By theorem 1195 w f p q cos 10 pounds50 feet cos

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Unformatted text preview: dinates) (0, −2) and must open upwards. With d = 4, we have a focal 4 diameter of 2d = 8, so the parabola contains the points (±4, 0). We graph r = 1−sin(θ) below. 12 4 2. We first rewrite r = 3−cos(θ) in the form found in Theorem 11.12, namely r = 1−(1/3) cos(θ) . 1 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 3 center of the ellipse is the midpoint of the vertices, which in this case is 2 , 0 .8 We know 3 3 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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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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