Thomas Earl Ricks

Mathematics Education

 

Assignment # 6

 

ÒGeometric Construction of a ParabolaÓ

 


 

The Parabola

 

We are all fairly familiar with the quadratic equation of the form:

 

 

This equation, when graphed, forms a parabola, such as

 

 

But there is also a geometric interpretation for a parabola.

 

A parabola is the set of points equidistant from a given line (called the directrix) and a given point not on the line (called the focus).

 

We will now go through a step-by-step construction of a parabola using this definition by using GeometerÕs Sketchpad.

 


 

GSP Construction of a Parabola

 

We start with a given line (the directrix) and a point not on the line (the focus).  We will color both of these bright green for reference.

 

 

 

We now wish to find a way using GSP to identify all the points that are equidistant from the line and focus point.

 

We could identify, for example, the most obvious point equidistant from the line and focus, which lies halfway in between:

 

 

But how to find the others?

 

Remember, we want to find all the points that are the same distance from focus as from the line, like so:

 

 

The distances (represented by the two black arrows) should be the same.

 

We may first notice that the distance from the line to our point we are trying to find will always lie along the perpendicular through the point to the line, like so:

 

 

But where should our point be?  Just being on the perpendicular is not enough.  Where on the perpendicular should our point lie so it is equidistant from focus to directrix?

 

We can solve this problem by relying on some elementary geometric principles.

 

We first choose a point on our directrix for which our perpendicular line will pass through:

 

 

 

We then construct a line segment from our focus to this base point (we will also forget about the point below the focus for now, so as to not confuse our diagram):

 

We then find the midpoint of this segment and draw a perpendicular through it:

 

 

We mark the point of intersection of this perpendicular with the other perpendicular, like so:

 

Does this intersection point look like it is equidistant from the focus and the directrix?

 

Is it?

 

Can you prove it?

 

We will now provide a quick justification for why this intersection point is equidistant from the directrix and focus.

 

Note that the line segment from the focus to the base point of the perpendicular to the directrix is bisected, which means the thick orange line segment is congruent to the thick purple line segment:

 

And the thick tan segment that is perpendicular to these two segments is congruent to itself:

And the angle formed between the tan segment and the orange is the same as the angle between the tan segment and the purple, since the tan is perpendicular, which means both angles are ninety degrees (represented by the thick grey right angle sign):

 

 

We therefore have two triangles that are congruent to each other by the Side-Angle-Side theorem (SAS).  Thus the orange triangle:

 

is congruent to the purple triangle:

 

by the SAS theorem.  This means that all corresponding parts of these congruent triangles are the same.  Therefore the orange line segment from the focus to our intersection point is congruent (or the same length) as the line segment from our base point on the directrix to the intersection point:

 

We have therefore developed a construction on GSP that allows us to find a point equidistant from the directrix and the focus. 

 

If we move our base point on the directrix, we can find other points (light blue in the diagram below) that are equidistant, like so:

 

Therefore, we simply animate the base point on the directrix in GSP, and trace the intersection point, to obtain a red trace of the points equidistant from the focus to the directrix.  This trace represents a parabola:

 

 

Now we can have all sorts of fun and experiment and explore!

 

For example, what happens if the focus point is brought closer to the directrix?

 

We get a steeper, or narrower parabola.

 

What if the directrix line is turned like so?

We get a parabola that is turned!

 

 

(By the way, running the animation for longer allows more points to be shown on the trace.)

 

See if you can experiment and discover!

 

Have fun!

 


 

Here is a similar investigation using a circle as a path instead of a directrix line.

 

With the focus inside the circle but not at the center, we get an ellipse:

 

With the focus at the center we get a smaller circle:

With the focus outside the circle, we get a hyperbola:

 

 

Notice the ÒsteepnessÓ of the hyperbola changes with how close the focus is to the circle:

 

 

 


 

Click here for more information about parabolas.

 


Home

Jim WilsonÕs Homepage

UGA Homepage