Use the form [tex] \left(x-h\right)^2=\frac{1}{4p}\left(y-k \right) [/tex] Then we have the vertex is [tex] (h,k) [/tex] and the distance from the vertex to the focus is p. Therefore we have [tex]\left(x-5\right)^2=-\frac{1}{4}\left(y+2\right)[/tex]. It is negative because the parabola is pointing downwards.
