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**Unformatted text preview: **ola at (0, 4). Since one focus is at (0, 0), which is 4 units away from the center, we
know the other focus is at (0, 8). According to Theorem 11.12, the conjugate axis has a length
√
(2)(6)
of √2ed 1 = √22 −1 = 4 3. Putting this together with the location of the vertices, we get that
e2 −
2
the asymptotes of the hyperbola have slopes ± 2√3 = ±
√ is (0, 4), the asymptotes are y = ± 3
3x √ 3
3. Since the center of the hyperbola + 4. We graph the hyperbola below.
y
8
7
6
5
4
y=3
2
1 −5 −4 −3 −2 −1 r= 1 2 6
1+2 sin(θ) 3 4 5 x 838 Applications of Trigonometry In light of Section 11.6.1, the reader may wonder what the rotated form of the conic sections would
look like in polar form. We know from Exercise 10a in Section 11.5 that replacing θ with (θ − φ) in
an expression r = f (θ) rotates the graph of r = f (θ) counter-clockwise by an angle φ. For instance,
4
to graph r = 1−sin 4θ− π all we need to do is rotate the graph of r = 1−sin(θ) , which we obtained in
( 4)
Example 11.6.4 number 1, counter-clockwise by π radians, as shown below.
4 3
2
1
−4 −3...

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