Explanation Formula used: General form of hyperbola is given by ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 , where ( h , k ) is the centre, a is length of semi transverse axis (if a >0) and b is length of semi conjugate axis. Vertices of hyperbola are ( h + a , k ) and ( h – a , k ) Foci of hyperbola are ( h + c , k ) and ( h – c , k ) where c = a 2 + b 2 The equations for the asymptotes are y = ± ( b a ) ( x − h ) + k Calculation: Changing the given equation in to general form 3 x 2 − 27 y 2 = 12 x 2 2 2 − y 2 ( 2 3 ) 2 = 1 ð Centre of hyperbola = (0,0), a = 2 and b = 0...

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 12.5, Problem 32E

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The vertices, foci, center and asymptotes of the hyperbola $3 x^{2} - 27 y^{2} = 12$ ?

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Chapter 12.5, Problem 31E

Chapter 12.5, Problem 33E

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Chapter 12 Solutions

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Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

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The vertices, foci, center and asymptotes of the hyperbola 3 x 2 − 27 y 2 = 12 ?

Solution Summary: The author explains how to determine the vertices, foci, center and asymptotes of the hyperbola.

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The vertices, foci, center and asymptotes of the hyperbola $3 x^{2} - 27 y^{2} = 12$ ?

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Chapter 12 Solutions

The vertices, foci, center and asymptotes of the hyperbola 3 x 2 − 27 y 2 = 12 ?

Solution Summary: The author explains how to determine the vertices, foci, center and asymptotes of the hyperbola.

Videos

The vertices, foci, center and asymptotes of the hyperbola 3x2−27y2=12?

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Chapter 12 Solutions

The vertices, foci, center and asymptotes of the hyperbola $3 x^{2} - 27 y^{2} = 12$ ?