Explanation Given: x 2 − 4 y 2 − 4 x = 0 Concept used: For the Hyperbola ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 . Center = ( h , k ) , Vertices = ( h − a , k ) and ( h + a , k ) , Transverse axis : y = k and the asymptotes are y − k = ± b a ( x − h ) . Rewriting the given equation in standard form we get ( x − 2 ) 2 4 − y 2 1 = 1

Precalculus: A Unit Circle Approach, Books a la Carte Edition plus MyMathLab with Pearson eText -- Access Card Package (2nd Edition)

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ISBN: 9780321869401

Author: J. S. Ratti

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Chapter 8.4, Problem 52E

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For the given equation find the center, vertices, transverse axis and asymptotes and graph the hyperbola.

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Chapter 8.4, Problem 51E

Chapter 8.4, Problem 53E

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Chapter 8 Solutions

Precalculus: A Unit Circle Approach, Books a la Carte Edition plus MyMathLab with Pearson eText -- Access Card Package (2nd Edition)

Ch. 8.2 - Prob. 1E Ch. 8.2 - Prob. 2E Ch. 8.2 - Prob. 3E Ch. 8.2 - Prob. 4E Ch. 8.2 - Prob. 5E Ch. 8.2 - Prob. 6E Ch. 8.2 - Prob. 7E Ch. 8.2 - Prob. 8E Ch. 8.2 - Prob. 9E Ch. 8.2 - Prob. 10E

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For the given equation find the center, vertices, transverse axis and asymptotes and graph the hyperbola.

Solution Summary: The author explains how the hyperbola is graphed using the given equation, which includes the center, vertices, transverse axis and asymptotes.

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For the given equation find the center, vertices, transverse axis and asymptotes and graph the hyperbola.

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Chapter 8 Solutions