Find the center, foci, and asymptotes of the hyperbola.
-$\frac { ( y - 1 ) ^ { 2 } } { 16 } - \frac { ( x - 1 ) ^ { 2 } } { 9 } = 1$
A) C: $( 1,1 ) ; \mathrm { F } : ( - 2,1 ) , ( 4,1 ) ; \mathrm { A } : \mathrm { y } = \frac { 16 } { 9 } \mathrm { x } - \frac { 7 } { 9 } , \mathrm { y } = - \frac { 16 } { 9 } \mathrm { x } - \frac { 7 } { 9 }$
B) C: $( 1,1 ) ; F : ( 1 , - 4 ) , ( 1,6 ) ; A : y = \frac { 4 } { 3 } x + \frac { 1 } { 3 } , y = - \frac { 4 } { 3 } x - \frac { 7 } { 3 }$
C) $\mathrm { C } : ( 1,1 ) ; \mathrm { F } : ( 1 , - 2 ) , ( 1,4 ) ; \mathrm { A } : \mathrm { y } = \frac { 16 } { 9 } \mathrm { x } + \frac { 1 } { 9 } , \mathrm { y } = - \frac { 16 } { 9 } \mathrm { x } - \frac { 7 } { 9 }$
D) C: $( 1,1 ) ; F : ( 1,6 ) , ( 1 , - 4 ) ; \mathrm { A } : \mathrm { y } = \frac { 4 } { 3 } \mathrm { x } - \frac { 1 } { 3 } , \mathrm { y } = - \frac { 4 } { 3 } \mathrm { x } + \frac { 7 } { 3 }$

Question

Quizplus · Accepted Answer

The center of the hyperbola is given by the point (h, k), which is (1, 1) for this equation. The distance between the center and the foci along the y-axis is given by $\sqrt{16 + 9} = 5$, so the foci are at (1, 1 ± 5) = (1, 6) and (1, -4). The asymptotes for a hyperbola of the form $\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1$ are given by $y = \pm \frac{a}{b}(x - h) + k$, which for this hyperbola are $y = \pm \frac{4}{3}(x - 1) + 1$, simplifying to $y = \frac{4}{3}x - \frac{1}{3}$ and $y = -\frac{4}{3}x + \frac{7}{3}$.

Find the Center, Foci, and Asymptotes of the Hyperbola (y−1)216−(x−1)29=1\frac { ( y - 1 ) ^ { 2 } } { 16 } - \frac { ( x - 1 ) ^ { 2 } } { 9 } = 116(y−1)2​−9(x−1)2​=1

Find the Center, Foci, and Asymptotes of the Hyperbola $\frac { ( y - 1 ) ^ { 2 } } { 16 } - \frac { ( x - 1 ) ^ { 2 } } { 9 } = 1$