Find an equation for the hyperbola described.
-Vertices $\left( \frac { 1 } { 2 } , - 3 \right)$ and $\left( - \frac { 9 } { 2 } , - 3 \right) ;$ asymptotes $\mathrm { y } + 3 = \pm \frac { 6 } { 5 } ( \mathrm { x } + 2 )$

A) $\frac { ( x + 2 ) ^ { 2 } } { 9 } - \frac { 4 ( y + 3 ) ^ { 2 } } { 25 } = 1$

B) $\frac { 4 ( \mathrm { x } + 2 ) ^ { 2 } } { 25 } - \frac { ( \mathrm { y } + 3 ) ^ { 2 } } { 9 } = 1$

C) $\frac { ( y + 3 ) ^ { 2 } } { 9 } - \frac { 4 ( x + 2 ) ^ { 2 } } { 25 } = 1$

D) $\frac { 4 ( x - 2 ) ^ { 2 } } { 25 } - \frac { ( y - 3 ) ^ { 2 } } { 9 } = 1$

Question

Find an equation for the hyperbola described.
-Vertices $\left( \frac { 1 } { 2 } , - 3 \right)$ and $\left( - \frac { 9 } { 2 } , - 3 \right) ;$ asymptotes $\mathrm { y } + 3 = \pm \frac { 6 } { 5 } ( \mathrm { x } + 2 )$

A) $\frac { ( x + 2 ) ^ { 2 } } { 9 } - \frac { 4 ( y + 3 ) ^ { 2 } } { 25 } = 1$

B) $\frac { 4 ( \mathrm { x } + 2 ) ^ { 2 } } { 25 } - \frac { ( \mathrm { y } + 3 ) ^ { 2 } } { 9 } = 1$

C) $\frac { ( y + 3 ) ^ { 2 } } { 9 } - \frac { 4 ( x + 2 ) ^ { 2 } } { 25 } = 1$

D) $\frac { 4 ( x - 2 ) ^ { 2 } } { 25 } - \frac { ( y - 3 ) ^ { 2 } } { 9 } = 1$

Quizplus · Accepted Answer

The vertices indicate a horizontal hyperbola centered at $(-4, -3)$. The distance between the vertices is 5, so $2a = 5$ and $a = \frac{5}{2}$. The slope of the asymptotes $\pm \frac{6}{5}$ gives $\frac{b}{a} = \frac{6}{5}$, so $b = 3$. The equation is $\frac{(x + 4)^2}{(\frac{5}{2})^2} - \frac{(y + 3)^2}{3^2} = 1$, which simplifies to choice B.

Find an Equation for the Hyperbola Described (12,−3)\left( \frac { 1 } { 2 } , - 3 \right)(21​,−3)

Find an Equation for the Hyperbola Described $\left( \frac { 1 } { 2 } , - 3 \right)$