Write an equation for the hyperbola.
-foci at $( 10 - \sqrt { 53 } , 7 ) , ( 10 + \sqrt { 53 } , 7 ) ; e = \frac { \sqrt { 53 } } { 7 }$
A) $\frac { ( y + 7 ) ^ { 2 } } { 49 } - \frac { ( x + 10 ) ^ { 2 } } { 4 } = 1$
B) $\frac { ( y + ) ^ { 2 } } { 49 } - \frac { ( x + 7 ) ^ { 2 } } { 4 } = 1$
C) $\frac { ( x - 10 ) ^ { 2 } } { 49 } - \frac { ( y - 7 ) ^ { 2 } } { 4 } = 1$
D) $\frac { ( y - 7 ) ^ { 2 } } { 49 } - \frac { ( x + 10 ) ^ { 2 } } { 4 } = 1$

Question

Quizplus · Accepted Answer

The center of the hyperbola is the midpoint of the foci, which is $(10, 7)$. The eccentricity $e = \frac{\sqrt{53}}{7}$ indicates that the distance from the center to each focus along the major axis is $\sqrt{53}$. Since the foci have the same y-coordinate, the major axis is horizontal, making this a horizontal hyperbola. The standard form of a horizontal hyperbola is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, where $(h, k)$ is the center. The distance from the center to a focus is $a \cdot e$, so $a = \frac{\sqrt{53}}{e} = 7$. Thus, $a^2 = 49$. The correct equation is $\frac{(x-10)^2}{49} - \frac{(y-7)^2}{4} = 1$.

Write an Equation for the Hyperbola (10−53,7),(10+53,7);e=537( 10 - \sqrt { 53 } , 7 ) , ( 10 + \sqrt { 53 } , 7 ) ; e = \frac { \sqrt { 53 } } { 7 }(10−53​,7),(10+53​,7);e=753​​

Write an Equation for the Hyperbola $( 10 - \sqrt { 53 } , 7 ) , ( 10 + \sqrt { 53 } , 7 ) ; e = \frac { \sqrt { 53 } } { 7 }$