Find the standard form of the equation of the hyperbola with the given characteristics.
vertices: (3, 3), (3, 7)
Foci: (3, 1), (3, 9)

A) $\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x - 3 ) ^ { 2 } } { 12 } = 1$
B) $\frac { ( y - 5 ) ^ { 2 } } { 4 } + \frac { ( x - 3 ) ^ { 2 } } { 12 } = 1$
C) $\frac { ( x - 5 ) ^ { 2 } } { 4 } - \frac { ( y - 3 ) ^ { 2 } } { 12 } = 1$
D) $\frac { ( y + 5 ) ^ { 2 } } { 4 } - \frac { ( x - 3 ) ^ { 2 } } { 12 } = 1$
E) $\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x + 3 ) ^ { 2 } } { 12 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics. 
vertices: (3, 3), (3, 7)
Foci: (3, 1), (3, 9)

A) $\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x - 3 ) ^ { 2 } } { 12 } = 1$
B) $\frac { ( y - 5 ) ^ { 2 } } { 4 } + \frac { ( x - 3 ) ^ { 2 } } { 12 } = 1$
C) $\frac { ( x - 5 ) ^ { 2 } } { 4 } - \frac { ( y - 3 ) ^ { 2 } } { 12 } = 1$
D) $\frac { ( y + 5 ) ^ { 2 } } { 4 } - \frac { ( x - 3 ) ^ { 2 } } { 12 } = 1$
E) $\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x + 3 ) ^ { 2 } } { 12 } = 1$

Quizplus · Accepted Answer

The vertices and foci suggest the hyperbola is vertical. The center is midway between both vertices and foci, at (3, 5). The distance between vertices (3, 3) and (3, 7) is 4, so $2a = 4$, giving $a^2 = 4$. The distance between foci (3, 1) and (3, 9) is 8, so $2c = 8$, giving $c^2 = 16$. Using $c^2 = a^2 + b^2$, we find $b^2 = 12$. The correct equation is $\frac{(y-5)^2}{4} - \frac{(x-3)^2}{12} = 1$.

Find the Standard Form of the Equation of the Hyperbola (y−5)24−(x−3)212=1\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x - 3 ) ^ { 2 } } { 12 } = 14(y−5)2​−12(x−3)2​=1

Find the Standard Form of the Equation of the Hyperbola $\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x - 3 ) ^ { 2 } } { 12 } = 1$