There is no "canonic" way to give a Cartesian system of equations for a parabola in 3D space. The simplest and oldest way is that of giving a parabola as intersection between a plane and a cone, see here for an example.
On the other hand, the locus of points whose distance from a given line (directrix) is the same as their distance from a given point (focus) is a parabolic cylinder, so you may find more natural to give the parabola as the intersection between this cylinder and the plane of focus and directrix.
Here's a picture of the situation (rotated by $90^\circ$), from which, as a bonus, we may deduce the parabola-ness of the curve from the parameterization (without relying on "knowing" it corresponds to $y=\frac{1}{4a}x^2$):
Point $F$ (the parabola's ostensible focus) is $a$ units "up" from origin $O$. Point $P$ is parameterized by $P=(2at,at^2)$, so $|OP'|=2at$ and $|PP'|=at^2$. Let $M$ be the midpoint of $\overline{OP'}$, and let $D$ be the point where the extension of $\overline{PP'}$ meets the ostensible directrix $a$ units "down" from $O$.
Clearly, $\triangle OFM\cong\triangle P'DM$. Also, $\triangle OFM\sim\triangle P'MP$ by the key proportionality of corresponding sides
$$\frac{|OM|}{|OF|}=\frac{at}{a}=t=\frac{at^2}{at}=\frac{|PP'|}{|MP'|} \tag{1}$$
But, then $\triangle MFP$ must be a right triangle whose legs are in the same proportion
$$\frac{|MP|}{|MF|}=t \tag{2}$$
so that $\angle MFP\cong\angle OFM\cong\angle MDP$, and we deduce $\overline{PF}\cong\overline{PD}$: therefore, $P$ lies on the parabola with focus $F$ and directrix through $D$. $\square$
To the question at hand ... By the Reflection Property of Parabolas, the bisector of $\angle FPD$ is tangent to the parabola at $P$. Thus, $\overleftrightarrow{MP}$ is a tangent line, and right-hand half $(1)$ shows that $t$ is its slope. (This is consistent with
@mathlove's comment and @ChiefVS's answer that, in the original horizontally-opening context, the slope of the tangent is $1/t$.)
Of course, the left-hand half of $(1)$, as well as $(2)$, give other interpretations of $t$.
Best Answer
Bézier curves are a convenient way to produce parameterizations of parabolas: a quadratic Bézier is a (part of a) parabola. If $P_0$ and $P_2$ are points on the parabola and $P_1$ the intersection of the tangents at those points, the quadratic Bézier curve they define is given by $$\phi:t\mapsto(1-t)^2P_0+2t(1-t)P_1+t^2P_2.\tag{1}$$ (The parameter $t$ is usually taken to range from $0$ to $1$ for a Bézier patch.)
We can reproduce your parametrization by taking the vertex $P_0(0,0)$ and an end of the latus rectum $P_2(p,2p)$ as the points on the parabola. (Here I use the conventional name $p$ for this parameter instead of the $a$ in the question.) The tangent at the end of the latus rectum meets the parabola’s axis at a 45° angle, so our third control point will be $P_1(0,p)$. Plugging these into (1) we get $$(1-t)^2(0,0)+2t(1-t)(0,p)+t^2(p,2p)=(pt^2,2pt),$$ as required. As described here, parametrization of a parabola by a pair of quadratic polynomials has a nice symmetry about the vertex. Choosing the vertex as our first control point makes this symmetry quite simple.
To obtain the corresponding parameterization for a general parabola, you can either rotate and translate these three points to match the position and orientation of the given parabola, or compute them from other information that you have about the parabola. For example, if we have a parabola with vertex $P_0(x_0,y_0)$, focal length $p$ and axis direction $\theta$, we will have $P_1=P_0+(-p\sin\theta,p\cos\theta)$ and $P_2=P_0+(p\cos\theta-2p\sin\theta,2p\cos\theta+p\sin\theta)$, which gives the parameterization $$\begin{align}x&=x_0-2pt\sin\theta+pt^2\cos\theta \\ y&= y_0+2pt\cos\theta+pt^2\sin\theta.\end{align}$$
I’ll leave working out this parameterization for the general-form equation to you. As a hint, remember that for the parabola $y=ax^2+bx+c$, $p={1\over4a}$ and that a parabola’s vertex is halfway between its focus and directrix.