Janet Kaplan

Constructing a Parabola

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We learn early on in our study of Algebra that a **parabola **is a graph of a quadratic
equation of the form y
= ax^{2} + 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 Geometer’s 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**.

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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 “Animate” 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.

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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.

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