Find the vertices and asymptotes of the hyperbola. $16 y ^ { 2 } - 9 x ^ { 2 } = 144$
A)vertices: (0, ±3) asymptote: $y = \pm \frac { 3 } { 4 } x$
B)vertices: (0, ±3) asymptote: $y = \pm \frac { 4 } { 3 } x$
C)vertices: (±3, 0) asymptote: $y = \pm \frac { 3 } { 4 } x$
D)vertices: (±3, 0) asymptote: $y = \pm \frac { 4 } { 3 } x$
E)vertices: (±3, 4) asymptote: $y = \pm \frac { 3 } { 4 } x$

Question

Find the vertices and asymptotes of the hyperbola.  $16 y ^ { 2 } - 9 x ^ { 2 } = 144$ 
A)vertices: (0, ±3) asymptote: $y = \pm \frac { 3 } { 4 } x$
B)vertices: (0, ±3) asymptote: $y = \pm \frac { 4 } { 3 } x$
C)vertices: (±3, 0) asymptote: $y = \pm \frac { 3 } { 4 } x$
D)vertices: (±3, 0) asymptote: $y = \pm \frac { 4 } { 3 } x$
E)vertices: (±3, 4) asymptote: $y = \pm \frac { 3 } { 4 } x$

Quizplus · Accepted Answer

The equation of the hyperbola is in the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where $a^2 = 16$ and $b^2 = 9$. Therefore, $a = 4$ and $b = 3$. The vertices are at $(0, \pm a)$, which are $(0, \pm 4)$. The asymptotes are $y = \pm \frac{a}{b} x$, which are $y = \pm \frac{4}{3} x$.

Find the Vertices and Asymptotes of the Hyperbola 16y2−9x2=14416 y ^ { 2 } - 9 x ^ { 2 } = 14416y2−9x2=144

Find the Vertices and Asymptotes of the Hyperbola $16 y ^ { 2 } - 9 x ^ { 2 } = 144$