Parabola Construction
by Emily Kennedy

A parabola consists of all points which are equidistant from a line d (the directrix) and a point F (the focus), where the focus does not lie on the directrix.

Given a directrix and a focus, let's construct a point which lies on the parabola.

Given any point X on d, I claim that the following construction generates a point P which is equidistant from F and d:

 Given: A line d, a point F (not on d), and a point X on d. 1. Draw . 2. Let M be the midpoint of . Draw h, the perpendicular bisector of . 3. Draw k, the line which is perpendicular to d and goes through X. 4. Let P be the point at which h and k intersect. (This point exists because F is not on d.)

In GSP, if I animate X and trace the point P, I will trace a parabola.

Now I will prove that the point P is equidistant from d and F.

The distance from P to d is the length of the segment which is perpendicular to d and whose endpoints are P and some point on d. That is, the distance from P to d is precisely the length of .

And the distance from P to F is simply the length of .

So I need to show that and are the same length.

h is the perpendicular bisector of , so we must have .

Also, since h is a perpendicular bisector, we must have .

And , clearly.

So, by the SAS congruence postulate, we have .

Thus, , and so their lengths are equal.

Therefore, P is equidistant from d and P.

Q.E.D.

We can use this construction to determine the equation describing a parabola with focus (a, b) and a horizontal directrix described by y = y0.