1. Conic Sections in Polar Coordinates

2. Focus-Directrix Definitions of the Conic Sections
Let F be a fixed point, the focus, and let D be fixed line, the directrix, in a plane. A conic section, or conic, is the set of all points P in the plane such that
where e is a fixed positive number, called the eccentricity.
If e = 1, the conic is a parabola.
If e < 1, the conic is an ellipse.
If e > 1, the conic is a hyperbola. PF PD
= e

3. Focus-Directrix Definitions of the Conic SectionsP F
˝/2
Directrix
Focus at pole
e = 1
Parabola O D P F
˝/2
Directrix
Focus at pole
e < 1
Ellipse O D P
Directrix
Focus at pole F
˝/2
e > 1
Hyperbola D´ P´

4. Standard Forms of the Polar Equations of Conics
Let the pole be a focus of a conic section of eccentricity e with the directrix |p| units form the focus. The equation of these 4 conics are listed below. 1.
Directrix: x = p
Polar axis
or x-axis
 = /2
or y-axis
Focus at pole
Directrix: x = - p 2.

5. Standard Forms of the Polar Equations of Conics
Let the pole be a focus of a conic section of eccentricity e with the directrix |p| units form the focus. The equation of these 4 conics are listed below. 3.
Directrix: y = p
Polar axis
or x-axis
 = /2
or y-axis
Focus at pole
Directrix: y = - p 4.

6. Graphing the Polar Equation of a Conic
If necessary, write the equation in one of the standard forms.
Use the standard form to determine values for e and p. Use the value of e to identify the conic.
Use the appropriate figure for the standard form of the equation shown on the previous slide to help guide the graphing process.

7. Text Example Graph the polar equation: Solution
Step 1 Write the equation in one of the standard forms. The equation is not in standard form because the constant term in the denominator is not 1.
Solution
To obtain 1 in this position, divide the numerator and denominator by 2.
The equation in standard form is
ep = 4
e = 1/2

8. Text Example cont. Solution
Step 2 Use the standard form to find e and p, and identify the conic. The voice balloons show that
e = 1/2 and ep = 1/2 p = 2.
Thus, e = 1/2 and p = 4. Because e = 1/2 < 1, the conic is an ellipse.
Solution
Step 3 Use the figure for the equation’s standard form to guide the graphing process. The figure for the conic’s standard form is shown below. We have symmetry with respect to the polar axis. One focus is at the pole and a directrix is x =4, located four units to the right of the pole.
Directrix: x = p
Polar axis
or x-axis
 = /2
or y-axis
Focus at pole

9. Text Example cont. Solution
The last figure indicated that the major axis is on the polar axis. Thus, we find the vertices by selecting 0 and  for . The corresponding values for r are 4/3 and 4, respectively. The figure on the right shows the vertices (4/3, 0) and (4, ).
Solution
You can sketch the upper half of the ellipse by plotting some points from  = 0 to  = . Using symmetry with respect to the polar axis, you can sketch the lower half. 3 4 5  6 4 2 2 3 7
11
Directrix
Focus
Vertex (4, )
Vertex
(4/3, 0)
3.5
3.1
2.7 2 r
5 /6
3/4
2 /3
/2 

10. Example Graph the polar equation:
Solution:
e=1 and ep = 3 so p=3 the shape is a parabola, the focus is at the pole and the directrix is y=3,
the vertex is at =/2.

11. Example cont.
Graph the polar equation:

12. Conic Sections in Polar Coordinates