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 at (0,±3) indicate that the major axis is along the y-axis, so $a=3$ and $a^2=9$. The foci at (0,±7) mean $c=7$ and $c^2=49$. For a hyperbola, $c^2=a^2+b^2$, so solving for $b^2$ gives $b^2=c^2-a^2=49-9=40$. Thus, the correct equation is $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1$.

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 $- \frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 40 } = 1$