The Bisector of a Point and a Line

Last time, we were able to define a line as the set of points that are equidistant from two points. Today, we will ask a slightly different question: what shape do we get by taking the set of points equidistant from a point and a line?

To start, we will look at the set of points equidistant from a point P, which we will call the focus point, and the line y = 0, commonly known as the x-axis. We do this because the distance between a point and the x-axis is simply the absolute value of the y-coordinate of that point. Our starting point is therefore:

sqrt[(X - P_x)² + (Y - P_y)²] = |Y|

We apply a liberal portion of algebra to this equation:

(X - P_x)² + (Y - P_y)² = Y²
(X - P_x)² + y² - 2P_yY + P_y² = y²
(X - P_x)² + P_y² = 2P_yY

At this point, we will consider a special case: that of the focus point being on the x-axis. Then, P_y = 0, and the equation simplifies to (X - P_x)² = 0, which is only true when x = P_x. The shape we get back, then, is the vertical line containing the focus point.

Let us then proceed, assuming that P_y ≠ 0:

(X - P_x)² + P_y² = 2P_yY
X² + (-2P_x) X + (P_x² + P_y²) = 2P_yY

The answer to our original question is thus shown to be a parabola, Y = Ax² + Bx + C, with:

D = 2P_y
A = 1 / D
B = -2P_x / D
C = (P_x² + P_y²) / D

A common question to ask of a parabola is what its roots are. If you think about the problem for a moment, the answer is self-evident. If you prefer not to think, you could just evaluate the determinant of the parabola:

det = B² - 4AC
= [4P_x² - 4(P_x² + P_y²)] / 4 P_y²
= -4 P_y² / 4 P_y²
= -1

A negative determinant tells us that the parabola has no real roots, which we expect. (Recall that we are looking for points the same distance away from a point P and the x-axis. If the parabola had a root, then there exists a point x = (P_x, 0) on the parabola, which must be distance zero from our original anchor point , which is a contradiction of the assumption that the anchor point has a nonzero y-coordinate.)

Let us now reverse the question. Say we are given a parabola Y = Ax² + Bx + C. Can we find the focus point for that parabola? The answer is sometimes. It is easy to solve for the points P_x and P_y in terms of A and B:

P_y = 1 / 2A
P_x = DB / -2 = -B / 2A

The problem is that this approach is only valid for parabolas with a determinant of -1. We will solve this problem in the next post.