## Calculus 8th Edition

$\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$
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, $\dfrac{5y^2}{72}-\dfrac{5x^2}{8}=1$