Transforming quadratic parametric curve to implicit form I have an algebraic parametric curve
$$ \mathbf{p}(t) = (x(t), y(t)) $$
where $x$ and $y$ are both polynomials of degree $\leq p$. Now, I want to find the implicit form $f(x, y) = 0$.
A document I'm reading claims without argumentation that there is always an equivalent implicit formulation of the form
$f(x, y) = 0 $, where $f$ is a polynomial of degree $\deg(f) \leq p$.
This is not obvious to me. I found some documents stating similar results. For example, in [1]:

Algebraic geometry provides us with the following key facts about algebraic curves: e.g. [2]:
Every plane parametric curve can be expressed as an implicit curve. Some, but not all, implicit curves can be expressed as parametric curves.

Unfortunately, I can't get my hands on a copy of [2].
[This question][3] considers the case of a quadratic parametric curve.
I was wondering if there is a simple proof for the statement every algebraic parametric curve on the plane can be expressed as an implicit curve. Moreover, I am curious if $x$ and $y$ being polynomials of degree $\leq p$ implies that $\deg(f) \leq p$ as well.
Since I don't know the nature of the answer, pointing me to a resource might be more appropriate.
[1]: Conversion methods between parametric and implicit curves and surfaces, Christoph M. Hoffmann, from http://graphics.stanford.edu/courses/cs348a-17-winter/Handouts/a228715.pdf
[2]: R.J. Walker, Algebraic Curves, Springer Verlag, New York, 1978.
[3]: Parabola in parametric form
 A: To begin with, you are dealing with the "great old theory" of elimination.
In this case, the answer is : yes, it is always possible to eliminate parameter $t$ between polynomial expressions and obtain an implicit  polynomial expression. 
Let us show it on an example, with a parametric curve defined by these equations :
$$\tag{1}\begin{cases}x=t^2+t+1\\y=t^3-1\end{cases}$$
If you use a Computer Algebra System such as Mathematica, elimination process is implemented under the following form: Eliminate[{x==t^2+t+1,y==t^3-1},t] giving the following implicit equation : 
$$\tag{2}x^3-3x^2-3xy-y^2=0$$
as its result.
But there is more to say. You need for this to be familiar with resultants (see remark below) (https://en.wikipedia.org/wiki/Resultant). The theorem you are looking for is the fact that the implicit equation can be obtained as the resultant of the $\color{red}{2}$nd degree polynomial $t^2+t+(1-x)$ (I was tempted to write $=0$...) and $\color{red}{3}$rd degree polynomial $t^3-(1+y)$ which the $5 \times 5$ parametric determinant (5 = $\color{red}{2+3}$):
$$d=\left|\begin{array}{ccccc}1& 1& (1-x)& 0 &0 \\
   0& 1& 1& (1-x)& 0\\
   0& 0& 1& 1& (1-x)\\
   1& 0& 0& -(1+y)& 0\\
   0& 1& 0& 0& -(1+y)\end{array}\right|=0$$
(one writes $\color{red}{3}$ times the $\color{red}{2}$nd degree polynomial, then $\color{red}{2}$ times the $\color{red}{3}$rd degree one, with a right shift for every "carriage return").
The expansion of determinant $d$ gives back (up to a certain unimportant constant factor) implicit equation (2).
This example can evidently been extended to a general case. 
Remark 1: The excellent reference (Cox, Little, O'Shea "Ideals, Varieties and Algorithms") given by @Jan-Magnus Økland can be found online. See pages 155-162 devoted to Sylvester's resultant.
Remark 2: the main idea behind the concept of resultant is that it expresses a necessary and sufficient condition between parameter(s) present in coefficients of 2 equations for these equations to have at least a common root. Let us take an example : consider a quadratic polynomial $at^2+bt+c$ (with $a\ne0$) and its derivative $2at+b$. They have a common root $t_0$ which is a double root iff the following resultant is zero:
$$\delta=\left|\begin{array}{ccc}a& b& c \\
   2a& b& 0\\
   0& 2a& b\end{array}\right|=-a(b^2-4ac)$$
This value is $0$ iff $b^2-4ac=0$, which is the classical criteria for a quadratic equation to have a double root. We have thus found back the concept of discriminant by using the resultant!
A: Hint Another idea based on my comment which may help. The tangent vectors along the parametric curve (using same notation as Jean Marie) will be given by
$$\left[\begin{array}{c}\frac{\partial x}{\partial t}\\\frac{\partial y}{\partial t}\end{array}\right]$$
These should be orthogonal to the gradient of the polynomial level-set you are seeking. This should give you a set of quite specific equations as the only thing remaining should be constant (quite literally the particular level you want to set your curve on).
