Assignment 6

Exploring how to derive the algebraic formula of a quadratic from the geometric definition of a parabola.


A parabola is defined to be the set of points that are equidistant from a given line and a fixed point. The line is called the directrix and the fixed point is called the focus. We are going to assume that the focus is not on the directrix.


Let's begin by looking at how to set up this problem. First, we need to select a line and a fixed point not on the line then connect a point on the line to the fixed point. As shown below.

Now we want to construct the midpoint of the segment and the perpendicular bisector to the segment. We construct the midpoint because we are talking about distance relationships and the perpendicular bisector to the segment because this is the tangent line to the parabola.

Now let's use the technology to trace the tangent line as our point on the line moves back and forth. To see the animation click here and you will go to a GSP file. Click on the animate button in the upper left corner. To see the results look below.

The trace of this looks like it leaves a shape like a parabola. Looking at all of the tangent lines is called looking at the envelope of the parabola. This leaves us with the sense of what the shape of a parabola is but but doesn't give us any way to describe the particular points that lie on the parabola.

How can we find the point on the perpendicular bisector of the segment that actually lies on the parabola? Since distance from a point to a line is measured along the perpendicular. Let's construct the segment perpendicular to the given line through the point D and see where it intersects the perpendicular bisector.



In the diagram, the intersection point of BD with the perpendicular bisector is the point B. Because CB is congruent to CB, AC is congruent to CD, and angle ACB and DCB are both right angles triangle ACB is congruent to triangle DCB by SAS. Since the distance between a point and a line is measured along the perpendicular and AB is congruent to BD, we know that the distance from the given line to the point B is equal to the distance from the point B to the point A.

Now, let's use GSP to trace this point and see if we get the actual points of the parabola. Click here if you want to see the animation. Look below for the results of the animation.

Indeed, the point that we have chosen gives the parabola that corresponds with the envelope of the parabola we saw in the previous problem.

How do we go about getting the algebraic equation of this graph? To do this we need to introduce the Cartesian Coordinate system. Let's look at a setup which is on the Cartesian Coordinate system and see if we can derive the more traditional algebraic understanding of a parabola.

We have oriented our parabolic construction on the Cartesian Coordinate system so that the directrix is on the x-axis and the focal point is on the y-axis. This should give us an orientation we are used to for the parabola as well as simplify the variable situation when we try to describe the parabola algebraically. It is important to note that you don't have to put your directrix and focal point in this orientation but this author did it in the hope of simplfying the algebra a bit.

To start with let's find the equation of the line M that P and F lie on. We know the coordinates of P = (p, 0) and the coordinates of F = (0, f). As you can see in this graphical depiction P, and F have numerical values that we could estimate but we want to look at the general case so we will leave them as having variable coordinates.

The equation of line M is y= (-f/p)x + f.

Now let's describe the equation of line L. We know the slope of line L is p/f because L is perpendicular to M. Let's try and find the y-intercept. We also know the point (p/2, f/2) lies on line L because it intersects the midpoint of the segment PF. So with a little substitution we get that the equation of line L is y = (p/f)x +(f/2 - p^2/2f).

Now how do we find the points that lie on the parabola. We construct the perpendicular through point P which happens to be always equal to x = p.

Then we look for the intersection of the equations x = p and y = (p/f)x +(f/2 - p^2/2f) and this should give us the locus of points of the parabola. Through simplification we get y = p^2/2f +f/2 so the relationship of these points to y is a quadratic equation. If we consider that the value p is a varying x value we can rewrite the equation as y = x^2/2f + f/2 and this gives us the more standard notation. This is how the connection is made between the geometric definition of the parabola and the algebraic association of parabolas with quadratic equations. I will leave you with a graph of the parabola given by this particular construction set on the Cartesian Coordinate system. Use the graphical values on the coordinate system to see if the equation looks right.

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