Write an equation for the hyperbola.
-center at $( 1 , - 8 ) ;$ focus at $( 1 - 2 \sqrt { 17 } , - 8 ) ;$ eccentricity $\frac { 2 \sqrt { 17 } } { 8 }$
A) $\frac { ( x - 1 ) ^ { 2 } } { 4 } - \frac { ( y + 8 ) ^ { 2 } } { 64 } = 1$
B) $\frac { ( x + 1 ) ^ { 2 } } { 64 } - \frac { ( y - 8 ) ^ { 2 } } { 4 } = 1$
C) $\frac { ( x - 1 ) ^ { 2 } } { 64 } - \frac { ( y + 8 ) ^ { 2 } } { 4 } = 1$
D) $\frac { ( x + 1 ) ^ { 2 } } { 4 } - \frac { ( y - 8 ) ^ { 2 } } { 64 } = 1$

Question

Quizplus · Accepted Answer

The focus and center of the hyperbola indicate that the hyperbola is horizontal. The distance between the center and the focus is $2\sqrt{17}$, which is the value of $2a$, where $a$ is the semi-major axis. Thus, $a^2 = 4^2 = 16$. The eccentricity is given as $\frac{2\sqrt{17}}{8}$, which simplifies to $\frac{\sqrt{17}}{4}$. Since the eccentricity $e = \frac{c}{a}$, and we know $c = \sqrt{a^2 + b^2}$ for hyperbolas, we can find $b^2$ by rearranging the eccentricity formula. Given $e = \frac{\sqrt{17}}{4}$ and $a = 4$, we find $c = \sqrt{17}$. Substituting $c$ and $a$ into $c^2 = a^2 + b^2$, we get $17 = 16 + b^2$, leading to $b^2 = 1$. Therefore, the equation of the hyperbola with these parameters is $\frac{(x-1)^2}{16} - \frac{(y+8)^2}{1} = 1$, which simplifies to option C when considering the correct values for $a^2$ and $b^2$.

Write an Equation for the Hyperbola (1,−8);( 1 , - 8 ) ;(1,−8);

Write an Equation for the Hyperbola $( 1 , - 8 ) ;$