Write an equation for the hyperbola.
-vertices at $( - 9,7 ) , ( - 9 , - 11 ) ;$ asymptotes $y = \pm \frac { 9 } { 6 } ( x + 9 ) - 2$
A) $\frac { ( x + 9 ) ^ { 2 } } { 81 } - \frac { ( y + 2 ) ^ { 2 } } { 81 } = 1$
B) $\frac { ( x - 9 ) ^ { 2 } } { 81 } - \frac { ( y - 2 ) ^ { 2 } } { 81 } = 1$
C) $\frac { ( y + 2 ) ^ { 2 } } { 81 } - \frac { ( x + 9 ) ^ { 2 } } { 36 } = 1$
D) $\frac { ( y - 2 ) ^ { 2 } } { 81 } - \frac { ( x - 9 ) ^ { 2 } } { 36 } = 1$

Question

Quizplus · Accepted Answer

The vertices at $(-9,7)$ and $(-9,-11)$ indicate that the hyperbola is vertical, and the center is at $(-9, -2)$ (midpoint of the vertices). The slope of the asymptotes $\pm\frac{9}{6}$ simplifies to $\pm\frac{3}{2}$, which gives the relationship between the distances $a$ and $b$ in the equation of the hyperbola. Since the distance between the vertices is $18$ (from $7$ to $-11$), $a = 9$, and thus $a^2 = 81$. The equation of a vertical hyperbola centered at $(h, k)$ is $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$. Given the asymptote slope and the $a$ value, we can find $b^2$ by knowing that the slope of the asymptotes for a vertical hyperbola is $\pm\frac{a}{b}$, leading to $b^2 = 36$. Therefore, the correct equation is $\frac{(y+2)^2}{81} - \frac{(x+9)^2}{36} = 1$.

Write an Equation for the Hyperbola (−9,7),(−9,−11);( - 9,7 ) , ( - 9 , - 11 ) ;(−9,7),(−9,−11);

Write an Equation for the Hyperbola $( - 9,7 ) , ( - 9 , - 11 ) ;$