I was wondering what the focus point and directrix of a general parabola 
I’m working through Adams Calculus a Complete Course ninth edition; on page 19 the parabola with focus point $F(0,p)$ and directrix $y=-p$ is given as $4py = x^2$, the vertex is $(0,0)$.
First realise that $4p(y-k) = (x-h)^2$ has $(h,k)$ as vertex. Writing $ax^2+bx+c$ in the form of $4p(y-k)=(x-h)^2$ will help us find its focus point, directrix and vertex, all expressed in $a, b$ and $c$.
$$
\begin{aligned}
y &=a x^{2}+b x+c \\
y-c &=a x^{2}+b x \\
\frac{y-c}{a} &=x^{2}+\left(\frac{b}{a}\right) x \\
\frac{y-c}{a}+\left(\frac{b}{2 a}\right)^{2} &=x^{2}+\left(\frac{b}{a}\right) x+\left(\frac{b}{2 a}\right)^{2} \\
\frac{1}{a} \left(y-c+\frac{b^{2}}{4 a}\right) &=\left(x+\frac{b}{2 a}\right)^{2} \\
4 \cdot \underbrace{\frac{1}{4 a}}_{p} (y-\underbrace{\left(c-\frac{b^{2}}{4 a}\right)}_{k}) &=(x-\underbrace{\left(\frac{-b}{2 a}\right)}_{h})^{2}
\end{aligned}
$$
If you don’t understand the adding of $\left(\frac{b}{2a}\right)^2$, just think of $x^2+6x+9 = (x+3)^2$, where $9 = \left(\frac{6}{2}\right)^2$.
The focus point is:
$
\begin{aligned}
(h, k+p) &= \left(-\frac{b}{2 a}, c-\frac{b^{2}}{4 a}+\frac{1}{4 a}\right) \\
&= \left(-\frac{b}{2 a}, c+\frac{1-b^{2}}{4 a}\right)
\end{aligned}
$
The directrix is:
$
\begin{aligned}
y &= k-p \\
&= c-\frac{b^{2}}{4 a}-\frac{1}{4 a} \\
&= c-\frac{1+b^{2}}{4 a}
\end{aligned}
$
The vertex is: $(h,k) = \left(-\frac{b}{2 a}, c-\frac{b^{2}}{4 a}\right)$.
The distance between the focal point and the vertex is $p$, which is $\frac{1}{4a}$. Why would you want to know this? Because you might want to know where to put the pan in the solar oven, or the light bulb in a parabola shaped mirror. The funny thing is that a mathematician might be interested in it just because it is possible to calculate this distance, so for no practical reason.
Just in case that you wonder if it is wright that $4p(y-k) = (x-h)^2$ is a parabola with $(h,k)$ as vertex. Let us take focus point $(h,k+p)$ and line $y=k-p$, now let us see where the points $(x,y)$ are that have the same distance to the line and the focus point:
$
\require{cancel}
\begin{aligned}
\sqrt{(x-x)^{2}+(y-(k-p))^{2}} &= \sqrt{(x-h)^{2}+(y-(k+p))^{2}} \\
\left(\sqrt{(x-x)^{2}+(y-(k-p))^{2}}\right)^2 &= \left(\sqrt{(x-h)^{2}+(y-(k+p))^{2}}\right)^2 \\
(x-x)^{2}+(y-(k-p))^{2} &= (x-h)^{2}+(y-(k+p))^{2} \\
0^{2}+(y-(k-p))^{2} &= (x-h)^{2}+(y-(k+p))^{2} \\
\cancel{y^2}-2(k-p)y+(k-p)^2 &= (x-h)^{2}+\cancel{y^2}-2(k+p)y+(k+p)^2 \\
-2(k-p)y+(k-p)^2 &= (x-h)^{2}-2(k+p)y+(k+p)^2 \\
-\cancel{2ky} +2py +\cancel{k^2}-2kp+\cancel{p^2} &= (x-h)^{2} -\cancel{2ky} -2py +\cancel{k^2} +2kp+\cancel{p^2} \\
2py -2kp &= (x-h)^{2} -2py +2kp \\
2py +2py -2kp -2kp &= (x-h)^{2} \\
4py – 4kp &= (x-h)^{2} \\
4p(y – k) &= (x-h)^{2}
\end{aligned}
$