Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
Vertices: (0,±3);focies: (0,±7)

A) $- \frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1$
B) $\frac { y ^ { 2 } } { 40 } - \frac { x ^ { 2 } } { 9 } = 1$
C) $\frac { y ^ { 2 } } { 9 } + \frac { x ^ { 2 } } { 40 } = 1$
D) $\frac { y ^ { 2 } } { 40 } - \frac { x ^ { 2 } } { 9 } = - 1$
E) $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. ​
Vertices: (0,±3);focies: (0,±7)
​ ​
A) $- \frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1$
B) $\frac { y ^ { 2 } } { 40 } - \frac { x ^ { 2 } } { 9 } = 1$
C) $\frac { y ^ { 2 } } { 9 } + \frac { x ^ { 2 } } { 40 } = 1$
D) $\frac { y ^ { 2 } } { 40 } - \frac { x ^ { 2 } } { 9 } = - 1$
E) $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1$

Quizplus · Accepted Answer

The vertices and foci indicate a vertical hyperbola centered at the origin. The distance from the center to a vertex is $a=3$, so $a^2=9$. The distance from the center to a focus is $c=7$, so $c^2=49$. For a hyperbola, $c^2=a^2+b^2$, so $b^2=c^2-a^2=49-9=40$. The equation of a vertical hyperbola is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, thus $\frac{y^2}{9}-\frac{x^2}{40}=1$.

Find the Standard Form of the Equation of the Hyperbola −y29−x240=1- \frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1−9y2​−40x2​=1

Find the Standard Form of the Equation of the Hyperbola $- \frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1$