Write the standard form of the equation of the hyperbola subject to the given conditions.
-Vertices: $( - 2,5 ) , ( 10,5 ) ;$ eccentricity $\frac { 5 } { 3 }$
A) $\frac { ( x - 4 ) ^ { 2 } } { 36 } - \frac { ( y - 5 ) ^ { 2 } } { 64 } = 1$
B) $\frac { ( x + 4 ) ^ { 2 } } { 36 } - \frac { ( y + 5 ) ^ { 2 } } { 64 } = 1$
C) $\frac { ( y - 5 ) ^ { 2 } } { 64 } - \frac { ( x - 4 ) ^ { 2 } } { 36 } = 1$
D) $\frac { ( y - 4 ) ^ { 2 } } { 36 } - \frac { ( x - 5 ) ^ { 2 } } { 64 } = 1$

Question

Quizplus · Accepted Answer

The vertices indicate the hyperbola is horizontal, so the $x$-term is positive. The center is the midpoint of the vertices, $(\frac{-2+10}{2}, \frac{5+5}{2}) = (4, 5)$. The distance between the vertices is 12, so $2a = 12$ and $a = 6$, making $a^2 = 36$. The eccentricity $e = \frac{5}{3}$ gives $c = ae$, so $c = 10$ and $c^2 = 100$. Using $c^2 = a^2 + b^2$, we find $b^2 = 64$. Thus, the correct equation is $\frac{(x-4)^2}{36} - \frac{(y-5)^2}{64} = 1$.

Write the Standard Form of the Equation of the Hyperbola (−2,5),(10,5);( - 2,5 ) , ( 10,5 ) ;(−2,5),(10,5);

Write the Standard Form of the Equation of the Hyperbola $( - 2,5 ) , ( 10,5 ) ;$