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

Section 10.5 Conic Sections 679
(_5, 0)
3
y=_ x
4
y
3
y= x
4
(_4, 0) (4, 0)
0 (5, 0) x
SOLUtion If we divide both sides of the equation by 144, it becomes
x 2
16 2 y 2
9 − 1
which is of the form given in (7) with a − 4 and b − 3. Since c 2 − 16 1 9 − 25,
the foci are s65, 0d. The asymptotes are the lines y − 3 4 x and y − 2 3 4 x. The graph is
shown in Figure 14.
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FIGURE 14
9x 2 2 16y 2 − 144
Example 5 Find the foci and equation of the hyperbola with vertices s0, 61d and
asymptote y − 2x.
SOLUtion From (8) and the given information, we see that a − 1 and ayb − 2. Thus
b − ay2 − 1 2 and c 2 − a 2 1 b 2 − 5 4 . The foci are s0, 6s5 y2d and the equation of the
hyperbola is
y 2 2 4x 2 − 1
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Shifted Conics
As discussed in Appendix C, we shift conics by taking the standard equations (1), (2),
(4), (5), (7), and (8) and replacing x and y by x 2 h and y 2 k.
Example 6 Find an equation of the ellipse with foci s2, 22d, s4, 22d and vertices
s1, 22d, s5, 22d.
SOLUtion The major axis is the line segment that joins the vertices s1, 22d, s5, 22d
and has length 4, so a − 2. The distance between the foci is 2, so c − 1. Thus
b 2 − a 2 2 c 2 − 3. Since the center of the ellipse is s3, 22d, we replace x and y in (4)
by x 2 3 and y 1 2 to obtain
as the equation of the ellipse.
sx 2 3d 2
4
1
sy 1 2d2
3
− 1
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y
3
y-1=_ 2(x-4)
(4, 4)
Example 7 Sketch the conic 9x 2 2 4y 2 2 72x 1 8y 1 176 − 0 and find its foci.
SOLUtion We complete the squares as follows:
4sy 2 2 2yd 2 9sx 2 2 8xd − 176
4sy 2 2 2y 1 1d 2 9sx 2 2 8x 1 16d − 176 1 4 2 144
(4, 1)
0 x
(4, _2)
3
y-1= 2(x-4)
FIGURE 15
9x 2 2 4y 2 2 72x 1 8y 1 176 − 0
4sy 2 1d 2 2 9sx 2 4d 2 − 36
sy 2 1d 2
2
9
sx 2 4d2
4
This is in the form (8) except that x and y are replaced by x 2 4 and y 2 1. Thus
a 2 − 9, b 2 − 4, and c 2 − 13. The hyperbola is shifted four units to the right and one
unit upward. The foci are (4, 1 1 s13 ) and (4, 1 2 s13 ) and the vertices are s4, 4d
and s4, 22d. The asymptotes are y 2 1 − 6 3 2 sx 2 4d. The hyperbola is sketched in
Figure 15.
− 1
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