Calc 2 Lecture Notes
Section 9.7
Page 1 of 11
Section 9.7: Conic Sections in Polar Coordinates
Big idea: The ellipse, hyperbola, and parabola all have the same equation in polar coordinates
that is parameterized by a single constant called the eccentricity. This is significant for physics,
since one can calculate the eccentricity for any object moving under the influence of a “central
force,” which means that the trajectory of that object can be predicted easily.
Big skill:. You should be able to plot the conic sections given in polar form, and convert the
rectangular forms of their equations to polar and parametric forms.
Theorem 7.1: Eccentricity of the conic sections.
The set of all points whose distance to the focus is the product of the eccentricity e and the
distance to a directrix is:

An ellipse for 0 < e < 1.

A parabola if e = 1.

A hyperbola if e > 1.
A circle has eccentricity 0 (because a = b…).
Practice: Show that theorem 7.1 is true, and then convert the Cartesian equation to polar form.
Assume that the focus is at the origin, and the directrix is at x = d > 0 (Note: shifting the origin of
the coordinate system to the focus is a common practice for many central force problems,
because this is the location of the central force; i.e., the sun or something like that).
Calc 2 Lecture Notes
Section 9.7
Page 2 of 11
x2  y 2  e  d  x 
x 2  y 2  e2  d  x 
2
x 2  y 2  e2 d 2  2e2 dx  e 2 x 2
1  e  x
2
2
 2e 2 dx  y 2  e 2 d 2
For e = 1 (parabola):
1  e2  x2  2e2dx  y 2  e2d 2
2dx  y 2  d 2
x
y2 d

2d 2
For 0 < e < 1 (ellipse):
 1 – e2 > 0…
1  e  x
2

1  e   x
2

2
2

 2e2 dx  y 2  e 2 d 2
2e 2 d
1  e2

x   y 2  e2 d 2

2

 e2 d   e4 d 2
2e 2 d
2
1  e   x  1  e2 x   1  e2    1  e2  y 2  e2d 2

 

2
e2 d 2 1  e 2  e 4 d 2
 2 e2 d 
2
1  e   x  1  e2   y  1  e2  1  e2


2
2
2
 2 e2 d 
e2 d 2
2
1  e   x  1  e2   y  1  e2


2
2
 2 e2 d 
x 

1  e2 
y2


1
2
2
 ed 
 ed 

2 


2
 1 e 
 1 e 
Calc 2 Lecture Notes
For e > 1 (hyperbola):
 1 – e2 < 0…
Section 9.7
1  e  x
2
2
Page 3 of 11
 2e 2 dx  y 2  e 2 d 2
 2 2e 2 d  2
x   y  e2d 2
  e  1  x  2
e 1 

2

 e2 d   e4 d 2
2e 2 d
2
2
 y 2  e2d 2
x 2   2
  e  1  x  2


1

e
1

e
e 1
 


2
e 2 d 2  e 2  1 e 4 d 2
 2 e2 d 
2
 2
  e  1  x  2   y 
e 1
e2  1
e 1 

2
2
2

e2 d 2
e2 d 
  e  1  x 2  2   y 2   2
e 1
e 1 

2
2
 2 e2 d 
x  2 
e 1 
y2

1

2
2
 ed 
 ed 
 2 

 2
 e 1 
 e 1 
x2  y 2  e  d  x 
r 2  e  d  r cos   
r  ed  er cos  
r 1  e cos     ed
r
ed
1  e cos  
Calc 2 Lecture Notes
Section 9.7
Polar form of the conic sections in this orientation: r 
Page 4 of 11
ed
e cos    1
Practice:
Find the polar equations for the conic sections with focus at (0, 0), directrix x = 2, and
eccentricities of e = 0.4, e = 0.8, e = 1, e = 1.2, e = 2. Then graph the equations.
Calc 2 Lecture Notes
Section 9.7
Page 5 of 11
For a parabola, the eccentricity is e  1 .
The rectangular coordinate equation for this orientation is x  
r = 1.00*1/(1.00cos(t)+1); 0.000000
y
<= t <= 6.283190






x
1 2 d
y 
2d
2
Calc 2 Lecture Notes
Section 9.7
Page 6 of 11
For an ellipse, the range of the eccentricity is 0 < e < 1.
The closer e is to zero, the closer the ellipse is to a circle. The closer e is to 1, the more the
ellipse stretches out.
The rectangular coordinate equation for this orientation is
 x  c
a2
2

y2
 1 , where c 2  a 2  b 2 ,
2
b
c
b2
b2 a 2  c 2
which implies that e   1  2 and the directrix is at x  d 

a
a
c
c
r = 0.75*1/(0.75cos(t)+1); 0.000000
y
<= t <= 6.283190





x

 ed
 1  e 2  a
 ed

b
 1  e 2
 ed  a 1  e 2 


2
2
2
 ed   b 1  e 
 a 2 1  e 2   b 2 1  e 2 
2
b2
a2
b2
2
e  1 2
a
1  e2 
  b2  
 b2  2
2
 1  2  d  b 1  1  2  
 a 
  a 
2
 2 e2 d 
x 

1  e2 
y2


1
2
2
 ed 
 ed 

2 


2
 1 e 
 1 e 
b4
a2

a 2 a 2  b2
b4
b4
2
d  2 2  2
a b
c
d2 

Calc 2 Lecture Notes
Section 9.7
Page 7 of 11
For a hyperbola, the eccentricity has value e > 1.
The rectangular coordinate equation for this orientation is
 x  c
a2
2

y2
 1 , where c 2  a 2  b 2 ,
b2
c
b2
b2 a 2  c 2
which implies that e   1  2 , and that the directrix is at x  d 

a
a
c
c
Calc 2 Lecture Notes
Section 9.7
Page 8 of 11
Theorem 7.2: Polar equations for conic sections with different directrixes.
The conic section with eccentricity e > 0, focus (0, 0) and the indicated directrix has the polar
equation:
ed

, if the directrix is the line x = d > 0.
r
e cos    1

r
ed
, if the directrix is the line x = d < 0.
e cos    1

r
ed
, if the directrix is the line y = d > 0.
e sin    1
Calc 2 Lecture Notes

r
Section 9.7
ed
, if the directrix is the line y = d < 0.
e sin    1
Practice: Graph and interpret the following conic sections:
4
4
3
; r
; r
r
cos    4
4sin    1
2sin    / 4  2
Page 9 of 11
Calc 2 Lecture Notes
Section 9.7
Page 10 of 11
Comparison of representations for the conic sections:
Parabola
Circle
Rectangular Representation: y  a  x  h   k
2
Polar Representation: r 
d
Polar Representation: r  R
sin    1
xt
Parametric Representation:
Rectangular Representation:
2
2
 x  h   y  k   R2
y  a t  h  k
Parametric Representation:
x  R cos  t   h
2
y  R sin  t   k
for 0  t  2
1 t2
h
2
1

t
Or
for   t  
2t
yR
k
1 t2
xR
Ellipse
Hyperbola
Rectangular Representation:
Rectangular Representation:
 x  h
 x  h
a2
2
y k

b2
2
1
Polar Representation: r 
for 0 < e < 1
2
a2
ed
e cos    1
y k

b2
2
1
Polar Representation: r 
for e > 1
ed
e cos    1
Calc 2 Lecture Notes
Parametric Representation:
Section 9.7
x  a cos  t   h
y  b sin  t   k
Page 11 of 11
Parametric Representation:
x  a cosh  t   h
y  b sinh  t   k
for 0  t  2
for   t   (right branch only)
1 t2
h
2
1

t
Or
for   t  
2t
yb
k
1 t2
1 t2
h
2
1

t
Or
for   t  
2t
yb
k
1 t2
xa
xa
Or
x  a sec  t   h
y  b tan  t   k
for 0  t  2
Show that rotating the graph of the unit hyperbola by 45 results in the graph of the reciprocal
0.5
function y 
.
x