Answer
$\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$
Work Step by Step
The general equation for the hyperbola is $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ ...(1)
Here, vertices: $(0, \pm a); Foci: F(0,\pm c)$ and $c^2=a^2+b^2$
and the asymptotes are: $y=\pm \dfrac{a}{b}x$
We are given that the Foci: $F(0, \pm 4)$
This gives: $c=4$
Now, $c^2=a^2+b^2 \implies (4)^2=a^2+b^2$
and the asymptotes are: $y=\pm \dfrac{a}{b}x=3x$
or, $\dfrac{a}{b}=3 \implies a=3b$
Now, $16=9b^2+b^2 \implies b^2=\dfrac{8}{5}$
Therefore, $a^2=16-b^2 \implies a^2=16-\dfrac{8}{5}=\dfrac{72}{5}$
Thus, the equation (1) becomes:
$\dfrac{y^2}{72/5}-\dfrac{x^2}{8/5}=1$
Hence, the general equation for the Hyperbola is
$\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$
