Write an equation for the hyperbola.
-foci at $( 6 - 2 \sqrt { 29 } , 9 ) , ( 6 + 2 \sqrt { 29 } , 9 )$; eccentricity $\frac { 2 \sqrt { 29 } } { 10 }$
A) $\frac { ( x - 6 ) ^ { 2 } } { 100 } - \frac { ( y - 9 ) ^ { 2 } } { 16 } = 1$
B) $\frac { ( y + ) ^ { 2 } } { 100 } - \frac { ( x + 9 ) ^ { 2 } } { 16 } = 1$
C) $\frac { ( y + 9 ) ^ { 2 } } { 100 } - \frac { ( x + 6 ) ^ { 2 } } { 16 } = 1$
D) $\frac { ( y - 9 ) ^ { 2 } } { 100 } - \frac { ( x + 6 ) ^ { 2 } } { 16 } = 1$

Question

Quizplus · Accepted Answer

The center of the hyperbola is the midpoint of the foci, which is $(6, 9)$. The distance between the foci is $4\sqrt{29}$, so $2a = 4\sqrt{29}$ giving $a = 2\sqrt{29}$. The eccentricity $e = \frac{2\sqrt{29}}{10} = \frac{\sqrt{29}}{5}$, and since $e = \frac{c}{a}$, we find $c = \sqrt{29}$. Using $c^2 = a^2 + b^2$ for hyperbolas, we find $b^2 = 16$. Since the foci are horizontal, the $x$ term is positive, and the center $(6, 9)$ matches choice A.

Write an Equation for the Hyperbola (6−229,9),(6+229,9)( 6 - 2 \sqrt { 29 } , 9 ) , ( 6 + 2 \sqrt { 29 } , 9 )(6−229​,9),(6+229​,9)

Write an Equation for the Hyperbola $( 6 - 2 \sqrt { 29 } , 9 ) , ( 6 + 2 \sqrt { 29 } , 9 )$