To determine The vertices and foci of the hyperbola x 2 36 + y 2 13 = 1 . Explanation Definition used: Equation of a Hyperbola: Center at ( 0 , 0 ) ; Transverse axis along the x-axis: An equation of the hyperbola with the center at ( 0 , 0 ) , foci at ( − c , 0 ) and ( c , 0 ) , and vertices at ( − a , 0 ) and ( a , 0 ) is x 2 a 2 − y 2 b 2 = 1 where b 2 = c 2 − a 2 . Calculation: The given equation of the hyperbola is x 2 36 + y 2 13 = 1 . Compare the given equation x 2 36 + y 2 13 = 1 with the general equation of hyperbola x 2 a 2 − y 2 b 2 = 1 , the center of the hyperbola is ( 0 , 0 ) and the value of a 2 = 36 and b 2 = 13 . Obtain the value of a and b from a 2 = 36 and b 2 = 13

Basic Technical Mathematics with Calculus (11th Edition)

11th Edition

ISBN: 9780134437736

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Maximization

Have you ever seen a bridge in the shape of a parabola? Where is its maximum on the bridge? The maximum of that bridge is the highest point on that bridge. The process of finding the maximum of the bridge here is called the maximization.

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Chapter 21.6, Problem 1PE

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The vertices and foci of the hyperbola x236+y213=1.

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Chapter 21.5, Problem 56E

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Chapter 21 Solutions

Basic Technical Mathematics with Calculus (11th Edition)

Ch. 21.1 - Find the distance between (−2, 6) and (5, −3). Ch. 21.1 - Prob. 2PE Ch. 21.1 - Prob. 1E Ch. 21.1 - Prob. 2E Ch. 21.1 - Prob. 5E Ch. 21.1 - In Exercises 5–14, find the distance between the...Ch. 21.1 - Prob. 7E Ch. 21.1 - Prob. 8E Ch. 21.1 - Prob. 9E Ch. 21.1 - Prob. 10E

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