Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (-10,-2), (0,-2)
Asymptotes: y = ± $\frac { 4 } { 5 }$ x

A) $\frac { ( x + 5 ) ^ { 2 } } { 25 } - \frac { ( y + 2 ) ^ { 2 } } { 16 } = 0$
B) $\frac { ( x + 5 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( y + 5 ) ^ { 2 } } { 25 } - \frac { ( x + 2 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( y - 2 ) ^ { 2 } } { 25 } - \frac { ( x + 2 ) ^ { 2 } } { 16 } = 1$
E) $\frac { ( x + 5 ) ^ { 2 } } { 25 } - \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (-10,-2), (0,-2)
Asymptotes: y = ± $\frac { 4 } { 5 }$ x
​
A) $\frac { ( x + 5 ) ^ { 2 } } { 25 } - \frac { ( y + 2 ) ^ { 2 } } { 16 } = 0$
B) $\frac { ( x + 5 ) ^ { 2 } } { 25 } + \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( y + 5 ) ^ { 2 } } { 25 } - \frac { ( x + 2 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( y - 2 ) ^ { 2 } } { 25 } - \frac { ( x + 2 ) ^ { 2 } } { 16 } = 1$
E) $\frac { ( x + 5 ) ^ { 2 } } { 25 } - \frac { ( y + 2 ) ^ { 2 } } { 16 } = 1$

Quizplus · Accepted Answer

The center of the hyperbola is the midpoint of the vertices, which is $(-5, -2)$. The distance between the vertices is 10 units, and since the vertices are on the x-axis, this distance represents $2a$, where $a = 5$. The slope of the asymptotes $\pm \frac{4}{5}$ gives the relationship between $a$ and $b$ in $b/a = 4/5$, leading to $b = 4$. Thus, the equation is $\frac{(x+5)^2}{25} - \frac{(y+2)^2}{16} = 1$.

Find the Standard Form of the Equation of the Hyperbola 45\frac { 4 } { 5 }54​

Find the Standard Form of the Equation of the Hyperbola $\frac { 4 } { 5 }$