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**Unformatted text preview: **11.12 the major axis should have length 2ed
1−e2 = (2)(4)
1−(1/3)2 = 9. 11.6 Hooked on Conics Again 837 (3, 0), even though we are not asked to do so. Finally, we know from Theorem 11.12 that the
√
4
length of the minor axis is √2ede2 = √1−(1/3)2 = 6 3 which means the endpoints of the minor
1−
√
12
axis are 3 , ±3 2 . We now have everything we need to graph r = 3−cos(θ) .
2
y 4
3
3 2 2 1 1
−4 −3 −2 −1
−1 1 2 3 −3 −2 −1
−1 4 1 2 3 4 5 6 x −2 −2 −3 −3 x = −12 −4 y = −4 r= 4
1−sin(θ) r= 12
3−cos(θ) 6
3. From r = 1+2 sin(θ) we get e = 2 > 1 so the graph is a hyperbola. Since ed = 6, we get
d = 3, and from the form of the equation, we know the directrix is y = 3. This means the
transverse axis of the hyperbola lies along the y -axis, so we can ﬁnd the vertices by looking
π
where the hyperbola intersects the y -axis. We ﬁnd r π = 2 and r 32 = −6. These two
2
points correspond to the rectangular points (0, 2) and (0, 6) which puts the center of the
hyperb...

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