Let \(F=(h,c)\) be the focus and \(y=d\) be the directrix of the parabola.

Let \(V=(h,k)\) be the vertex with the parabola defined by \(y=\dfrac{1}{4f}(x-h)^2+k\), with focal length \(\vert f \vert\).

\[c=k+f\] \[d=k-f\]

On the one hand, adding \(k=c-f\) and \(k=d+f\) gives \(2k=c+d\), so

\[k = \dfrac{c+d}{2}.\]

On the other hand, adding \(f=c-k\) and \(f=k-d\) gives \(2f=c-d\), so

\[f=\dfrac{c-d}{2}.\]

Therefore, given the focus

\[F=(h,c)\]

and the directrix

\(y=d\),

we have found that

\[V=(h,k) = \left(h,\dfrac{c+d}{2} \right),\] \[f = \dfrac{c-d}{2},\]

and the vertex form of the parabola is

\[y=a(x-h)^2+k,\]

for \(a=\dfrac{1}{4f}\).

Focus and directrix to vertex form