Constructing a Parabola
††††††††††† We learn early on in our study of Algebra that a parabola is a graph of a quadratic equation of the form y = ax2 + bx + c.
Representing all the solutions to the equation, its structure and location on the Cartesian plane is determined by the coefficients of the terms for that particular equation.
††††††††††† But there is another way of defining a parabola, a more geometric way. And this is the focus of this exploration using Geometerís Sketchpad.
††††††††††† Geometrically speaking, a parabola is a locus of points equidistant from a fixed point, known as the focus, and a line called the directrix.
We construct a parabola as follows:
1) Begin by constructing a point for the focus.† Construct a line, the directrix, assuming the focus is not on the line.
We want to find the locus of points that are equidistant from the focus and the directrix.
2) Choose any point along the directrix and select it. Then construct a perpendicular line to the directrix at that point.
3) Construct a segment from the focus to the selected point on the directrix, and then construct its midpoint.
4) Construct a perpendicular bisector to the segment, as well as its intersection with the perpendicular line from the directrix. This is the Locus.
At this point our GSP construction looks like this:
5) Dragging the selected point along the directrix will show the movement of the intersection (locus point). Add the trace feature to the locus and it becomes clear that the locus of points is a parabola.
Click HERE to see a fully constructed sketch of the parabola. Click the ďAnimateĒ button within the sketch for a beautiful picture!
Notice that, as the trace indicates, the line going through the intersection and the midpoint of the base is tangent to the parabola.
What is really happening?
Remember that the definition of a parabola is the locus of points that are equidistant from the focus and the directrix.† Notice that the intersection (x,y) of the perpendicular to the directrix and the perpendicular bisector creates a vertex of an isosceles triangle, the base of which is the segment from (0,a) to (x,-a).† That is, the distance from (0,a) to (x,y) is equal to the distance from (x,y) to (x,-a).
As we drag the selected point along the directrix, the particular distances change, but the relationship remains the same. The intersection at (x,y) is always at a point equidistant from the focus and the directrix.