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

Section 10.6 Conic Sections in Polar Coordinates 683
y
polar axis. If the point P has polar coordinates sr, d, we see from Figure 1 that
P
l (directrix)
| PF | − r | Pl | − d 2 r cos
F
r
¨
r cos ¨
x=d
x
Thus the condition | PF | y | Pl | − e, or | PF | − e | Pl | , becomes
2
r − esd 2 r cos d
C
d
If we square both sides of this polar equation and convert to rectangular coordinates,
we get
x 2 1 y 2 − e 2 sd 2 xd 2 − e 2 sd 2 2 2dx 1 x 2 d
FIGURE 1
or s1 2 e 2 dx 2 1 2de 2 x 1 y 2 − e 2 d 2
After completing the square, we have
2D
e 2 2
d
Sx 1 1 y 2
1 2 e 1 2 e − e 2 d 2
3
2 s1 2 e 2 d 2
If e , 1, we recognize Equation 3 as the equation of an ellipse. In fact, it is of the form
where
4
sx 2 hd 2
1 y 2
a 2 b − 1 2
h − 2
e 2 d
1 2 e a 2 − e 2 d 2
2 s1 2 e 2 d b 2 − e 2 d 2
2 1 2 e 2
In Section 10.5 we found that the foci of an ellipse are at a distance c from the center,
where
5
c 2 − a 2 2 b 2 − e 4 d 2
s1 2 e 2 d 2
This shows that c − e 2 d
1 2 e 2 − 2h
and confirms that the focus as defined in Theorem 1 means the same as the focus defined
in Section 10.5. It also follows from Equations 4 and 5 that the eccentricity is given by
e − c a
If e . 1, then 1 2 e 2 , 0 and we see that Equation 3 represents a hyperbola. Just as we
did before, we could rewrite Equation 3 in the form
and see that
sx 2 hd 2
2 y 2
a 2 b − 1 2
e − c a
where c 2 − a 2 1 b 2 n
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