Explanation Calculation: Consider the equation of the hyperbola, x 2 − 9 y 2 + 36 y − 72 = 0 Group the x and y terms separately, x 2 − 9 y 2 − 36 y = 72 x 2 − 9 y 2 − 4 y = 72 Complete the square, x 2 − 9 y 2 − 4 y + 4 + 9 4 = 72 x 2 − 9 y − 2 2 = 72 − 36 x 2 − 9 y − 2 2 = 36 Write the above equation in the standard form of the equation of hyperbola with horizontal transverse axis, that is, x − h 2 a 2 − y − k 2 b 2 = 1 . x 2 − 9 y − 2 2 = 36 x − 0 2 36 − y − 2 2 4 = 1 x − 0 2 6 2 − y − 2 2 2 2 = 1 Now, compare with the standard equation, h = 0 , k = 2 , a = 6 and b = 2 Since c 2 = a 2 + b 2 , c = a 2 + b 2 = 6 2 + 2 2 = 2 10

College Algebra Real Mathematics Real People Edition 7

18th Edition

ISBN: 9781305752368

Author: Ron Larson

Publisher: CENGAGE C

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Chapter 4.4, Problem 74E

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To calculate: The center, foci, and vertices of the hyperbola x2−9y2+36y−72=0, and also, plot the graph using the asymptotes as an aid.

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Chapter 4.4, Problem 73E

Chapter 4.4, Problem 75E

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College Algebra Real Mathematics Real People Edition 7

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The center, foci, and vertices of the hyperbola x 2 − 9 y 2 + 36 y − 72 = 0 , and also, plot the graph using the asymptotes as an aid.

Solution Summary: The author calculates the center, foci, and vertices of the hyperbola using the asymptotes as an aid.

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To calculate: The center, foci, and vertices of the hyperbola x2−9y2+36y−72=0, and also, plot the graph using the asymptotes as an aid.

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Chapter 4 Solutions