Write an equation for the hyperbola.
-foci at $( 8 - 2 \sqrt { 29 } , 9 ) , ( 8 + 2 \sqrt { 29 } , 9 ) ; e = \frac { 2 \sqrt { 29 } } { 10 }$
A) $\frac { ( y + 9 ) ^ { 2 } } { 100 } - \frac { ( x + 8 ) ^ { 2 } } { 16 } = 1$

B) $\frac { ( y + ) ^ { 2 } } { 100 } - \frac { ( x + 9 ) ^ { 2 } } { 16 } = 1$

C) $\frac { ( y - 9 ) ^ { 2 } } { 100 } - \frac { ( x + 8 ) ^ { 2 } } { 16 } = 1$

D) $\frac { ( x - 8 ) ^ { 2 } } { 100 } - \frac { ( y - 9 ) ^ { 2 } } { 16 } = 1$

Question

Write an equation for the hyperbola.
-foci at $( 8 - 2 \sqrt { 29 } , 9 ) , ( 8 + 2 \sqrt { 29 } , 9 ) ; e = \frac { 2 \sqrt { 29 } } { 10 }$
A) $\frac { ( y + 9 ) ^ { 2 } } { 100 } - \frac { ( x + 8 ) ^ { 2 } } { 16 } = 1$

B) $\frac { ( y + ) ^ { 2 } } { 100 } - \frac { ( x + 9 ) ^ { 2 } } { 16 } = 1$

C) $\frac { ( y - 9 ) ^ { 2 } } { 100 } - \frac { ( x + 8 ) ^ { 2 } } { 16 } = 1$

D) $\frac { ( x - 8 ) ^ { 2 } } { 100 } - \frac { ( y - 9 ) ^ { 2 } } { 16 } = 1$

Quizplus · Accepted Answer

The foci are horizontally aligned (same y-coordinate), indicating a horizontal hyperbola. The center is at (8, 9), and the distance between each focus and the center is $2\sqrt{29}$, which is 2a. The eccentricity $e = \frac{c}{a}$ gives $a = 10$, so $a^2 = 100$. The equation format for a horizontal hyperbola is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, where (h, k) is the center. Since $c^2 = a^2 + b^2$ and $c = 2\sqrt{29}$, $b^2 = 16$. Thus, the correct equation is $\frac{(x-8)^2}{100} - \frac{(y-9)^2}{16} = 1$.

Write an Equation for the Hyperbola (8−229,9),(8+229,9);e=22910( 8 - 2 \sqrt { 29 } , 9 ) , ( 8 + 2 \sqrt { 29 } , 9 ) ; e = \frac { 2 \sqrt { 29 } } { 10 }(8−229​,9),(8+229​,9);e=10229​​

Write an Equation for the Hyperbola $( 8 - 2 \sqrt { 29 } , 9 ) , ( 8 + 2 \sqrt { 29 } , 9 ) ; e = \frac { 2 \sqrt { 29 } } { 10 }$