**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
âˆš
(2)(6)
of âˆš2ed 1 = âˆš22 âˆ’1 = 4 3. Putting this together with the location of the vertices, we get that
e2 âˆ’
2
the asymptotes of the hyperbola have slopes Â± 2âˆš3 = Â±
âˆš is (0, 4), the asymptotes are y = Â± 3
3x âˆš 3
3. Since the center of the hyperbola + 4. We graph the hyperbola below.
y
8
7
6
5
4
y=3
2
1 âˆ’5 âˆ’4 âˆ’3 âˆ’2 âˆ’1 r= 1 2 6
1+2 sin(Î¸) 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 Î¸ with (Î¸ âˆ’ Ï†) in
an expression r = f (Î¸) rotates the graph of r = f (Î¸) counter-clockwise by an angle Ï†. For instance,
4
to graph r = 1âˆ’sin 4Î¸âˆ’ Ï€ all we need to do is rotate the graph of r = 1âˆ’sin(Î¸) , which we obtained in
( 4)
Example 11.6.4 number 1, counter-clockwise by Ï€ radians, as shown below.
4 3
2
1
âˆ’4 âˆ’3 âˆ...

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