Explanation Formula used: The hyperbola ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 has foci ( h , ( k ± c ) ) , where c 2 = a 2 + b 2 , vertices ( h , ( k ± a ) ) , and asymptotes ( y − k ) = ± ( a b ) ( x − h ) . Calculation: The given equation of hyperbola is, 9 y 2 − 4 x 2 − 36 y − 8 x = 4 Rewrite the equation as follows. 9 y 2 − 4 x 2 − 36 y − 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 On further simplifications, 9 ( y − 2 ) 2 − 4 ( x + 1 ) 2 = 36 9 ( y − 2 ) 2 36 − 4 ( x + 1 ) 2 36 = 36 ( y − 2 ) 2 4 − ( x + 1 ) 2 9 = 1 Compare this equation with the standard equation of hyperbola. ( y − k ) 2 a 2 − ( x − h ) 2 b 2 = 1 ( y − ( − 2 ) ) 2 4 − ( x − ( 1 ) ) 2 9 = 1 Therefore, the center of the hyperbola ( h , k ) is ( − 1 , 2 ) . Here a 2 = 4 and. b 2 = 9 . Then, a 2 = 4 a = 4 a = 2 b 2 = 9 b = 9 b = 3 By the above mentioned formula of ellipse, the vertices is said to be ( h , ( k ± a ) )

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Chapter 10.5, Problem 24E

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The vertices, foci and asymptotes of the hyperbola 9y2−4x2−36y−8x=4 and sketch it.

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The vertices, foci and asymptotes of the hyperbola 9 y 2 − 4 x 2 − 36 y − 8 x = 4 and sketch it.

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The vertices, foci and asymptotes of the hyperbola 9y2−4x2−36y−8x=4 and sketch it.

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