The Conic Sections in Polar Coordinates

This activity is an interactive study of the polar form of the equation for a conic section. Readers should be familiar with polar coordinates and triangle trigonometry. This activity is also a vehicle for the introduction of the Geogebra, though no prior experience with the Geogebra is assumed.

You can also download a Sketchpad Version of this activity.

The Definition

We begin by introducing the focus-directrix definition of the conic section. The directrix is a line. The focus is a point, usually chosen so that it does not lie on the directrix.

Let's begin with a focal point `F` at the origin of our coordinate system and a vertical line (called the directrix) k units from the origin, as shown in Figure 1.

Coordinate system with focus at origin and vertical directrix k units to the right of the origin.

Coordinate system with focus at origin and vertical directrix k units to the right of the origin.

Our task is to locate all points `P` such that the distance from `F` to `P` is a constant multiple of the distance from `P` to the directrix, the line `x=k`. In symbols, we want to locate all points `P` satisfying the relation

`FP = e PD`,

where `e` is a proportionality constant called the eccentricity of the conic section, and `D` is the point on the directrix closest to the point `P`. For example, in Figure 2, the ellipse shown is the set of all points `P` satisfying the relation `FP = 0.53 PD`.

An image showing all points whose ratio of distance to origin divided by distance to directrix is 0.53.

All points `P` on the ellipse satisfy the relation `FP=0.53 PD`.

In Figure 2, the eccentricity of the ellipse is `e = 0.53`, a number less than unity (`e<1`).; All ellipses have eccentricity less than one.

For purposes of discussion, we need to draw the auxiliary line segment `PB` and label some existing line segments, angles, and points. This we do in Figure 3.

Labeling the image for discussion. The focus F is at the origin, the point P is on the ellipse, the point D is on the vertical directrix x=k. We also drop a vertical segment from P to B on the x-axis, mark angle BFP as theta, and mark the radial length FP as r.

Adding auxiliary line and labels to ease upcoming discussion.

Let `r` and `theta` represent the segment `FP` and the angle `BFP`, respectively. Because triangle `BFP` is a right triangle,

`FB = r cos theta`.

Next, segment `PD` has length equal to the difference of segment `FC` and `FB`; i.e.,

`PD = FC - FB`.

Consequently,

`PD = k - r cos theta`.

Thus, `FP = e PD` becomes

`r = e (k - r cos theta)`.

Solving this last equation for `r`, we arrive at the equation of the conic section in polar form; i.e.,

`r = (ek)/(1 + e cos theta)`.

Geogebra

Geogebra is an open source clone (plus some of its own fine bells and whistles) of the Geometer's Sketchpad. Geogebra can be downloaded at the following URL:

http://www.geogebra.org/cms/

Alternately, providing you have the proper Java Runtime installed, you can start up Geogebra directly from the above website by clicking on the "Start Geogebra" link. In a sense, this is a better starting technique as you will always have the latest update of the software.

When you start Geogebra, you obtain a window similar to that in Figure 4.

An image showing Geogebra's startup window.

An image showing Geogebra's startup window.

The menus across the top of the Geogebra window contain extensive commands for creating dynamic, geometrical objects. We begin by selecting View->Axes from the menu system, which toggles on and off the axes (Note: We use the notation View->Axes to indicate to our readers to first select the main menu item View, followed by the sub-menu item Axes.) Secondly, we use the mouse to drag the separation bar between the algebra window and sketch window a bit to the left. The result is shown in Figure 5.

Turn off the axes and enlarge the sketch window.

Turn off the axes and enlarge the sketch window.

Geogebra's toolbar contains some of the most commonly used construction tools. Each has a little down-arrow in the lower-right corner of the tool icon. When you click this down-arrow, a drop-down list expands, showing a variety of extra tools. Before continuing with this activity, take some time to play with the tools and the menu items, enough to familiarize yourself with the Geogebr environment.

The Basic Sketch

Select the Segment between two points tool and create a line segment, as shown in Figure 6. Use the New Point tool to construct a point on the segment, near the position `C` indicated in Figure 6. Use the Midpoint or Center tool, then select the segment to create the midpoint of the segment at `D`. At this point, your sketch should resemble that in Figure 6.

Line segment AB with midpoint at D and auxiliary point at C.

Line segment `AB` with midpoint at `D` and auxiliary point at `C`.

We are now going to change some of the labels in Figure 6. Select the Move-tool from Geogebra's toolbar. Click and drag each label in the sketch into a more favorable position. Next, double-click the label `A` with the Move tool. Change the label in the resulting dialog box to the letter `F`, because this is the point we will use as the focus of our conic section. Similarly, change the letter `D` to the letter `M`, a much more satisfactory letter to represent the midpoint of the segment `FB`. Your sketch should closely resemble that of Figure 7.

Name the point at A to F, as it is the focus. Name the midpoint M.

Let `F` represent the focus and `M` the midpoint of segment `FB`.

Select the Circle with center through point tool from the toolbar. Single-click the focus at `F` (you must let up the mouse after the single-click --- don't hold it down), then move the mouse and click in the vicinity of point `A` in Figure 8. You can resize the circle later by dragging the point `A`. Select the New Point tool from the toolbar and construct a point on the circle. Select the Move tool (the selection arrow), double-click the label of the newly created point on the circle, and change the label to the letter Q, as shown in Figure 8.

Draw a circle with center at the focus F, then use the point tool to place a point Q on the circle.

Construct a circle with center at `F` and construct a point `Q` on the circle.

Select the Ray through two points tool. Click the point `F`, then the point `Q` to construct a ray at F that emanates outward through the point Q, as shown in Figure 9. You may need to adjust the position of the label Q with the Move tool.

Draw the ray FQ.

Construct the ray `FQ`.

Select the Perpendicular line tool, click the point `B`, then click the segment `FB`. This should create a line through `B`, perpendicular to the segment `FB`, as shown in Figure 10.

Draw a line through B perpendicular to segment FB.

Draw a line through `B` perpendicular to segment `FB`.

Finding the Eccentricity

The point `C` in Figure 10 is a point on the conic section. At least, that's our plan. Consequently, `FC = e CB`; i.e., the ratio `FC//CB` equals `e`, the eccentricity of the conic section. We need to calculate this ratio, but first we must measure the segments `FC` and `CB`. Select the Distance or length tool (on the fourth icon from the right end of the toolbar). Click the points `F` and `C`, which will calculate the distance between the points `F` and `C`. In a similar manner, calculate the distance between the points `C` and `B`. When this is accomplished, the measures of the segments `FC` and `CB` will appear in the Algebra window as distanceFC and distanceCB, respectively, similar to that shown in Figure 11. Note: The actual numbers may differ, due to your personal configuration of Geogebra. Don't worry about numbers; everyone will have different numbers at this point. Continue to the next step.

Measure the distance between F and C, then between C and B.

Measure the distance between `F` and `C`, then between `C` and `B`.

Computing the Eccentricity

We need to computer the eccentricity, in this case defined as `e=FC//CB`. In the Input window, as shown in Figure 12, enter the expression e=distanceFC/distanceCB.

Use the Input window to calculate FC/CB, which is the eccentricity.

Use the Input window to calculate `FC//CB`, which is the eccentricity.

When you press Enter, the result of this calculation in the Input window is placed in the algebra window, in our case as `e=3.79`, as shown in Figure 13. Note: Again, don't worry about the number you get for your ratio. Plunge ahead with the activity.

The resulting eccentricity appears in the Sketchpad window.

The resulting eccentricity appears in the Algebra window.

Computing the Radial Length

Recall that the polar form of the conic section is

`r = (ek)/(1 + e cos theta)`.

Recall that `k` represents the distance of the directrix from the origin (See Figure 3), so we'll have to find the distance `k=FB`. Similarly, we'll need to find the measure of angle `theta` (again, see Figure 3).

Thus, we need two additional measurements, the length of segment `FB` and the measure of angle `\angle BFQ`. Use the Distance or length tool to measure `FB`. Then set k=distanceFB in the Input box as shown in Figure 14. (Turns out this is a bug/feature in Geogebra and `k` winds up being a "free object", which is not what we want. A better move here is to type the following in the Input box, then hit Enter: k=Distance[F,B]. This makes `k` a dependent object, which is what we need.)

Measure the length of segment FB and measure angle BFQ.

Measure the length of segment `FB`.

Pressing Enter executes the calculation in the Input window and the result shows up in the Algebra window, in our case `k=5.56` as seen in Figure 15. Next, use the Angle measurement tool (found in the fourth icon from the right on the Geogebra toolbar), then click `B`, `F`, and `Q`, in that order. The result shows up in the Algebra window, in our case as `alpha=41.65^\circ` as seen in Figure 15. Note: Again, your numbers will differ. Don't worry about them and continue with the activity.

Measure the length of segment FB and measure angle BFQ.

Measure angle `\angle BFQ`.

Calculating the Radial Length

We're now ready to calculate the radial length using our formula

`r = (ek)/(1 + e cos theta)`.

In the Input window, enter the expression r=e*k/(1+e*cos(α)), as shown in Figure 16.

Calculating the radial length r.

Calculating the radial length `r`.

Some comments are in order.

  1. Simply type cos( to get the cosine shown in Figure 16.
  2. Getting the `alpha` is a bit trickier, unless you know what to look for. Note that to the right of the Input window, there are three drop-down lists. The second one contains the Greek characters. Select `alpha` from this drop down list.

Pressing Enter executes the calculation in the Input window and places the result in the Algebra window, in our case `r=5.5`, as shown in Figure 17. Note: Again, don't worry about getting different numbers. Plunge ahead with the activity.

Calculate the radial length.

Calculate the radial length.

Defining the Conic

Select the Circel with center and radius tool, then click the point `F`. A dialog box will pop up asking for the radius. Enter `r` in this box and press Apply. The resulting circle is shown in Figure 18.

Create a circle of radius r centered at F.

Create a circle of radius `r` centered at `F`.

Select the Intersect two objects tool (from the second icon from the left), then click the newly formed circle, then the ray `FQ`. Use the Move tool to double-click the label on the new point of intersection and change the label to `P`, as shown in Figure 19.

Create an intersection point at P.

Create an intersection point at `P`.

Dynamic Geometry

Now for the really fun part. Use the Move tool to drag the point `Q` around the circumference of the circle and watch the action of the point `P`. It is important that you make one complete revolution about the circle. Note: Remember, the ratio `FC//CB` is the eccentricity. In Figure 19, the point `C` lies to the right of the midpoint. Therefore, the ratio `FC//CB` is larger than unity and the conic is a hyperbola.

It gets better. Select the Locus tool (on the fourth menu from the left), select point `P`, then the point `Q`, in that order. Select Construct-->Locus. Again, the eccentricity is greater than one, so the locus of the point `P`, as the point `Q` travels about the circle, is the hyperbola shown in Figure 20.

Construct the locus of point P and point Q moves around the circle.

Construct the locus of point `P` and point `Q` moves around the circle.

Further Exploration

The theory says that the conic drawn depends on the eccentricity.

  1. If the eccentricity is less than one, the conic will be an ellipse.
  2. If the eccentricity equals one, the conic will be a parabola.
  3. If the eccentricity is greater than one, the conic will be a hyperbola.

Now you know why we created the point `C` on the segment `FB`. It controls the eccentricity. Use the Move tool to move the point `C` to various positions on the line segment `FB` and observe how your sketch changes.

Formatting

You can hide parts of your sketch that you don't want your readers to see. Simply right-click (or ctrl-click on the Mac) the object you want to hide and uncheck Show object. Use this utility to hide objects in Figure 20 that you find distracting to the concepts you are trying to portray.

Homework

Save your sketch as conic.ggp in your home directory. See me in my office for a grade when you have finished with this activity. Alternatively, you can email me the file conic.ggp. If it runs on my machine, you receive full credit.