Parametric Curve Derivative Formula Misunderstanding I was reading this site 
on determine the derivative and tangents to a parametric curve. However, he begins by saying:

"suppose that we were able to eliminate the parameter from the parametric form and write the parametric equations in the form $y=F(x$)".

I thought the whole point of parametric equations was that this isn't usually possible (e.g. a circle), and so his argument that follows from there is thus redundant, or am I missing something? 
 A: I agree that the explanation is a little confusing. In my opinion, it's clearer when thinking in terms of linear approximations. Since linear approximations are so fundamental (and often neglected in single variable calculus courses) to understanding more advanced calculus, I will go over them here in case the details are hazy.
Imagine some curvy function, $f(x)$, and pick a point in your mind at $(a, f(a))$. Now imagine a tangent line $T(x)$ at that point. For values of $x$ close to $x=a$, the tangent line is a good approximation for $f(x)$. However, as we go farther away from $x=a$, the tangent line diverges from $f(x)$ and becomes a worse approximation.
Call the change in height (relative to the point we picked above) in the line $\Delta l$. The slope of the tangent line is $\frac{\Delta l}{\Delta x} = f'(a)$, which can be rewritten as $\Delta l = f'(a)\Delta x$. So we can think of $\Delta l$ as a function of $\Delta x$, such that we choose any $\Delta x$ we want, and get a corresponding $\Delta l$.
Now, let $\Delta y$ be the change in the value of $f(x)$, relative to the point we chose earlier. We can say that $\Delta l \approx \Delta y$ for small values of $\Delta x$. To be exact, $\epsilon = \lvert \Delta l - \Delta y\rvert$, where $\epsilon$ is the error. Here is an image to summarize the state of affairs so far:

Again, because $\Delta y \approx \Delta l$ for small $\Delta x$, we have that $$\Delta y \approx f'(a)\Delta x$$
(Notice that, when we make $\Delta x$ infinitely small by taking $\lim_{\Delta x \to 0}$, we have $dy=f'(a)dx$. We chose $x=a$, but we can in fact choose any $x$ and arrive at $dy=f'(x)dx$, which is the definition of the differential.)
So, to summarize, we now have a function that approximates changes in $f$ near some point $x$, given a change in $x$. That is, $\Delta y \approx f'(x)\Delta x$. It bears repeating that this is a function of $\Delta x$. We choose a fixed value of $x$ (therefore $f'(x)$ is fixed at that point), and let $\Delta x$ vary.
(We can arrive at the chain rule by taking one step further. Assume that $x$ is now a function of $u$. Then, $\Delta x \approx x'(u)\Delta u$, which we can plug into the linearization of $f$ and get $\Delta y \approx f'(x)x'(u)\Delta u$. Dividing both sides by $\Delta u$ and taking $\lim_{\Delta u \to 0}$ gives the single variable chain rule.)
Now, consider a parametric curve $x = x(t)$ and $y = y(t)$. You are entirely right that $y$ is often not a valid function of $x$, since parametric curves can easily fail the vertical line test as many times as they want. Despite this, we can find a slope of the curve. Simply pick an arbitrary point (x, y) on the curve (which has a unique value of $t$ associated with it). Now, we can approximate the slope near this point by computing $\frac{\Delta y}{\Delta x}$. Making $\Delta t$ smaller makes the slope closer to the slope AT $(x,y)$.
What are $\Delta x$ and $\Delta y$? Well, we can use the linear approximations $$\Delta x \approx x'(t)\Delta t$$ $$\Delta y \approx y'(t)\Delta t$$
Then, $$\frac{\Delta y}{\Delta x} \approx \frac{y'(t)\Delta t}{x'(t)\Delta t}$$
From here, we can cancel the $\Delta t$s so that $$\frac{\Delta y}{\Delta x} \approx \frac{y'(t)}{x'(t)}$$
Then we take $\lim_{\Delta t \to 0}$ and get $$\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$$
As we make $\Delta t$ smaller, $\Delta y$ and $\Delta x$ get smaller as well, until they eventually become $dy$ and $dx$.
Although I prefer that explanation because it gets more to the heart of what $\frac{dy}{dx}$ truly means when $x=x(t)$ and $y=y(t)$, I will try to answer your question more directly.
It is always true that if $y$ is a function of $x$, and $x$ is a function of $t$, then $\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}$ from the chain rule (provided all the usual conditions are satisfied). You can chop up the parametric curve into a bunch of valid functions of $x$, and put them in a large piecewise function that describes the entire curve. Then, if you're interested in the slope at some point $(x(t_0), y(t_0))$, choose the "piece" on which the point lies, and apply the chain rule to that.
A: It seems to be only a theoretical assumption aimed to derive the given formula for $\frac{dy}{dx}$. Once we have that formula, we don't need to eliminate the parameter but we can use directly the formula to find the derivative from the parametric equation.
A: It is possible to write parametric equations in normal Cartesian coordinates. But in some cases its hard. 
The reason the person writing the notes says so is because he wants to show how one would find $F'(x)= \frac{dy}{dt}\cdot\frac{dt}{dx}$.
His argument  basically says that if $x = f(t), y =g(t)$ and we write the function in terms of $y =F(x)$, we can substiute back $x,y$ to get;
$g(t) = F(f(t))$
differentiating wrt $t$ gives; $g'(t) =F'(f(t))\cdot f'(t) $
$\implies \frac{dy}{dt} =F'(x)\cdot \frac{dx}{dt}$
