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3. The parabola

3.1 Exploring the parabola as locus

You will already have studied some of the properties of the parabola in algebra. In this section we will study the parabola as a special case of the locus of a conic section.

Let’s start with a matric examination question, which shows that knowledge about the locus definition of a parabola is useful.

Solve this problem from the Mathematics HG, Nov 2002 National Paper 2:

Show that the equation of the locus of a point P(x, y), which is equidistant from the line y = -1 and the point M(1;2), has the form ay = bx2 + cx + d.

From the Cartesian equation it is clear that the locus is a parabola. But can you explain why this kind of condition always produces a parabola, and not some other curve?

What is variable and what is constant in the condition? Can you generalize the condition of the locus that will always produce a parabola?

Click here for a discussion:  

A parabola is a conic, and therefore satisfies the general conic locus definition of all points P such that the its distance PF from a fixed point F (the focus) is in a constant ratio (e) to its distance PD to a fixed line (the directrix), i.e. PF = e x PD. For a parabola, e = 1, which means PF = PD, i.e. P is equidistant from the point and from the line.

We will now construct a parabola from this definition. Open the Exploring the parabola applet:

In the applet, click on “Show construction lines”, and make sure that you understand the construction:

  • Draw a segment from the focus F to any point D on the directrix.
  • Draw the perpendicular bisector of FD.
  • Draw a perpendicular to the directrix at D to intersect the perpendicular bisector of FD in P.
Then all points P satisfy the condition that PF = PD.

In the applet, drag or animate D along the directrix, causing P to move and trace a locus of points …
How would you describe this locus?

Stop the animation by clicking on the Animate button again.
Now move the focus to another position. Clear any traces by clicking on the red X.
Animate again ...
What influences the shape of the locus – when is it “thin’, when is it “fat”?

Move the directrix by dragging A and/or B to any position ... Clear any traces and animate again ...

The curves certainly look like parabolas! But we know that we cannot trust our perceptions, so we must prove it is a parabola. The applet shows that PF = PD … Now prove geometrically that the construction implies that PF = PD, therefore that the locus satisfies the definition of a parabola, and therefore that the locus indeed is a parabola.