# Calculus 2 Lecture Notes, Section 9.7 - DOC

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```					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
 2e2 dx  y 2  e 2 d 2

For e = 1 (parabola):
1  e2  x 2  2e2 dx  y 2  e2d 2
2dx  y 2  d 2
y2 d
x        
2d 2

For 0 < e < 1 (ellipse):
 1 – e2 > 0…
1  e  x2       2
 2e 2 dx  y 2  e 2 d 2
               2e 2 d     
1  e   x
2            2

1  e2
x   y 2  e2d 2
                          
     2e 2 d      e2 d   e4 d 2
2

1  e   x  1  e2 x   1  e2    1  e2  y 2  e2d 2
2

2

                         
e 2 d 2 1  e 2  e 4 d 2
2
 2 e2 d 
1  e   x  1  e2   y  1  e2  1  e2
2                    2

            
2
 2 e2 d            e2 d 2
1  e   x  1  e2   y  1  e2
2                    2

            
2
 2 e2 d 
x           
     1  e2        y2
2
           2
1
 ed            ed 
     2               
 1 e          1 e 
2
Calc 2 Lecture Notes                                   Section 9.7                          Page 3 of 11

For e > 1 (hyperbola):
 1 – e2 < 0…
1  e  x
2       2
 2e 2 dx  y 2  e 2 d 2
 2 2e 2 d 
  e  1  x  2
2
x   y 2  e2d 2
    e 1 
    2e 2 d    e2 d   e4 d 2
2

  e  1  x  2
2         2
x 2   2         y 2  e2d 2
    e 1      e 1   e 1
                       
e 2 d 2  e 2  1 e 4 d 2
2
 2 e2 d 
  e  1  x  2   y 
2                 2
 2
    e 1           e2  1        e 1
2
       e2 d        e2 d 2
  e  1  x 2  2   y 2   2
2

      e 1         e 1
2
 2 e2 d 
x  2 
     e 1       y2
2
          2
1
 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
ed
r
1  e cos  
Calc 2 Lecture Notes                    Section 9.7                           Page 4 of 11

ed
Polar form of the conic sections in this orientation: r 
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 .
1 2 d
The rectangular coordinate equation for this orientation is x        y 
2d     2
r = 1.00*1/(1.00cos(t)+1); 0.000000 <= t <= 6.283190
y


x
                              


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.
 x  c
2
y2
The rectangular coordinate equation for this orientation is                 1 , where c2  a2  b2 ,

a2     b2
c     b2                                        b2 a 2  c2
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 <= t <= 6.283190
y


x
                              



     ed
           a
   1  e2
     ed
            b

    1  e2
 ed  a 1  e 2 


 ed   b 1  e 
2   2        2

 a 2 1  e 2   b 2 1  e 2 
2

b2
1  e2 
a2
b2
e  1 2
2

a
 b2                b2  
 1  2  d 2  b 2 1  1  2  
 a                 a 
2
 2 e2 d                                           b4     a2
x                                            d2     2
     1  e2        y2                             a 2 a  b2
             1        
 ed 
2
 ed 
2
b4     b4
     2 
d2  2       2
                                a  b2 c
 1 e          1 e 
2
Calc 2 Lecture Notes                  Section 9.7                                Page 7 of 11

For a hyperbola, the eccentricity has value e > 1.
 x  c
2
y2
The rectangular coordinate equation for this orientation is               1, where c2  a2  b2 ,

a2      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
      r                 , if the directrix is the line x = d > 0.
e cos    1

ed
       r                  , if the directrix is the line x = d < 0.
e cos    1

ed
       r                  , if the directrix is the line y = d > 0.
e sin    1
Calc 2 Lecture Notes                        Section 9.7                 Page 9 of 11

ed
       r                  , 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      4 sin    1      2sin    / 4   2
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              Rectangular Representation:
2

 x  h   y  k             R2
2             2

d
Polar Representation: r                                     Polar Representation: r  R
sin    1

xt                                                               x  R cos  t   h
Parametric Representation:
Parametric Representation:
y  a t  h  k
2                                                 y  R sin  t   k
for 0  t  2

1 t2
xR     h
Or     1 t2    for   t  
2t
yR       k
1 t2

Ellipse                                                      Hyperbola

Rectangular Representation:                                  Rectangular Representation:
 x  h         y k                                       x  h          y k
2              2                                             2               2

              1                                                           1
a2              b2                                             a2            b2

ed                                                                ed
Polar Representation: r                                     Polar Representation: r 
e cos    1                                                    e cos    1
for 0 < e < 1                                                for e > 1
Calc 2 Lecture Notes                     Section 9.7                             Page 11 of 11

x  a cos  t   h                                      x  a cosh  t   h
Parametric Representation:                         Parametric Representation:
y  b sin  t   k                                      y  b sinh  t   k
for 0  t  2                                     for   t   (right branch only)

1 t2                                              1 t2
xa        h                                        xa      h
Or      1 t2    for   t                      Or      1 t2    for   t  
2t                                                 2t
y b       k                                      y b       k
1 t2                                              1 t2

x  a sec  t   h
Or                         for 0  t  2
y  b tan  t   k

Show that rotating the graph of the unit hyperbola by 45 results in the graph of the reciprocal
0.5
function y      .
x

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