Answer
$\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$
Work Step by Step
The standard form for the Hyperbola is $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$
with Vertices:$(0, \pm a); Foci: F(0,\pm c); c^2=a^2+b^2$
and the asymptotes are: $y=\pm \dfrac{a}{b}x$
Given: Foci: $F(0, \pm 4)$
and $Foci: F(0,\pm c)= F(0, \pm 4) \implies c=4$
and $ (4)^2=a^2+b^2$
and the asymptotes are: $y=\pm \dfrac{a}{b}x=3x$
or, $ a=3b$
Now, $16=9b^2+b^2 $
and $b^2=\dfrac{8}{5}$
Therefore,
$a^2=16-\dfrac{8}{5}=\dfrac{72}{5}$
The standard form for the Hyperbola is $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$
$\dfrac{y^2}{(72/5)}-\dfrac{x^2}{(8/5)}=1$
Hence, $\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$
