Page 84 - Calculus for the Clueless, Calc II

Example 15—

Given vertices (2,3) and (12,3) and one focus (11,3), find the equation of the ellipse.

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Two vertices give the center; (12 + 2)/(2,3) = (7,3). F(11,3). (x - 7) /a + (y - 3) /b = 1. a = 12 - 7 = 5. c = 11 -
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7 = 4. a - b = c . 5 - b . b = 9 (no need for b). (x - 7) /25 + (y - 3) /9 = 1.
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Example 16—

Find the equation of the hyperbola with vertices (0,±6) and asymptotes y = ± (3/2)x.

V(0,±6) says the center is (0,0) and its shape is y /36 - x /a = 1. The slope of the asymptotes is 3/2 = square
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root of the number under y over the square root of the number under the x term. So 3/2 = 6/_. So _= 4. So a 2
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= 4 = 16. The equation is y /36 - x /16 = 1.
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This kind of question is shorter in length, but it does take practice. So practice!!