Explanation Given: Center at ( − 3 , 1 ) ; focus at ( − 3 , 6 ) ; vertex at ( − 3 , 4 ) . Formula used: Center Transverse axis Foci Vertices Equation Asymptotes ( h , k ) Parallel to y -axis ( h , k ± c ) ( h , k ± a ) ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 ; b 2 = c 2 − a 2 x − h = ± b a ( y − k ) Calculation: The center of the hyperbola is ( h , k ) = ( − 3 , 1 ) ; h = − 3 , k = 1 . Focus: ( − 3 , 6 ) = ( h , k ± c ) ; c = 5 . Vertex: ( − 3 , 4 ) = ( h , k ± a ) ; a = 3 ... To determine c. To graph: The equation ( y − 1 ) 2 9 − ( x + 3 ) 2 16 = 1 .

Precalculus Enhanced with Graphing Utilities (7th Edition)

7th Edition

ISBN: 9780134119281

Author: Michael Sullivan, Michael Sullivan III

Publisher: PEARSON

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Chapter 10.4, Problem 42SB

To determine

a & b

To find: An equation of the hyperbola described.

To determine

c.

To graph: The equation $\frac{{(y - 1)}^{2}}{9} - \frac{{(x + 3)}^{2}}{16} = 1$ .

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Chapter 10.4, Problem 41SB

Chapter 10.4, Problem 43SB

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Precalculus Enhanced with Graphing Utilities (7th Edition)

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a &amp; b To find: An equation of the hyperbola described.

Solution Summary: The author explains the hyperbola's center, focus, and vertex, which are parallel to y -axis.