http://www.ck12.org Chapter 10. Polar and Parametric Equations
Solution: The circle in blue has a center at 90◦and has a diameter of 2. Its equation isr=2 cos(θ− 90 ◦).
The red ellipse appears to have center at (2, 0) witha=4 andc=2. This means the eccentricity ise=^12. In order
to write the equation in polar form you still need to findk.
k=ac^2 −c=^422 − 2 = 8 − 2 = 6
Thus the equation for the ellipse is:
r= 1 − 216 ··cos^12 (θ)
Example B
Identify the center, foci, vertices and equations of the directrix lines for the following conic:
r= 4 − 5 ·cos^20 (θ− 34 π)
Solution: First the polar equation needs to be in graphing form. This means that the denominator needs to look like
1 −e·cos(θ−β).
r= 4 − 5 ·cos^20 (θ− 3 π
4)·
(^14)
14 =
5
1 −^54 ·cos(θ−^34 π)=4 ·^54
1 −^54 ·cos(θ−^34 π)
e=^54 , k= 4 , β=^34 π= 135 ◦Using this information and the relationships you were reminded of in the guidance, you can set up a system and
solve foraandc.