Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
Vertices: $( 0 , \pm 6 )$ ; asymptotes: $y = \pm 6 x$
A) $\frac { y ^ { 2 } } { 1 } - \frac { x ^ { 2 } } { 36 } = 1$
B) $\frac { y ^ { 2 } } { 36 } + \frac { x ^ { 2 } } { 1 } = 1$
C) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 1 } = 1$
D) $\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 1 } = 1$
E) $\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 1 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. 
Vertices: $( 0 , \pm 6 )$ ; asymptotes: $y = \pm 6 x$ 
A) $\frac { y ^ { 2 } } { 1 } - \frac { x ^ { 2 } } { 36 } = 1$
B) $\frac { y ^ { 2 } } { 36 } + \frac { x ^ { 2 } } { 1 } = 1$
C) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 1 } = 1$
D) $\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 1 } = 1$
E) $\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 1 } = 1$

Quizplus · Accepted Answer

The vertices at $(0, \pm 6)$ indicate that the major axis is along the y-axis, and the distance from the center to the vertices is 6, so $a = 6$, which means $a^2 = 36$. The slopes of the asymptotes $\pm 6x$ suggest that $b/a = 6$, and since $a = 6$, $b = 36$. Therefore, the equation is $\frac{y^2}{36} - \frac{x^2}{1} = 1$, matching option D.

Find the Standard Form of the Equation of the Hyperbola (0,±6)( 0 , \pm 6 )(0,±6)

Find the Standard Form of the Equation of the Hyperbola $( 0 , \pm 6 )$