Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
Vertices: $( \pm 4,0 )$ ; foci: $( \pm 8,0 )$
A) $\frac { y ^ { 2 } } { 48 } - \frac { x ^ { 2 } } { 16 } = 1$
B) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 48 } = 1$
C) $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 48 } = 1$
D) $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 48 } = 1$
E) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 48 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. 
Vertices: $( \pm 4,0 )$ ; foci: $( \pm 8,0 )$ 
A) $\frac { y ^ { 2 } } { 48 } - \frac { x ^ { 2 } } { 16 } = 1$
B) $\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 48 } = 1$
C) $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 48 } = 1$
D) $\frac { y ^ { 2 } } { 16 } + \frac { x ^ { 2 } } { 48 } = 1$
E) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 48 } = 1$

Quizplus · Accepted Answer

The standard form of a hyperbola with a horizontal transverse axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where $2a$ is the distance between the vertices, and $2c$ is the distance between the foci. Given vertices at $(\pm 4,0)$, we have $a=4$, so $a^2=16$. The foci are at $(\pm 8,0)$, so $c=8$ and $c^2=64$. Using the relationship $c^2=a^2+b^2$, we find $b^2=c^2-a^2=64-16=48$. Therefore, the equation is $\frac{x^2}{16} - \frac{y^2}{48} = 1$.

Find the Standard Form of the Equation of the Hyperbola (±4,0)( \pm 4,0 )(±4,0)

Find the Standard Form of the Equation of the Hyperbola $( \pm 4,0 )$