Parabola
with Circular Directrix

Erin
Mueller

Construct the locus of points equidistant from a fixed
point F and a circle. In other words, repeat the parabola construction but use
a circle as the ÒdirectrixÓ. Let F be any point in the plane other than the
center of the circle. Assume F is not on the circle; it can be either inside or
outside.

Below is a picture of what occurs when we trace the locus
of points equidistant from a fixed point (F) and a circle.

As we can see from above, we have a hyperbola when our
focus point (F) is outside of the circle. Below, we can see all of the hidden
parts of our picture to get a good look at what is REALLY happening.

Our intersection point (B) is being traced to form the hyperbola
as point (A) goes around our circle. When this happens, the two lines that
intersect to form (B) are also being moved. When these two lines become
parallel, we have our asymptotes. Only as long as these two line intersect,
will we have a point to trace that will become a hyperbola. Notice that our
focus (F) and the center of the circle are the two foci points of the
hyperbola.

Now look what happens when we trace the midpoint of
segment AF. What shape do you think we will see?

If you guessed a circle, you would be correct. The center
of this new circle is also the center of our hyperbola.

Now what if the focus (F) were inside the circle.

We now see an ellipse. The foci points of the ellipse are
the same as for the hyperbola; (F) and the center of the circle. Using what we know about ellipses (the
fact that the total distance from F(1) to F(2) to B is constant), we can see
how the intersection point (B) would remain inside the circle.

Again, we can also make a circle similar to the circle
formed inside of the hyperbola. Only this time, the circle will be tangent to
the outside of the ellipse.

Pretty neat huh?