Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin.
Vertices: (0,±5);focies: (0,±6)

A) $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 11 } = - 1$
B) $\frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 11 } = 1$
C) $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 11 } = 1$
D) $\frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 11 } = - 1$
E) $- \frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 11 } = - 1$

Question

Find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. ​
Vertices: (0,±5);focies: (0,±6)
​
A) $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 11 } = - 1$
B) $\frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 11 } = 1$
C) $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 11 } = 1$
D) $\frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 11 } = - 1$
E) $- \frac { y ^ { 2 } } { 25 } + \frac { x ^ { 2 } } { 11 } = - 1$

Quizplus · Accepted Answer

The standard form of a hyperbola with a vertical transverse axis is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, where $2a$ is the distance between the vertices, and $c$ is the distance from the center to each focus. Given vertices at (0, ±5), $a = 5$, so $a^2 = 25$. With foci at (0, ±6), $c = 6$, so $c^2 = 36$. Using $c^2 = a^2 + b^2$, we find $b^2 = 36 - 25 = 11$. Thus, the correct equation is $\frac{y^2}{25} - \frac{x^2}{11} = 1$.

Find the Standard Form of the Equation of the Hyperbola y225−x211=−1\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 11 } = - 125y2​−11x2​=−1

Find the Standard Form of the Equation of the Hyperbola $\frac { y ^ { 2 } } { 25 } - \frac { x ^ { 2 } } { 11 } = - 1$