What is the method to find the parametric equations for all types of parabolas (in both directions)?
So if I had $2$ points then:
Parametrise from $A$ to $B$ where $A = \left( \frac{3}{\sqrt2} , 9 \right)$ and $B = ( 0, 0 )$.
What would $x(t)$ and $y(t)$ be?
Is there a general way to solve all parabolic parametric equations? If yes, then how?
Probably the simplest approach is to let $x=t$ and $z=2t^2$, so we can easily eliminate $t$ and get $z=2x^2$. Now we just let $t$ take values on the interval $\left[0,\frac{3}{\sqrt{2}}\right]$. But... this isn't quite right.
We want to start at $\frac{3}{\sqrt{2}}$ and end at $0$. The above parameterization does the opposite. To reverse it, let's let $x=\frac{3}{\sqrt{2}}-t$, and $z=2\left(\frac{3}{\sqrt{2}}-t\right)^2$. We can still eliminate $t$ to get $z=2x^2$, but now, when $t=0$, we're at the $x$-value $\frac{\sqrt{3}}{2}$, and when $t=\frac{\sqrt{3}}{2}$ at the end of its interval, we're at the $x$-value $x=0$.
There are infinitely many other ways to parameterize this. Sometimes, people like to have $t$ come from the interval $[0,1]$; we could do that. Alternatively $x$ could depend on $t$ in some other, non-linear fashion, as long as we can still eliminate $t$ and get $z=x^2$, we can do a lot of different things.
In fact, let $f(t)$ be any function that decreases monotonically on the interval $[a,b]$, with $f(a)=\frac{\sqrt{3}}{2}$ and $f(b)=0$. Then we can define a parameterization by $x(t)=f(t)$ and $z(t)=2(f(t))^2$, where we get the desired curve by letting $t$ run from $a$ to $b$. In my second paragraph above, I did this, using a linear function for $f$.
Even that description doesn't cover all the ways this problem could be solved. It's very open-ended.