To find this locus of points, we can construct a line perpendicular to the Directrix and passing through some arbitrary point A on the Directrix. Now notice that the shortest distance between any point on this line and the Directrix is equivalent to the distance between that point and A.

Next, we construct a line segment between the Focus, which is some arbitrary point not located on the directrix, and A. We then construct a perpendicular bisector to that segment. This perpendicular bisector will intersect with the line through A which is perpendicular to the Directrix. Let us call this point of intersection B.

Now the Segment between A and Focus is the base of an isosceles triangle, the other two sides of which are AB and BFocus. We know that this is an isosceles triangle because B is a point on the perpendicular bisector of AFocus. Then B is equidistant from A and Focus, which is equivalent to saying that B is equidistant from the Focus and the Directrix.

Therefore, we expect B to trace out a parabola as A moves along the Directrix.

Below is a model which was created using Geometer's Sketchpad software. Point A and the Focus can be moved by clicking and dragging the points. The path of point B will be traced, and the tracing can be cleared by clicking the red "X" at the bottom of the applet.