Explanation Given: ( x + 2 ) 2 9 − ( y − 3 ) 2 4 = 1 Concept and Formula used: There are two standard forms of ellipse given by ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 with focus ( h ± c , k ) , where c = a 2 + b 2 , asymptotic lines are y − k = ± b a ( x − h ) and vertices are ( h ± a , k ) . ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 with focus ( h , k ± c ) , where c = a 2 − b 2 , asymptotic lines are y − k = ± a b ( x − h ) and vertices are ( h , k ± a ) . Calculation: As given ( x + 2 ) 2 9 − ( y − 3 ) 2 4 = 1 we have ( x − { − 2 } ) 2 9 −

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

Publisher: PEARSON

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Chapter 8, Problem 29RE

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The graph, vertices, foci and asymptote of the given hyperbola is to be determined.

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Chapter 8, Problem 28RE

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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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The graph, vertices, foci and asymptote of the given hyperbola is to be determined.

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To find: The graph, vertices, foci and asymptote of the given hyperbola is to be determined.

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Chapter 8 Solutions

To find:

The graph, vertices, foci and asymptote of the given hyperbola is to be determined.