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2 b 2 a = 18 → b 2 = 9a 2c = 12 → c = 6 Since c 2 = a 2 + b 2 , a 2 + b 2 = c 2 a 2 + 9a = 6 2 a 2 + 9a − 36 = 0 (a + 3)(a − 12) = 0 a = −3 ,12 Since a is a distance, then a = 12. The length of the transverse axis is 2a = 24 . ▣ 6. Find the distance between the foci of the curve 9x 2 + 4y 2 − 36x − 8y + 4 = 0. [SOLUTION] From completing the squares, 9x 2 − 36x + 4y 2 − 8y = −4 9(x 2 − 4x) + 4(y 2 − 2y) = −4 9(x 2 − 4x + 4) + 4(y 2 − 2y + 1) = 36 9(x − 2) 2 + 4(y − 1) 2 = 36 (x − 2) 2 4 + (y − 1) 2 9 = 1 Since it is an ellipse, then a 2 = 9 → a = 3, b 2 = 4 → b = 2. From ellipse, c = √a 2 − b 2 c = √9 − 5 = √5 The distance between the foci is 2c = 2√5 . ▣ 7. A parabolic arc spans a width of 40 ft with a 20 ft wide road passing under the bridge. The minimum vertical clearance over the roadway must be 10 ft. What is the height of the smallest such arch that can be used? [SOLUTION] Using the squared property of a parabola, x1 2 y1 = x2 2 y2 ( 40 2 ) 2 h = ( 20 2 ) 2 h − 10 h = 13.333 ft

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