Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (5,-6), (5,-2)
Foci: (5,-7), (5,-1)

A) $\frac { ( y + 4 ) ^ { 2 } } { 4 } - \frac { ( x - 5 ) ^ { 2 } } { 5 } = 1$
B) $\frac { ( y + 4 ) ^ { 2 } } { 4 } + \frac { ( x - 5 ) ^ { 2 } } { 5 } = 1$
C) $\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x + 4 ) ^ { 2 } } { 5 } = 1$
D) $\frac { ( y + 5 ) ^ { 2 } } { 4 } - \frac { ( x - 4 ) ^ { 2 } } { 5 } = 1$
E) $\frac { ( y - 4 ) ^ { 2 } } { 4 } - \frac { ( x + 5 ) ^ { 2 } } { 9 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (5,-6), (5,-2)
Foci: (5,-7), (5,-1)
​
A) $\frac { ( y + 4 ) ^ { 2 } } { 4 } - \frac { ( x - 5 ) ^ { 2 } } { 5 } = 1$
B) $\frac { ( y + 4 ) ^ { 2 } } { 4 } + \frac { ( x - 5 ) ^ { 2 } } { 5 } = 1$
C) $\frac { ( y - 5 ) ^ { 2 } } { 4 } - \frac { ( x + 4 ) ^ { 2 } } { 5 } = 1$
D) $\frac { ( y + 5 ) ^ { 2 } } { 4 } - \frac { ( x - 4 ) ^ { 2 } } { 5 } = 1$
E) $\frac { ( y - 4 ) ^ { 2 } } { 4 } - \frac { ( x + 5 ) ^ { 2 } } { 9 } = 1$

Quizplus · Accepted Answer

The vertices and foci are aligned vertically, indicating a vertical hyperbola. The center is midway between both vertices and foci, at (5, -4). The distance between vertices (4 units) gives 2a, and between foci (6 units) gives 2c, leading to a^2 = 4 and c^2 = 9. Since c^2 = a^2 + b^2, b^2 = 5. The correct equation is $\frac{(y+4)^2}{4} - \frac{(x-5)^2}{5} = 1$.

Find the Standard Form of the Equation of the Hyperbola (y+4)24−(x−5)25=1\frac { ( y + 4 ) ^ { 2 } } { 4 } - \frac { ( x - 5 ) ^ { 2 } } { 5 } = 14(y+4)2​−5(x−5)2​=1

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