Problem:
Review the concept of eccentricity through examining the effects of changing the value of the variable e in the following equation.

Visualization:
We first recall that eccentricity is the fixed ratio of the distance from any point on the conic to the focus versus the distance from that point to the directrix.
Simply speaking, the eccentricity of a conic is the measure of its curvature.
We can also use eccentricity to determine if a conic will be a circle, ellipse, parabola, or hyperbola.
The general rule is that if e < 1, the conic is an ellipse
if e = 1, the conic is a parabola.
if e > 1, the conic is a hyperbola.
As an addition, we note that if e = 0, the conic is a circle.

We will now verify this through experimentation.

Using Derive:

  1. Prepare the program for graphics.
  2. If the first menu item is Algebra then press .

    Let us start at 0 and slowly increase e to 1.

  3. Press for Author.
  4. Enter the function
    r=1/(1 + 0 cos )
  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 changed back to rectangular.


  9. Now press and then to see the graph.
We see this is a circle
This is easily expected as our equation simplies to r=1, the equation of a polar circle with radius 1.

We now allow e to take increasing values from the open interval (0,1)

  1. Next press to return to the algebra window.
  2. Press for Author.
  3. Enter the function
    r=1/(1+ .01 cos )
  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+ .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 and then to see the new graph drawn.
  11. Next press to return to the algebra window.
  12. Press for Author.
  13. Enter the function
    r=1/(1+ .2 cos )
  14. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  15. Now press and then to see the new graph drawn.
  16. Next press to return to the algebra window.
  17. Press for Author.
  18. Enter the function
    r=1/(1+ .4 cos )
  19. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  20. Now press and then to see the new graph drawn.
  21. Next press to return to the algebra window.
  22. Press for Author.
  23. Enter the function
    r=1/(1+ .6 cos )
  24. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  25. Now press and then to see the new graph drawn.
  26. Next press to return to the algebra window.
  27. Press for Author.
  28. Enter the function
    r=1/(1+ .8 cos )
  29. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  30. Now press and then to see the new graph drawn.
  31. Next press to return to the algebra window.
  32. Press for Author.
  33. Enter the function
    r=1/(1+ .9 cos )
  34. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  35. Now press and then to see the new graph drawn.
  36. Next press to return to the algebra window.
  37. Press for Author.
  38. Enter the function
    r=1/(1+ .95 cos )
  39. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  40. Now press and then to see the new graph drawn.
    We now have the following graph
What we see is a pattern of ellipses becoming more enlongated or oblong until they approach a parabolic shape.
This confirms our hypothesis that when 0 < e < 1, the conic will be an ellipse.

Now we see what happens when e is exactly one.

  1. First, we want to eliminate most of the graphs.
  2. to do this we press to enter the delete menu.
  3. Then we press to select the "allbut" option.
  4. Now, we press and derive will erase all the graphs and then redraw the graph when e = .95
  5. Next press to return to the algebra window.
  6. Press for Author.
  7. Enter the function
    r=1/(1+ .95 cos )
  8. 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.
  9. Change the -3.1416 to -3 and then press
  10. Then change the 3.1416 to 3 and then press
    The graph should now draw fine and appear as follows.
This is a parabola with a curvature very similar to the ellipse with e = .95
This is exactly what we expected and confirms the hypothesis

Now we will let e take increasing values larger than 1.

  1. Next press to return to the algebra window.
  2. Press for Author.
  3. Enter the function
    r=1/(1+ 1.2 cos )
  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
  6. We will eventually need to decrease our endvalues, so let us do it now.
    This will help to save plotting time.
  7. Change the -3 to -2 and then press
  8. Then change the 3 to 2 and then press
  9. Now press to see the graph.
  10. Next press to return to the algebra window.
  11. Press for Author.
  12. Enter the function
    r=1/(1+ 1.5 cos )
  13. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  14. Now press and then to see the new graph drawn.
  15. Next press to return to the algebra window.
  16. Press for Author.
  17. Enter the function
    r=1/(1+ 2 cos )
  18. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  19. Now press and then to see the new graph drawn.
  20. Next press to return to the algebra window.
  21. Press for Author.
  22. Enter the function
    r=1/(1+ 2.4 cos )
  23. Press to get the plot window.
    Derive will automatically redraw any old graphs; so don't be surprised when the old graph(s) appear.
  24. Now press and then to see the new graph drawn.
    We now have the following graphed.
Although we don't see the other side, these shapes are actually hyperbolas.
Thus, we again confirm our hypothesis about the effect of our e value.

This document was originally written by Richard Rupp.