Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
Vertices: $( 0 , \pm 5 )$ ;foci: $( 0 , \pm 9 )$
A) $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 56 } = 1$
B) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 56 } = 1$
C) $\frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 56 } = 1$
D) $\frac { y ^ { 2 } } { 56 } - \frac { x ^ { 2 } } { 25 } = 1$
E) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 56 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. ​
Vertices: $( 0 , \pm 5 )$ ;foci: $( 0 , \pm 9 )$ ​
A) $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 56 } = 1$
B) $\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 56 } = 1$
C) $\frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 56 } = 1$
D) $\frac { y ^ { 2 } } { 56 } - \frac { x ^ { 2 } } { 25 } = 1$
E) $\frac { x ^ { 2 } } { 25 } + \frac { y ^ { 2 } } { 56 } = 1$

Quizplus · Accepted Answer

The standard form of a hyperbola with a vertical transverse axis is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where $2a$ is the distance between the vertices, and $c$ is the distance from the center to a focus. Given vertices at $(0, \pm 5)$, we have $a = 5$, so $a^2 = 25$. With foci at $(0, \pm 9)$, we have $c = 9$, so $c^2 = 81$. Using the relationship $c^2 = a^2 + b^2$, we find $b^2 = c^2 - a^2 = 81 - 25 = 56$. Therefore, the equation is $\frac{y^2}{25} - \frac{x^2}{56} = 1$.

Find the Standard Form of the Equation of the Hyperbola (0,±5)( 0 , \pm 5 )(0,±5)

Find the Standard Form of the Equation of the Hyperbola $( 0 , \pm 5 )$