Explanation Definition used: “ Equation of a Hyperbola: Center at ( 0 , 0 ) ; Transverse axis along the y-axis An equation of the hyperbola with the center at ( 0 , 0 ) , foci at ( 0 , − c ) and ( 0 , c ) , and vertices at ( 0 , − a ) and ( 0 , a ) is y 2 a 2 − x 2 b 2 = 1 where b 2 = c 2 − a 2 . The hyperbola has a transverse axis of length 2 a along y axis and a conjugate axis at ( ± b , 0 ) . Calculation: The given equation of the hyperbola is y 2 16 − x 2 4 = 1 . Compare the given equation y 2 16 − x 2 4 = 1 with the general equation of hyperbola y 2 a 2 − x 2 b 2 = 1 , the center of the hyperbola is ( 0 , 0 ) and the value of a 2 = 16 and b 2 = 4 . Obtain the value of a and b from a 2 = 16 and b 2 = 4 . a 2 = 16 and b 2 = 4 a = 16 and b = 4 a = 4 and b = 2 Since the value of a is 5, the vertices are ( 0 , − 4 ) and ( 0 , 4 ) . Thus, the vertices of the hyperbola is ( 0 , − 4 ) and ( 0 , 4 ) _

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ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 21.6, Problem 1E

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The vertices, end of conjugate axis and foci of the hyperbola $y216−x24=1$ and sketch the curve.

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Chapter 21.6, Problem 3PE

Chapter 21.6, Problem 2E

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Chapter 21 Solutions

Basic Technical Mathematics

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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The vertices, end of conjugate axis and foci of the hyperbola y 2 16 − x 2 4 = 1 and sketch the curve.

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