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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: 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!

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!

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: 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 $t_0$ 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!

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: 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!

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 = $\color{red}{2+3}$):

$$d=\left|\begin{array}{ccccc}1& 1& (1-x)& 0 &0 \\ 0& 1& 1& (1-x)& 0\\ 0& 0& 1& 1& (1-x)\\ 1& 0& 0& -(1+y)& 0\\ 0& 1& 0& 0& -(1+y)\end{array}\right|$$

(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 shift for every "carriage return").

The expansion of determinant $d$ gives back (up to a certain unimportant constant factor) implicit equation (2).

Remark: 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 $t_0$ 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!

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: 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 $t_0$ 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!