Explanation Given: The hyperbola equation is as follows. 9 y 2 − 4 x 2 − 36 y + 8 x = 4 (1). Rewrite the equation as below. 9 y 2 − 36 y − 4 x 2 + 8 x = 4 9 y 2 − 36 y + 36 − 36 − 4 x 2 + 8 x + 4 − 4 = 4 9 y 2 − 36 y + 36 − 4 x 2 + 8 x + 4 = 4 + 36 − 4 9 ( y 2 − 4 y + 4 ) − 4 ( x 2 + 2 x + 1 ) = 36 Factorize the equation as below. 9 ( y − 2 ) 2 − 4 ( x + 1 ) 2 = 36 Divide the equation by the value 36 on both the sides. 9 ( y − 2 ) 2 36 − 4 ( x + 1 ) 2 36 = 36 ( y − 2 ) 2 4 − ( x + 1 ) 2 9 = 1 (2). Then, compare the equation (2) with the standard equation of hyperbola. y 2 a 2 − x 2 b 2 = 1 Calculation: Compute the center of the hyperbola using the equation: ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 ( y − ( − 2 ) ) 2 a 2 − ( x − ( 1 ) ) 2 b 2 = 1 Therefore, the center of the hyperbola ( h , k ) is ( − 1 , 2 ) . Substitute the value of 4 for a 2 and 9 for b 2 in equation (2). a 2 = 4 a = 4 a = 2 b 2 = 9 b = 9 b = 3 Compute the vertices. v e r t i c e s = ( h , ( k ± a ) ) Substitute the value − 1 for h , 2 for k and 2 for a . v e r t i c e s = ( − 1 , ( 2 ( ± 2 ) ) ) v e r t i c e s = ( − 1 , ( 2 ( − 2 ) ) ) , ( − 1 , ( 2 ( + 2 ) ) ) v e r t i c e s = ( − 1 , 0 ) , ( − 1 , 4 ) Then, the vertices of the hyperbola are ( − 1 , 0 ) and ( − 1 , 4 )

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ISBN: 9781305266636

Author: James Stewart

Publisher: Cengage Learning

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Chapter 10.5, Problem 24E

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To Find: The vertices, foci and asymptotes of the hyperbola for the equation 9y2−4x2−36y+8x=4 .

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Chapter 10.5, Problem 23E

Chapter 10.5, Problem 25E

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The vertices, foci and asymptotes of the hyperbola for the equation 9 y 2 − 4 x 2 − 36 y + 8 x = 4 .

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