Answer
$\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$
Work Step by Step
Given: Foci: $F(0, \pm 4)$ This gives: $c=4$
Let us consider the standard equation for the Hyperbola: $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$
with vertices:V$(0, \pm a); Foci: F(0,\pm c); c^2=a^2+b^2$
and Asymptotes: $y=\pm \dfrac{a}{b}x$
Now, $c^2=a^2+b^2$
or, $(4)^2=a^2+b^2$
Asymptotes: $y=\pm \dfrac{a}{b}x=3x$
This implies that $a=3b$
and $16=9b^2+b^2$ or, $b^2=\dfrac{8}{5}$
Therefore, $a^2=16-b^2$
or, $a^2=16-\dfrac{8}{5}=\dfrac{72}{5}$
Hence, we have
$\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$
