Problem:
Determine if the following polar conic equations are parabolas, circles, ellipses, or hyperbolas.


Visualization:
Using Derive:

  1. Prepare the program for graphics.
  2. If the first menu item is Algebra then press .
  3. Press for Author.
  4. Enter the function
    r=3/(3-2cos )
  5. Press to get the plot window.

    Now we need to prepare Derive to graph in polar coordinates.

  6. Press to enter the options menu.
  7. Now press to enter the state menu.
  8. Finally, press and then to tell Derive to graph using continuous polar coordinates.
    Note: Derive will continue to graph in polar coordinates for the remainder
    of the session or until the state is chaged back to rectangular.


  9. Now press and then to see the graph.
We see this is an ellipse

We will now plot the remaining equations, pay attention to what shape each equation gives.

  1. Next press to return to the algebra window.
  2. Press for Author.
  3. Enter the function
    r=2.5
  4. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  5. Now press and then to see the new graph drawn.
  6. Next press to return to the algebra window.
  7. Press for Author.
  8. Enter the function
    r=1/(1=cos )
  9. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  10. Now press
    Derive can have troubles with parabolas/hyperbolas when their input value is pi.
    To compensate for this we will change the end values to -3 and 3.
  11. Change the -3.1416 to -3 and then press
  12. Then change the 3.1416 to 3 and then press
    The graph should now draw fine.
  13. Next press to return to the algebra window.
  14. Press for Author.
  15. Enter the function
    r= 1/(1-1.2 cos
  16. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  17. Now press
    Derive can have troubles with parabolas/hyperbolas when their input value is pi.
    To compensate for this we will change the end values to -2.5 and 2.5.
  18. Change the -3 to -2.5 and then press
  19. Then change the 3 to 2.5 and then press
    The graph should now draw fine.
    We now have the following graphed.

As a hint, the last was the hyperbola
To find out more about why we saw the shapes we did, see Eccentricity and Polar Conics

This document was originally written by Richard Rupp.