On The Parametric Equation Of A Parabola Let's look at a parabola with an equation $(y-k)^2=4a(x-h)$. I'm struggling to understand why its parametric equation would be $x=h+at^2$ and $y=k+2at$. Since it is being subtracted by $h$ and $k$ respectively, why would its $x$ and $y$ values increase/have $h$ and $k$ added to them instead? 

I understand that since each $x$-value is being subtracted by $h$, each $x$-value needs to be $h$ bigger to "achieve" the same y-value. But that also means that for each $y$-value, each $x$-value will be h smaller right? So how does subtracting by $h$ increase the value of $x$ by $h$, as shown in the parametric equation $x=h+at^2$? I ask the same question for $y=k+2at$.

Anyways, I have no idea what the "x" output the parametric equation gives out even means? Does the parametric equation give out the x-value which will be inputted into the equation? But then no matter where the graph is, that doesn't change what x-value would be inputted into the equation, right? So why would subtracting x by h change the parametric equation of x either, if x will remain unchanged?

Or does the parametric equation give out the x-value as in the x-value to be plotted on the graph? But then if so, I still don't get why the sign would be +h, instead of -h. Yes, in the graph, for every y-value, the x-value of a graph y^2=4a(xx-h) would be left of the original graph. But how do we know the parametric equation refers to the x-value for every given y-value. And if that is not what is referred to by the parametric equation, why are we subtracting?

Can someone explain this to me as simply as possible, since I'm still a beginner. I'm just putting this as a precaution, and try not to use Calculus in the response, since I haven't learnt it yet.
Let's start with the simplest case, in which $h$ and $k$ are both $0$.  Then the parabola is
$$y^2 = 4ax$$
and the parametric equation is
$$y=2at, \hskip{0.5in} x=at^2$$
and you can directly verify that 
$$y^2 = (2at)^2 = 4a^2t^2 = 4a(at^2) = 4ax$$
so everything works nicely.

Now let's consider the general case.  We have
$$(y-k)^2 = 4a(x-h)$$
If we introduce (temporarily) the new variables $u = x-h, v = y-k$ then the equation can be written as
$$v^2 = 4au$$
which has exactly the form of the simple case we already considered.  So we know the parametric equations for $u$ and $v$ are
$$v=2at, \hskip{0.5in} u=at^2$$
But now we remember that $u = x-h$ and $v = y-k$.  That means the parametric solutions are
$$y-k=2at,\hskip{0.5in}  x-h=at^2$$
which leads directly to the solution
$$y=k+2at,\hskip{0.5in}  x=h+at^2$$

Now let's look back and try to understand what's happening.  The main idea is that the "shift" induced by replacing $x$ and $y$ with $x-h$ and $y-k$, respectively, corresponds to a shift in the solution that also replaces $x$ and $y$ with $x-h$ and $y-k$.  The "plus" sign just comes from moving the negative terms to the other side of the equals sign.
Let us try
$$(y-k)^2=4a(x-h)=4a^2t^2,$$
then
$$y=k\color{red}\pm 2at,\quad x=h+at^2.\tag1$$
Formulas $(1)$ take in account that the parabola has two branches.

Therefore, the issue parametric function presents only one branch of parabola.