Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. $\text { Vertices: } ( \pm 4,0 ) ; \text { asymptotes: } y = \pm 6 x$
A) $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 576 } = 1$
B) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 576 } = 1$
C) $\frac { y ^ { 2 } } { 576 } - \frac { x ^ { 2 } } { 16 } = 1$
D) $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 576 } = 1$
E) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 576 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. $\text { Vertices: } ( \pm 4,0 ) ; \text { asymptotes: } y = \pm 6 x$ 
A) $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 576 } = 1$
B) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 576 } = 1$
C) $\frac { y ^ { 2 } } { 576 } - \frac { x ^ { 2 } } { 16 } = 1$
D) $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 576 } = 1$
E) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 576 } = 1$

Quizplus · Accepted Answer

The vertices indicate that the hyperbola is oriented along the x-axis, so the x-term is positive in the standard form. The distance from the center to a vertex is $a=4$, so $a^2=16$. The slope of the asymptotes is $\pm \frac{b}{a}$, and since the slope is $\pm 6$, $b=6a=24$, so $b^2=576$. Thus, the correct equation is $\frac{x^2}{16} - \frac{y^2}{576} = 1$.

Find the Standard Form of the Equation of the Hyperbola Vertices: (±4,0); asymptotes: y=±6x\text { Vertices: } ( \pm 4,0 ) ; \text { asymptotes: } y = \pm 6 x Vertices: (±4,0); asymptotes: y=±6x

Find the Standard Form of the Equation of the Hyperbola $\text { Vertices: } ( \pm 4,0 ) ; \text { asymptotes: } y = \pm 6 x$