Write an equation for the hyperbola.
-center at $( 5 , - 10 )$; focus at $( 5 - 4 \sqrt { 10 } , - 10 ) ; e = \frac { 4 \sqrt { 10 } } { 12 }$
A) $\frac { ( x + 5 ) ^ { 2 } } { 144 } - \frac { ( y - 10 ) ^ { 2 } } { 16 } = 1$

B) $\frac { ( x + 5 ) ^ { 2 } } { 16 } - \frac { ( y - 10 ) ^ { 2 } } { 144 } = 1$

C) $\frac { ( x - 5 ) ^ { 2 } } { 144 } - \frac { ( y + 10 ) ^ { 2 } } { 16 } = 1$

D) $\frac { ( x - 5 ) ^ { 2 } } { 16 } - \frac { ( y + 10 ) ^ { 2 } } { 144 } = 1$

Question

Write an equation for the hyperbola.
-center at $( 5 , - 10 )$; focus at $( 5 - 4 \sqrt { 10 } , - 10 ) ; e = \frac { 4 \sqrt { 10 } } { 12 }$
A) $\frac { ( x + 5 ) ^ { 2 } } { 144 } - \frac { ( y - 10 ) ^ { 2 } } { 16 } = 1$

B) $\frac { ( x + 5 ) ^ { 2 } } { 16 } - \frac { ( y - 10 ) ^ { 2 } } { 144 } = 1$

C) $\frac { ( x - 5 ) ^ { 2 } } { 144 } - \frac { ( y + 10 ) ^ { 2 } } { 16 } = 1$

D) $\frac { ( x - 5 ) ^ { 2 } } { 16 } - \frac { ( y + 10 ) ^ { 2 } } { 144 } = 1$

Quizplus · Accepted Answer

The center of the hyperbola is at $(5, -10)$, so the equation should have $(x-5)^2$ and $(y+10)^2$. The focus is to the left of the center, indicating a horizontal hyperbola, so the $x$ term should be positive and larger. The eccentricity $e = \frac{c}{a}$, where $c = 4\sqrt{10}$ and $e = \frac{4\sqrt{10}}{12}$, gives $a = 12$, so $a^2 = 144$. Since $c^2 = a^2 + b^2$, we find $b^2 = 16$.

Write an Equation for the Hyperbola (5,−10)( 5 , - 10 )(5,−10)

Write an Equation for the Hyperbola $( 5 , - 10 )$