Write the standard form of the equation of the hyperbola centered at the origin. Vertices: $(0,-3),(0,3)$ Asymptotes: $y=\frac{1}{4}x,y=-\frac{1}{4}x$

A) $\frac{y^2}{16}-\frac{x^2}{9}=1$
B) $\frac{y^2}{9}-\frac{x^2}{9/16}=1$
C) $(0,10)\left(\frac{3}{2},-\frac{31}{4}\right)$
D) $\frac{y^2}{9}-\frac{x^2}{144}=1$
E) $\frac{x^2}{9}-\frac{y^2}{144}=1$

Since the hyperbola is centered at the origin, the standard form of the equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $a$ represents the distance from the origin to the vertices, and $b$ represents the distance from the origin to the endpoints of the conjugate axis.

The distance between the center and each vertex is $3$, so $a=3$. The distance between the center and the endpoints of the conjugate axis is $b$, which can be found using the asymptotes. The slopes of the asymptotes are $\pm\frac{1}{4}$, so the slope of the line passing through the center and one endpoint of the conjugate axis is $\pm 4$. Using the distance formula, we have
$$b^2={\left(\frac{1}{2}\cdot \text{distance between endpoints}\right)}^2={\left(\frac{1}{\sqrt{1^2+4^2}}\cdot 6\right)}^2=\frac{36}{17}$$
Thus, the standard form of the equation is $\frac{x^2}{9}-\frac{y^2}{\frac{36}{17}}=1$, which can also be written as $\frac{x^2}{9}-\frac{y^2}{\frac{4}{17}}=1$. Multiplying both sides by $\frac{4}{17}$, we get $\frac{x^2}{\frac{36}{17}}-\frac{y^2}{4}=1$, which matches the equation in choice (D). Therefore, the answer is D.