Find the standard form of the equation of the parabola satisfying the given conditions.
Focus: $(2,4)$; Directrix: $x=-6$
The standard form of the equation is
(Type an equation. Simplify your answer.)

Step 1 ：Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix.

Step 2 ：Since the focus is at $(2,4)$ and the directrix is $x=-6$, the vertex of the parabola is at the midpoint of the focus and directrix, which is $\left(\frac{2-(-6)}{2},4\right) = (4,4)$.

Step 3 ：The parabola opens to the right if the focus is to the right of the directrix, and to the left if the focus is to the left of the directrix. Since $2 > -6$, the parabola opens to the right.

Step 4 ：The distance from the vertex to the focus or directrix is $|2-(-6)| = 8$, so $4p = 8$, where $p$ is the distance from the vertex to the focus or directrix. Solving for $p$ gives $p = 2$.

Step 5 ：Since the parabola opens to the right and the vertex is $(4,4)$, the standard form of the equation of the parabola is $(y-4)^2 = 4p(x-4)$.

Step 6 ：Substituting $p = 2$ into the equation gives $(y-4)^2 = 4*2(x-4)$.

Step 7 ：Simplifying the equation gives $(y-4)^2 = 8(x-4)$.

Step 8 ：Thus, the standard form of the equation of the parabola is \(\boxed{(y-4)^2 = 8(x-4)}\).