Explanation Given: The hyperbola equation is as follows. y 2 − 16 x 2 = 16 (1). Divide the equation (1) with the value 16 . y 2 16 − 16 x 2 16 = 16 16 y 2 16 − x 2 1 = 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 − 0 ) 2 16 + ( x − 0 ) 2 1 = 1 Therefore, the center of the hyperbola ( h , k ) is ( 0 , 0 ) _ . Substitute the value 16 for a 2 and 1 for b 2 in equation (2). a 2 = 16 a = 16 a = 4 b 2 = 1 b = 1 b = 1 Compute the vertices. v e r t i c e s = ( h , ( k ± a ) ) Substitute the value 0 for h , 0 for k and 4 for a . v e r t i c e s = ( 0 , ( ± 4 ) ) Therefore, the vertices of the hyperbola are ( 0 , ± 4 ) _ . Compute the value of c using the equation below. c 2 = a 2 + b 2 Substitute the value 16 for a and 1 for b . c 2 = ( 4 ) 2 + ( 1 ) 2 c 2 = 16 + 1 c = 17 Compute the foci

8th Edition

ISBN: 9781305266636

Author: James Stewart

Publisher: Cengage Learning

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Chapter 10.5, Problem 22E

To determine

To Find: The vertices, foci and asymptotes of the hyperbola for the equation y2−16x2=16.

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Single Variable Calculus

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The vertices, foci and asymptotes of the hyperbola for the equation y 2 − 16 x 2 = 16 .

Solution Summary: The author plots the vertices, foci, and asymptotes of the hyperbola for the equation y2-16x2.

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To Find: The vertices, foci and asymptotes of the hyperbola for the equation y2−16x2=16.

Chapter 10 Solutions