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

Section 10.6 Conic Sections in Polar Coordinates 685
x=_5
(directrix)
FIGURE 3
(2, π)
y
r= 10
3-2 cos ¨
focus
0 (10, 0) x
From Theorem 6 we see that this represents an ellipse with e − 2 10
3 . Since ed − 3 ,
we have
10 10
3
d −
e − 3
2 − 5
so the directrix has Cartesian equation x − 25. When − 0, r − 10; when − ,
r − 2. So the vertices have polar coordinates s10, 0d and s2, d. The ellipse is sketched
in Figure 3.
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Example 3 Sketch the conic r −
3
12
2 1 4 sin .
SOLUtion Writing the equation in the form
r −
6
1 1 2 sin
we see that the eccentricity is e − 2 and the equation therefore represents a hyperbola.
Since ed − 6, d − 3 and the directrix has equation y − 3. The vertices occur when
− y2 and 3y2, so they are s2, y2d and s26, 3y2d − s6, y2d. It is also useful
to plot the x-intercepts. These occur when − 0, ; in both cases r − 6. For additional
accuracy we could draw the asymptotes. Note that r l 6` when 1 1 2 sin l 0 1 or
0 2 and 1 1 2 sin − 0 when sin − 2 1 2 . Thus the asymptotes are parallel to the rays
− 7y6 and − 11y6. The hyperbola is sketched in Figure 4.
y
π
”6, 2’
π
”2, 2 ’
y=3 (directrix)
FIGURE 4
12
r −
2 1 4 sin
(6, π) 0 (6, 0)
focus
x
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When rotating conic sections, we find it much more convenient to use polar equations
than Cartesian equations. We just use the fact (see Exercise 10.3.73) that the graph of
r − f s 2 d is the graph of r − f sd rotated counterclockwise about the origin through
an angle .
11
10
r= 3-2 cos(¨-π/4)
Example 4 If the ellipse of Example 2 is rotated through an angle y4 about the
origin, find a polar equation and graph the resulting ellipse.
SOLUtion We get the equation of the rotated ellipse by replacing with 2 y4 in
the equation given in Example 2. So the new equation is
_5 15
_6
FIGURE 5
r= 10
3-2 cos ¨
r −
10
3 2 2 coss 2 y4d
We use this equation to graph the rotated ellipse in Figure 5. Notice that the ellipse has
been rotated about its left focus.
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