Find an equation for the conic that satisfies the given conditions.
hyperbola, foci $( 0 , \pm 6 )$, vertices $( 0 , \pm 3 )$
A) $6 x ^ { 2 } = y$
B) $\frac { ( x - 6 ) ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { 27 } = 1$
C) $\frac { 6 } { 27 } x ^ { 2 } = y$
D) $\frac { x ^ { 2 } } { 36 } + \frac { y ^ { 2 } } { 27 } = 1$
E)
$\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { 27 } = 1$

Question

Quizplus · Accepted Answer

The given hyperbola has its foci and vertices along the y-axis, indicating it opens up and down. The distance between the center and each focus (focal length, $c$) is 6, and the distance between the center and each vertex (vertex length, $a$) is 3. Using the relationship $c^2 = a^2 + b^2$, where $b^2$ is the distance squared from the center to the co-vertices, we find $b^2 = 36 - 9 = 27$. The standard form of a vertical hyperbola is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, leading to $\frac{y^2}{9} - \frac{x^2}{27} = 1$, or equivalently, $\frac{y^2}{36} - \frac{x^2}{27} = 1$.

Find an Equation for the Conic That Satisfies the Given (0,±6)( 0 , \pm 6 )(0,±6)

Find an Equation for the Conic That Satisfies the Given $( 0 , \pm 6 )$