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