ASSIGNMENT 6

BY

SHADRECK CHITSONGA

PARABOLAS

Most of the students are familiar with the curve known as the parabola. They have encountered so many times in Algebra. They know that given any quadratic function, the resulting graph is that of a parabola. But how many of them have been exposed to the geometric part of the parabola. Do students know how to define the parabola in terms of geometry? What is the relationship between the geometric definition and the algebraic definition? In this write I will discuss the algebraic and the geometric definitions of the parabola.

Let us consider the graph of shown in figure 1.

Figure 1

We can use the method of completing the square to help us analyze this graph.

Without go through all the necessary steps, we can see that

is equivalent to

This can be written as . This information tells us that the parabola has its vertex at . In general a parabola with its vertex at (h, k) is given by the formula where d is the distance from the directrix to the vertex of the parabola. In the graph the directrix is known shown. Not surprisingly it is not mentioned in algebra classes. Another aspect that is omitted in most algebra lessons is the focus (not shown in the graph above). The d is also the distance from the focus to the vertex. For the time being we will just mention that directrix will be the line

and the focus will be at . This will become clear, as we will at the geometric definition of a parabola.

GEOMETRY AND THE PARABOLA

Before we define the parabola geometrically let us first of all construct the parabola in GSP.

1. Draw any line and put a free point on the line. We will call this point D. Put another point above the line and call it F.

2.  Through D draw a perpendicular to the line drawn in 1.

3. Join DF and bisect in N.

4. Through draw a perpendicular to the line DF to meet the perpendicular drawn in (2) at G. Select the point D and G and go to construct menu and choose locus.

5. Refer to figure two (for steps 1-4)

Figure 2

Figure 3

Figure 3 shows the parabola as constructed in GSP. Remember we did not start with any equation; we just used lines and the locus. Why does this work?

Let us look at a few things from the diagram.

i. The distance from the focus to the vertex of the parabola is

the same as the distance from the vertex to the directrix.

ii. The distance from the focus to the point G on the curve is

the same as the distance from the point G to the directrix.

Let us now define the parabola in terms of locus:

A parabola is the locus of a point that is equidistant from a fixed point and a fixed line.

(The fixed point here is the focus and the fixed line is the directrix)

PROOF FOR CONSTRUCTION AND DEFINITION

We are going to use congruency.

Consider the diagram below which is extracted from the construction of the parabola.

Figure 4

In triangles FNG and DNG

GN=GN (common)

FN=ND (midpoint-construction)

Angle FND=Angle GND (Rt. Angle by construction)

Therefore triangles FNG and DNG are congruent by SAS.

If follows that FG=DG

Going back to our definition using locus we can see that Focus is fixed, D is any point on the directrix that is free to move, but the directrix itself is not free to move. The point G is the point that is equidistant from the fixed point F(focus) and the fixed line (directrix)

Connection between algebraic and geometric definitions.

Let us consider a simple case when we have a quadratic function , where a is a constant.

Figure 5

From the geometric definition of a parabola we know that FG and GD are equal in length.

Using the distance formula we see that

FG=

HG=

But from above we see that “m” is the distance from the focus to the directrix. The distance from the focus to the vertex is half this distance. The vertex and the focus are collinear. This then means that:

Going back to the algebraic representation of a parabola , we can write this as . We know that , which means that and this is the same quantity that we found geometrically.