameterize the point A as [t,0], where t ranges between -1 and 1. The point B takes two positions [0, sqrt(1-t^2)] and [0, -sqrt(1-t^2)] . Now the point C = [x,y] can be written as

C = s*A + (1-s)*B + r*perp(A-B),

where s and r are fixed numbers, and perp(A-B) is the vector A-B rotated 90 degrees counterclockwise (so if B = [0, sqrt(1-t^2)] then perp(A-B) = [sqrt(1-t^2), t] and if B = [0, -sqrt(1-t^2)] then perp(A-B) = [-sqrt(1-t^2), t] ).

To show that C traces out an ellipse, write parametric equations for C and eliminate the parameter:

x = s*t+r*sqrt(1-t^2), y = (1-s)*sqrt(1-t^2)+r*t

Case 1 . s <> r^2+s^2

To eliminate the parameter, solve the equations for t and sqrt(1-t^2) , and substitute those values into the equation t^2+sqrt(1-t^2)^2 = 1 .

t = (-x+x*s+r*y)/(-s+s^2+r^2), sqrt(1-t^2) = (-s*y+...

Making the substitution, we get


This quadratic equation in x and y graphs an ellipse (since it is not a hyperbola or parabola). Below is a typical locus where r = 2 and s = 1. Note that the foci have been inserted. See below for the computation.

[Maple Plot]

Here is a locus where the the "triangle" ABC is a segement: r = 0, s = 1.5

[Maple Plot]

To compute the foci of the ellipse, note that the ellipse (-x+x*s+r*y)^2/((-s+s^2+r^2)^2)+(-s*y+r*x)^2/((-s+s... is the image of the circle x^2+y^2 = -s+s^2+r^2 under the linear transformation defined by the matrix A := matrix([[-1+s, r], [r, -s]])

The eigenvectors and eigenvalues of A are u := vector([(-1/2+1/2*sqrt((2*s-1)^2+4*r^2)+s)/r, ... , ueig := -1/2+sqrt((2*s-1)^2+4*r^2)/2 , and

v := vector([(-1/2-1/2*sqrt((2*s-1)^2+4*r^2)+s)/r, ... , veig := -1/2-sqrt((2*s-1)^2+4*r^2)/2 . Since A is symmetric, we know the eigenvectors are perdicular and form the principal axes of the ellipse. The eigenvalues (in absolute value) are the major semi-axis M = 1/2+sqrt((2*s-1)^2+4*r^2)/2 in the director of v and minor semi-axis m = abs(-1/2+sqrt((2*s-1)^2+4*r^2)/2) in the direction of u. So the foci are F[1] = sqrt(M^2-m^2) u/abs(u) and F[2] = -sqrt(M^2-m^2) u/abs(u) .

Note that the ellipse is a circle exactly when (2*s-1)^2+4*r^2 = 0 , that is, when s = 1/2 and r = 0; geometrically, this says the only time the trace of C is a circle is when C is the midpoint of the segment AB. Here is the locus at that point.

[Maple Plot]

Case 2. s = s^2+r^2

At time t =1, the vertex C is at [s,r] and the segment AB is the unit segment [[0,0],[1,0]]. Note the point [s,r] is on the circle of radius 1/2 centered at [1/2,0] and so the triangle ABC is a right triangle with the vertex C opposite the hypotenuse. We can eliminate here the parameter t by multiplying x by r and y by s and subtracting, to get 0. So the vertex C traces out a segment on the line y = r/s x. With a little more work, one could show this segment to be a diameter of the unit circle.

Here is an animation showing the trace of C in magenta for all the exceptional values for r and s.

[Maple Plot]

Maple computations