Example of parameterizing equations
$$
\left\{\begin{array}{l}
x=\frac{-8 k}{4+k^{2}} \\
y=\frac{16-4 k^{2}}{4+k^{2}}
\end{array}\right.
$$

into a common equation, where $\mathrm{k}$ is the parameter.
$$
\begin{array}{l}
\text { Solution One (Single Substitution Method) } \\
\text { (1) Squaring (1) gives } \\
x^{2}=\frac{64 k^{2}}{\left(4+k^{2}\right)^{2}} \text {. } \\
\text { From (2), } k^{2}=\frac{4(4-y)}{y+4} \cdot(y+4 \neq 0) \\
\text { Substituting into (3) and simplifying gives } \\
\frac{x^{2}}{4}+\frac{y^{2}}{16}=1 \text {. } \\
\text { Solution One can be called the "Single Substitution Method", which involves using one } \\
\text { of the equations in the parametric equation to solve for the parameter (or a function of the } \\
\text { parameter) }[\mathrm{k}=\varphi(\mathrm{x}) \text { or } \mathrm{k}=\phi(\mathrm{y}) \text {; } \\
\mathrm{f}(\mathrm{k})=\varphi(\mathrm{x}) \text { or } \mathrm{f}(\mathrm{k})=\psi(\mathrm{y}) \mathrm{J} \text {, and then substituting it into the other equation of the parametric } \\
\text { equation. } \\
\text { Solution Two (Comprehensive Substitution Method) } \\
\text { From (2), } \mathrm{y}+4=\frac{32}{4+\mathrm{k}^{2}} \text {. } \\
\text { Dividing (1) by (3) gives } \frac{x}{y+4}=-\frac{k}{4} \text {, i.e., } k=-\frac{4 x}{y+4} \text {. } \\
\text { Substituting into (1) and simplifying gives } \\
\frac{x^{2}}{4}+\frac{y^{2}}{16}=1 \text {. } \\
\text { Compared to Solution One, this method involves using both equations of the parametric } \\
\text { equation to solve for } k \text { in terms of } x \text { and } y \text {, i.e., } k=F(x, y) \text {, } \\
\text { and then substituting it back into one of the equations (usually the simpler one) to obtain the } \\
\text { ordinary equation. Hence, it is called the "Comprehensive Substitution Method" or "Reverse Substitution Method". } \\
\text { Solution Three (Reciprocal Method) } \\
\text { (1) can be rewritten as } \frac{4}{k}+k=-\frac{8}{x} \text {. } \\
\text { Dividing (2) by (1) gives } \frac{4}{k}-k=-\frac{2 y}{x} \text {. } \\
\text { From (3) and (4), we get } \\
\frac{4}{k}=-\frac{y+4}{x} \text {, } \\
\mathrm{k}=\frac{y-4}{x} \text {. } \\
\text { Multiplying these two equations and simplifying gives } \frac{x^{2}}{4}+\frac{y^{2}}{16}=1 \text {. } \\
\text { Solution Three involves using the parametric equations to find two expressions, one containing } \\
\mathrm{k} \text { and the other containing } \frac{1}{k} \text {, and then multiplying them to eliminate the parameter, hence it is } \\
\text { called the "Reciprocal Method" or "Product Constant Method". } \\
\text { Solution Four (Determinant Method) } \\
\text { (1) and (2) can be written as } \\
\left\{\begin{array}{l}
\mathrm{xk}^{2}+8 \mathrm{k}=-4 \mathrm{x}, \\
(\mathrm{y}+4) \mathrm{k}^{2}=4(4-y) .
\end{array}\right. \\
\text { Using the knowledge of linear equations, we have } \\
D=\left|\begin{array}{ll}
x & 8 \\
y+1 & 0
\end{array}\right|=-8(y+4) \neq 0 . \\
D_{\mathrm{K} 2}=\left\{\begin{array}{cc}
-4: & 8 \\
4(4-y) & 0 \\
x & -4 x
\end{array}=-32(4-y),\right. \\
D_{\mathrm{R}}=\left|\begin{array}{ll}
x & -4 x \\
y+4 & 4(4-y)
\end{array}\right|=32 x_{0} \\
\text { Thus, } \square \\
\left\{\begin{array}{l}
k^{2}=\frac{4(4-y)}{4+y}, \\
k=-\frac{4 x}{y+4} .
\end{array}\right. \\
\text { From this, we get } \\
\frac{x^{2}}{4}+\frac{y^{2}}{16}=1 \text {. } \\
\text { Solution Four treats the two equations of the parametric equation as two unknowns, solves them using } \\
\text { a system of linear equations, and then transforms to eliminate the parameter, called the "Determinant Method". } \\
\text { It is easy to see that Solution Three is sometimes a special case of this method. } \\
\text { Solution Five (Sum Constant Method) } \\
\text { The original parametric equation can be rewritten as } \\
\int \frac{\mathrm{x}}{-2}=\frac{4 \mathrm{k}}{\mathrm{k}^{2}+4} \text {, } \\
\frac{y}{4}=\frac{4-k^{2}}{k^{2}+4} \text {. } \\
\text { (3) } \\
{ }^{2}+(4)^{2} \text { simplifying gives } \\
\frac{x^{2}}{4}+\frac{y^{2}}{16}=1 \text {. } \\
\text { Using the product of two equations with a constant sum to eliminate the parameter requires } \\
\text { careful observation, attention to coefficient and constant term transformations, to be effective. } \\
\text { If used properly, it can simplify the solution, hence it is called the "Sum Constant Method". Consider another example: } \\
\text { Eliminate the parameters } x_{1}, y_{1} \\
\mathrm{x}=\frac{\mathrm{x}_{1}{ }^{2}+4 \mathrm{p}^{2}}{2 \mathrm{x}_{1}}, \\
y=\frac{y_{1}{ }^{2}-4 p^{2}}{2 y_{1}} . \\
\text { where } \mathrm{y}_{1}{ }^{2}=2 \mathrm{px}_{1}, \mathrm{p} \neq 0 \text {. } \\
\text { Solution The given parametric equation can be written as } \\
x=\frac{y_{1}{ }^{4}+16 p^{4}}{4 \mathrm{p} T_{1}^{2}} \text {. } \\
y=\frac{v_{1}{ }^{2}-4 p^{2}}{2 y_{1}} \text {. } \\
\text { (4) }{ }^{2}+(3) \times(-p) \text { simplifying gives } \\
\text { Solution Six } \text { (Trigonometric Method) } \\
\text { Let } \mathrm{k}=2 \mathrm{~g} \frac{\alpha}{2} \text {, we get } \\
-\frac{x}{2}=\sin \alpha, \\
\frac{1}{4}=\cos \alpha \text {. } \\
\text { (3) }{ }^{2}+(4)^{2} \text { gives } \frac{x^{2}}{4}+\frac{y^{2}}{16}=1 \text {. } \\
\end{array}
$$

The trigonometric method involves making a trigonometric substitution for the parameter, and then using trigonometric formulas, especially those with constant terms (such as $\sin ^{2} \alpha+\cos ^{2} a=1, \sec ^{2} \theta-\operatorname{tg}^{2} \theta=1$).
$$