Find the standard-form equation of the hyperbola centered at the origin which satisfies the given conditions.
-Vertices at $( 0,4 )$ and $( 0 , - 4 )$; asymptotes $y = \frac { 2 } { 3 } x$ and $y = - \frac { 2 } { 3 } x$
A) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 36 } = 1$
B) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 9 } = 1$
C) $\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 16 } = 1$
D) $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 4 } = 1$

Question

Quizplus · Accepted Answer

The distance between the vertices is 8, so $2a = 8$, $a = 4$. The slope of the asymptotes is $\pm \frac { 2 } { 3 }$, so $b = 3$. The standard-form equation of the hyperbola is $\frac { y ^ { 2 } } { a ^ { 2 } } - \frac { x ^ { 2 } } { b ^ { 2 } } = 1$, so the equation of the hyperbola is $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 36 } = 1$.

Find the Standard-Form Equation of the Hyperbola Centered at the Origin