POLAR EQUATIONS OF CONICS
Conic sections can be described in a polar grid based on the following definition.
Definition
Let D denote a fixed line called the directrix. Let F denote a fixed point
called the focus. Let e denote a fixed positive number called the eccentricity. A conic is
then the set of points P   r,  in the plane such that
d ( P, F )
 e.
d ( P, D )
This results in 3 possible cases:

If e  1 , then the conic is a parabola.

If e  1 , then the conic is an ellipse.

If e  1 , then the conic is a hyperbola.
Note that in the case of an ellipse or a hyperbola, the major axis or transverse axis,
respectively, is a line passing through the focus perpendicular to the directrix. In both
cases the eccentricity F satisfies the equation
e
c
,
a
where c  d  C, F1   d C, F2  and a  d  C,V1   d  C,V2  .
If the focus F is located at the pole (remember, we are now working in the polar grid) and
we place the directrix D a distance of p units away from the pole, we then have the
following 4 cases:
Polar Equation of Conic
Description of Conic
r
ep
1  e cos 
D is located p units to the left of the pole.
r
ep
1  e cos 
D is located p units to the right of the pole.
r
ep
1  e sin 
D is located p units above of the pole.
r
ep
1  e sin 
D is located p units below of the pole.
Example
Identify the conic described by the polar equation r  2r cos  3  0 and find
the rectangular equation of this conic.
Answer
Here we have
r  2r cos  3  0
 r (1  2 cos  )  3  0
 r (1  2 cos  )  3
3
r
1  2 cos 
3
2 
2
r
1  2 cos 
Thus we infer that polar equation describes a hyperbola since e  2  1 . Moreover, we have
3
3
p  so the directrix is located
units to the left of the pole.
2
2
From this we conclude that its transverse axis is along the polar axis and one of its foci is at the
pole. To find its two vertices, set   0 and    in the equation of the hyperbola. [Do you
see why?] This yields the two points (3, 0) , or (3,  ) , and (1,  ) . [check this!] The center of
the hyperbola must then be located at the midpoint (2,  ) .
Now, to find the rectangular equation of this hyperbola, proceed as follows:
r  2r cos  3  0
 r  3  2r cos 
 r 2   3  2r cos  
 x2  y 2  3  2x 
2
2
 x 2  y 2  9  12 x  4 x 2
 3 x 2  12 x  y 2  9
 3( x 2  4 x)  y 2  9
 3( x  2)2  12  y 2  9
 3( x  2) 2  12  y 2  9
 3( x  2) 2  y 2  3
 ( x  2) 2 
( x  2) 2


12
y2
1
3
y2
 3
2
1
From this equation, we conclude that the hyperbola has its center at the point C  (2, 0) and
its transverse axis is along the x-axis. Moreover, we have a  1 , b  3 , and c  1  3  2 .
This implies that the hyperbola’s vertices are located at the points V1  (3, 0) and V1  (1,0) ,
and its foci are located at the points F1  (4, 0) and F2  (0, 0) .
[Compare these points in Cartesian coordinates with those found previously in polar
coordinates… They should be consistent!]