Find the standard form of the equation of the hyperbola with the given characteristics. vertices: (1,-2), (5,-2)
Asymptotes: $y = \frac { 5 } { 2 } x , y = - \frac { 5 } { 2 } x$
A) $\frac { ( y - 3 ) ^ { 2 } } { 4 } - \frac { ( x + 2 ) ^ { 2 } } { 25 } = 1$
B) $\frac { ( x - 3 ) ^ { 2 } } { 4 } + \frac { ( y + 2 ) ^ { 2 } } { 25 } = 1$
C) $\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y + 2 ) ^ { 2 } } { 25 } = 1$
D) $\frac { ( x - 3 ) ^ { 2 } } { 4 } - \frac { ( y + 2 ) ^ { 2 } } { 25 } = 1$
E) $\frac { ( x - 2 ) ^ { 2 } } { 4 } - \frac { ( y + 3 ) ^ { 2 } } { 25 } = 1$

Question

Quizplus · Accepted Answer

The center of the hyperbola is the midpoint of its vertices, which is $(\frac{1+5}{2}, -2) = (3, -2)$. The distance between the vertices (4 units) is $2a$, so $a=2$ and $a^2=4$. The slopes of the asymptotes ($\pm \frac{5}{2}$) give $b/a = 5/2$, so $b=5$ and $b^2=25$. Since the vertices are horizontal from each other, the $x$ term is positive, and the equation is of the form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, where $(h,k)$ is the center. Thus, the correct equation is $\frac{(x-3)^2}{4} - \frac{(y+2)^2}{25} = 1$.

Find the Standard Form of the Equation of the Hyperbola y=52x,y=−52xy = \frac { 5 } { 2 } x , y = - \frac { 5 } { 2 } xy=25​x,y=−25​x

Find the Standard Form of the Equation of the Hyperbola $y = \frac { 5 } { 2 } x , y = - \frac { 5 } { 2 } x$