Write an equation for the hyperbola.
-center at $( 6 , - 8 ) ;$ focus at $( 6 - \sqrt { 58 } , - 8 ) ; e = \frac { \sqrt { 58 } } { 7 }$
A) $\frac { ( x + 6 ) ^ { 2 } } { 49 } - \frac { ( y - 8 ) ^ { 2 } } { 9 } = 1$
B) $\frac { ( x + 6 ) ^ { 2 } } { 9 } - \frac { ( y - 8 ) ^ { 2 } } { 49 } = 1$
C) $\frac { ( x - 6 ) ^ { 2 } } { 9 } - \frac { ( y + 8 ) ^ { 2 } } { 49 } = 1$
D) $\frac { ( x - 6 ) ^ { 2 } } { 49 } - \frac { ( y + 8 ) ^ { 2 } } { 9 } = 1$

Question

Quizplus · Accepted Answer

The center of the hyperbola is at (6, -8), indicating that the terms inside the parentheses should be $(x-6)$ and $(y+8)$. The eccentricity $e = \frac{\sqrt{58}}{7}$ and the focus being at $(6 - \sqrt{58}, -8)$ indicates that the hyperbola is horizontal. The distance from the center to the focus (c) is $\sqrt{58}$, and since $e = \frac{c}{a}$, we find $a = 7$. Therefore, $a^2 = 49$ and $c^2 = 58$, leading to $b^2 = c^2 - a^2 = 58 - 49 = 9$. The correct equation is $\frac{(x-6)^2}{49} - \frac{(y+8)^2}{9} = 1$.

Write an Equation for the Hyperbola (6,−8);( 6 , - 8 ) ;(6,−8);

Write an Equation for the Hyperbola $( 6 , - 8 ) ;$