Is it possible to get a focus and directrix straight from the equation itself or through a formula?
For example, in $y = (x-2)^2 + 1$, you can tell from the equation that the vertex is $(2,1)$.
Or to find the midpoint of $(0,4)$ and $(3,6)$, you use $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ and get $(\frac{3}{2},5)$.
Can this be done with parabolas for a directrix and focus?
Since there seems to be no other answers yet I found this, under the section for a general parabola it seems that given the directix line: $$ ax+by+c=0 $$ and the focus at the point $(u,v)$, then the general form of the parabola can be written in this form: $$ \frac{\left(ax+by+c\right)^2}{{a}^{2}+{b}^{2}}=\left(x-u\right)^2+\left(y-v\right)^2 $$
You can see this is correct from this graph.
So for example you could have the focus at $(1,0)$ and the directix as the line $x+1=0$ therefore substituting into the form above: $$ \frac{(x+1)^2}{{1}^{2}+{0}^{2}}=\left(x-1\right)^2+\left(y-0\right)^2$$ which simplifies to give: $y^2=4x$ which if you plot you can check to see that it's correct.
Now I guess you could work in the other direction to manipulate an equation of the parabola into the above form. I would imagine that it could get rather complicated but there are some simplifications that you may be able to apply.
For example you can simplify the general form above into the more commonly seen form: $y^2=4px$ by making a few assumptions. Namely we let the directrix be the line $x+p=0$ and the focus be the point $(p,0)$. If you then apply these simplifications to the general form you get: $$ \frac{(x+p)^2}{1^2+1^2}=(x-p)^2+y^2 \;\; \text{which gives} \;\; y^2=4px$$
Indeed you can make any number of simplifications to see if the parabola you have fits that form, for example the right opening parabola with directix: $x+p=0$ and focus $(u,0)$ gives the general formula: $$y^2=(p+u) (p-u+2 x)$$
You can visualise these using a graphing calculator that allows you to change parameters, for example this link has the above example. The thing I'm looking into and I'll edit this answer to add it if I find out how the general form was derived since it isn't derived nor cited on the wiki page. Hopefully I interpreted your question correctly and this answers at least some of it.