James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

684 Chapter 10 Parametric Equations and Polar Coordinates
By solving Equation 2 for r, we see that the polar equation of the conic shown in Figure
1 can be written as
ed
r −
1 1 e cos
If the directrix is chosen to be to the left of the focus as x − 2d, or if the directrix is
cho sen to be parallel to the polar axis as y − 6d, then the polar equation of the conic is
given by the following theorem, which is illustrated by Figure 2. (See Exercises 21–23.)
y
x=d
directrix
x=_d
directrix
y
y=d
y
directrix
y
F
x
F
x
F
x
F
x
y=_d
directrix
(a) r=
ed
1+e cos ¨
(b) r=
ed
1-e cos ¨
(c) r=
ed
1+e sin ¨
(d) r=
ed
1-e sin ¨
FIGURE 2
Polar equations of conics
6 Theorem A polar equation of the form
r −
ed
1 6 e cos
or
r −
ed
1 6 e sin
represents a conic section with eccentricity e. The conic is an ellipse if e , 1,
a parabola if e − 1, or a hyperbola if e . 1.
Example 1 Find a polar equation for a parabola that has its focus at the origin and
whose directrix is the line y − 26.
SOLUTION Using Theorem 6 with e − 1 and d − 6, and using part (d) of Figure 2, we
see that the equation of the parabola is
r −
6
1 2 sin
n
Example 2 A conic is given by the polar equation
r −
10
3 2 2 cos
Find the eccentricity, identify the conic, locate the directrix, and sketch the conic.
SOLUTION Dividing numerator and denominator by 3, we write the equation as
r −
10
3
1 2 2 3 cos
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