Find the standard form of the equation of the hyperbola satisfying the given conditions. Foci at $(0,-4)$ and $(0,4)$; vertices at $(0,2)$ and $(0,-2)$

The equation is $\square$.

Step 1 ：The standard form of the equation of a hyperbola with its center at the origin (0,0) is given by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) if the transverse axis is along the x-axis, and \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) if the transverse axis is along the y-axis.

Step 2 ：In this case, the foci and vertices are along the y-axis, so the transverse axis is along the y-axis. The distance from the center to a focus is \( c \), and the distance from the center to a vertex is \( a \).

Step 3 ：Given that the foci are at (0,-4) and (0,4), the distance from the center to a focus is \( c = 4 \).

Step 4 ：Given that the vertices are at (0,2) and (0,-2), the distance from the center to a vertex is \( a = 2 \).

Step 5 ：We can find \( b \) using the relationship \( c^2 = a^2 + b^2 \).

Step 6 ：Let's calculate \( b^2 \) using this relationship. \( a = 2, c = 4 \), so \( b^2 = c^2 - a^2 = 12 \).

Step 7 ：The value of \( b^2 \) is 12. Now we can write the standard form of the equation of the hyperbola. Since the transverse axis is along the y-axis, the equation is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

Step 8 ：Final Answer: The equation of the hyperbola is \( \boxed{\frac{y^2}{4} - \frac{x^2}{12} = 1} \).