Translating standard form quadratics to focus-directrix form
Standard form
\[ax^2+bx+c\]Discriminant
\[D=b^2-4ac\] \[h=-\frac{b}{2a}\] \[k=-\frac{D}{4a}\]Vertex
\[V=(h,k)\] \[f=\frac{1}{4a}\]Focal length \(\vert f \vert\)
Focus
\[F=(h,k+f)\]Directrix
\[y=d, \qquad d=k-f\]Distance between \((x,y)\) and directrix is
\[|y-(k-f)|\]Distance between \((x,y)\) and focus is
\[\sqrt{(x-h)^2+(y-(k+f))^2}\]Therefore
\[\begin{align} (y-(k-f))^2&=(x-h)^2+(y-(k+f))^2\\ y^2 - 2(k-f)y + (k-f)^2&=(x-h)^2+y^2-2(k+f)y+(k+f)^2\\ y^2 - 2ky + 2fy + k^2 -2fk + f^2&=(x-h)^2+y^2-2ky-2fy + k^2 + 2fk + f^2\\ 2fy -2fk&=(x-h)^2 -2fy + 2fk\\ 4fy - 4fk&=(x-h)^2\\ 4f(y-k)&=(x-h)^2 \end{align}\]