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Unformatted text preview: dinates) (0, −2) and must open upwards. With d = 4, we have a focal
diameter of 2d = 8, so the parabola contains the points (±4, 0). We graph r = 1−sin(θ) below.
2. We ﬁrst rewrite r = 3−cos(θ) in the form found in Theorem 11.12, namely r = 1−(1/3) cos(θ) .
Since e = 3 satisﬁes 0 < e < 1, we know that the graph of this equation is an ellipse. Since
ed = 4, have d = 12 and, based on the form of the equation, we know the directrix is x = −12.
This means the ellipse has a major axis along the x-axis. We can ﬁnd the vertices of the ellipse
by ﬁnding the points of the ellipse which lie on the x-axis. We ﬁnd r(0) = 6 and r(π ) = 3
which correspond to the rectangular points (−3, 0) and (6, 0), so these are our vertices. The
center of the ellipse is the midpoint of the vertices, which in this case is 2 , 0 .8 We know
one focus is (0, 0), which is 2 from the center 2 , 0 and this allows us to ﬁnd the other focus
8 As a quick check, we have from Theorem...
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