A solution is a = 1, b = −2 we may take the equation asr = 1− 2 cos θ. The cartesian point(−3, 0) is either the polar point (3, π) or (−3, 0), and the cartesian point (−1, 0) is eitherthe polar point (1, π) or (−1, 0). The curve is not symmetric about the y-axis,so Theorem 11.1.1(a) eliminates the sine function, thus r = a ± b cos θ. ![]() (a) From (8-9), r = a ± b sin θ or r = a ± b cos θ. ![]() Janu11:47 L24-ch11 Sheet number 2 Page number 476 blackġ8. (a) r2 = x2 + y2 = 4 circle (b) y = 4 horizontal line Janu11:47 L24-ch11 Sheet number 1 Page number 475 blackĥ. ![]() Janu11:47 L24-ch11 Sheet number 1 Page number 475 black 475 CHAPTER 11 Analytic Geometry in Calculus EXERCISE SET 11.1 1.
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