(a) Explanation The calculation of center, vertices, foci and asymptotes of the hyperbola x 2 − 4 y 2 + 16 = 0 as follows. The equation of the hyperbola x 2 − 4 y 2 + 16 = 0 can be written in the standard form y 2 a 2 − x 2 b 2 = 1 as y 2 4 − x 2 16 = 1 . Now, compare the equation y 2 4 − x 2 16 = 1 with the hyperbola y 2 a 2 − x 2 b 2 = 1 , the value of a = 4 and b = 2 . The centre of the hyperbola C ( 0 , 0 ) , the vertices of the hyperbola y 2 a 2 − x 2 b 2 = 1 are V ( 0 , ± a ) ; thus, the vertices of the hyperbola x 2 − 4 y 2 + 16 = 0 are V 1 ( 0 , − 2 ) and V 2 ( 0 , + 2 ) . Note that the foci of the hyperbola y 2 a 2 − x 2 b 2 = 1 are F ( 0 , ± c ) where c = a 2 + b 2 . Substitute a = 2 and b = 4 in c , c = a 2 + b 2 = 2 2 + 4 2 = 20 = 2 5 Thus, the foci of the hyperbola of hyperbola x 2 − 4 y 2 + 16 = 0 are F 1 ( 0 , − 2 5 ) and F 2 ( 0 , 2 5 ) (b) To determine To sketch: The graph of the hyperbola x 2 − 4 y 2 + 16 = 0 .

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Author: Stewart

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Chapter 11, Problem 30RE

(a)

To determine

To find: The center, vertices, foci and asymptotes of the hyperbola x2−4y2+16=0 .

(b)

To determine

To sketch: The graph of the hyperbola x2−4y2+16=0 .

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Chapter 11, Problem 29RE

Chapter 11, Problem 31RE

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The center, vertices, foci and asymptotes of the hyperbola x 2 − 4 y 2 + 16 = 0 .

Solution Summary: The author explains the calculation of center, vertices, foci and asymptotes of the hyperbola x2-4y+16=0.

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To find: The center, vertices, foci and asymptotes of the hyperbola x2−4y2+16=0 .

To sketch: The graph of the hyperbola x2−4y2+16=0 .

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Chapter 11 Solutions