Explanation Given: The equations of hyperbola are, x 2 4 2 − y 2 3 2 = 1 ( x − 3 ) 2 4 2 − ( y − 1 ) 2 3 2 = 1 Approach: The following equation is the basic equation for a hyperbola that is not centered at the origin. ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 With the following properties, Vertices V = ( ± a + h , k ) …… ( 1 ) Asymptotes ( y − k ) = ± b a ( x − h ) …… ( 2 ) Foci F = ( ± c + h , k ) …… ( 3 ) Here, c 2 = a 2 + b 2 …… ( 4 ) Calculation: Consider the equation, x 2 4 2 − y 2 3 2 = 1 Convert the above equation into standard equation of hyperbola. ( x − 0 ) 2 4 2 − ( y − 0 ) 2 3 2 = 1 Compare the above equation of the hyperbola with the standard equation a hyperbola. a = 4 b = 3 h = 0 k = 0 Calculate the vertices of the hyperbola. Substitute 4 for a , 0 for h and 0 for k in equation ( 1 ) . V = ( ± 4 , 0 ) Calculate the asymptotes of the hyperbola. Substitute 4 for a , 3 for b , 0 for h and 0 for k in equation ( 2 ) . y = ± 3 4 x Calculate the value of c . Substitute 4 for a and 3 for b in equation ( 4 ) . c 2 = 4 2 + 3 2 c 2 = 25 c = ± 25 = ± 5 Calculate the foci of the hyperbola. Substitute 5 for c , 0 for h and 0 for k in equation ( 3 ) . F = ( ± 5 , 0 ) The vertices, foci and asymptotes of hyperbola x 2 4 2 − y 2 3 2 = 1 is shown in figure below, Figure ( 1 ) Consider the equation, ( x − 3 ) 2 4 2 − ( y − 1 ) 2 3 2 = 1 Compare the above equation of the hyperbola with the standard equation a hyperbola

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN: 9781305071742

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Cengage Learning

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Chapter 12.4, Problem 4E

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Algebra and Trigonometry (MindTap Course List)

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The vertices, foci and asymptotes of the hyperbola.

Solution Summary: The author explains the basic equation for a hyperbola that is not centered at the origin.

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To find: The vertices, foci and asymptotes of the hyperbola.

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Chapter 12 Solutions

To find:

The vertices, foci and asymptotes of the hyperbola.