Solve the problem.
-Find the vertices and asymptotes of the following hyperbola: $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 16 } = 1$
A) Vertices: $( 0,4 ) , ( 0 , - 4 ) ;$ Asymptotes: $y = \pm 4 x$
B) Vertices: $( 4,0 ) , ( - 4,0 )$; Asymptotes: $\mathrm { y } = \pm 1 \mathrm { x }$
C) Vertices: $( 0,16 ) , ( 0 , - 16 )$; Asymptotes: $y = \pm \frac { 1 } { 4 } x$
D) Vertices: $( 0,4 ) , ( 0 , - 4 ) ;$ Asymptotes: $y = \pm 1 x$

Question

Quizplus · Accepted Answer

The given equation is of a hyperbola centered at the origin with a vertical transverse axis. The vertices are found by setting $x = 0$ and solving for $y$, giving $(0, 4)$ and $(0, -4)$. The asymptotes for a hyperbola of the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ are $y = \pm \frac{a}{b}x$, and since $a^2 = b^2 = 16$, $a = b = 4$, the asymptotes are $y = \pm x$.

Solve the Problem y216−x216=1\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 16 } = 116y2​−16x2​=1

Solve the Problem $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 16 } = 1$