Find the standard form of the equation of the parabola satisfying the given conditions.
Focus: $(-1,6)$; Directrix: $y=2$

Step 1 ：Given the focus of the parabola at (-1,6) and the directrix at y=2.

Step 2 ：The standard form of the equation of a parabola with a vertical axis is \((x-h)^2=4p(y-k)\), where \((h,k)\) is the vertex of the parabola and \(p\) is the distance from the vertex to the focus or the directrix.

Step 3 ：The vertex is the midpoint between the focus and the directrix. Since the directrix is a horizontal line, the x-coordinate of the vertex is the same as the x-coordinate of the focus, and the y-coordinate of the vertex is the average of the y-coordinate of the focus and the y-coordinate of the directrix.

Step 4 ：Calculating the vertex, we get the x-coordinate as -1 and the y-coordinate as 4.0.

Step 5 ：We can find \(p\) by calculating the distance from the vertex to the focus or the directrix. Here, \(p\) is 2.0.

Step 6 ：Substituting these values into the standard form of the equation of a parabola, we get the equation as \((x + 1)^2 = 8.0*y - 32.0\).

Step 7 ：Simplifying the equation, we get \(y = 0.125*(x + 1)^2 + 4\).

Step 8 ：Final Answer: The standard form of the equation of the parabola satisfying the given conditions is \(\boxed{y = 0.125*(x + 1)^2 + 4}\).