Find the standard form of the equation of the hyperbola satisfying the given conditions. Foci at (0,-6) and (0,6); vertices at (0,2) and (0,-2)

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, since the foci and vertices are along the y-axis, the transverse axis is along the y-axis. Therefore, the equation of the hyperbola is of the form: \[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]

Step 3 ：The distance from the center to a focus is c, and the distance from the center to a vertex is a. Therefore, we have: \[c = 6\] and \[a = 2\]

Step 4 ：We also know that for a hyperbola, \[c^2 = a^2 + b^2\]. We can use this equation to solve for b.

Step 5 ：Substituting the values of a and c into the equation, we get \[b = \sqrt{c^2 - a^2} = \sqrt{6^2 - 2^2} = \sqrt{32} = 5.656854249492381\]

Step 6 ：Substituting the values of a and b into the equation of the hyperbola, we get \[\frac{y^2}{4} - \frac{x^2}{32} = 1\]

Step 7 ：\(\boxed{\frac{y^2}{4} - \frac{x^2}{32} = 1}\) is the standard form of the equation of the hyperbola.