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 hyperbola is vertical with center at the origin, vertices at (0, ±3), and foci at (0, ±6). The equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ with $a = 3$ and $c = 6$, so $b^2 = c^2 - a^2 = 36 - 9 = 27$.

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 $( 0 , \pm 6 )$