Construct a table of values like the table below.
| [latex]t[/latex] |
[latex]x=5\cos t[/latex] |
[latex]y=2\sin t[/latex] |
| [latex]\text{0}[/latex] |
[latex]x=5\cos \left(0\right)=5[/latex] |
[latex]y=2\sin \left(0\right)=0[/latex] |
| [latex]\text{1}[/latex] |
[latex]x=5\cos \left(1\right)\approx 2.7[/latex] |
[latex]y=2\sin \left(1\right)\approx 1.7[/latex] |
| [latex]\text{2}[/latex] |
[latex]x=5\cos \left(2\right)\approx -2.1[/latex] |
[latex]y=2\sin \left(2\right)\approx 1.8[/latex] |
| [latex]\text{3}[/latex] |
[latex]x=5\cos \left(3\right)\approx -4.95[/latex] |
[latex]y=2\sin \left(3\right)\approx 0.28[/latex] |
| [latex]\text{4}[/latex] |
[latex]x=5\cos \left(4\right)\approx -3.3[/latex] |
[latex]y=2\sin \left(4\right)\approx -1.5[/latex] |
| [latex]\text{5}[/latex] |
[latex]x=5\cos \left(5\right)\approx 1.4[/latex] |
[latex]y=2\sin \left(5\right)\approx -1.9[/latex] |
| [latex]-1[/latex] |
[latex]x=5\cos \left(-1\right)\approx 2.7[/latex] |
[latex]y=2\sin \left(-1\right)\approx -1.7[/latex] |
| [latex]-2[/latex] |
[latex]x=5\cos \left(-2\right)\approx -2.1[/latex] |
[latex]y=2\sin \left(-2\right)\approx -1.8[/latex] |
| [latex]-3[/latex] |
[latex]x=5\cos \left(-3\right)\approx -4.95[/latex] |
[latex]y=2\sin \left(-3\right)\approx -0.28[/latex] |
| [latex]-4[/latex] |
[latex]x=5\cos \left(-4\right)\approx -3.3[/latex] |
[latex]y=2\sin \left(-4\right)\approx 1.5[/latex] |
| [latex]-5[/latex] |
[latex]x=5\cos \left(-5\right)\approx 1.4[/latex] |
[latex]y=2\sin \left(-5\right)\approx 1.9[/latex] |
Plot the [latex]\left(x,y\right)[/latex] values from the table. See Figure 4.

Next, translate the parametric equations to rectangular form. To do this, we solve for [latex]t[/latex] in either [latex]x\left(t\right)[/latex] or [latex]y\left(t\right)[/latex], and then substitute the expression for [latex]t[/latex] in the other equation. The result will be a function [latex]y\left(x\right)[/latex] if solving for [latex]t[/latex] as a function of [latex]x[/latex], or [latex]x\left(y\right)[/latex] if solving for [latex]t[/latex] as a function of [latex]y[/latex].
[latex]\begin{align}&x=5\cos t \\ &\frac{x}{5}=\cos t&& \text{Solve for }\cos t. \\ &y=2\sin t&& \text{Solve for }\sin t. \\ &\frac{y}{2}=\sin t \end{align}[/latex]
Then, use the Pythagorean Theorem.
[latex]\begin{gathered} {\cos }^{2}t+{\sin }^{2}t=1\\ {\left(\frac{x}{5}\right)}^{2}+{\left(\frac{y}{2}\right)}^{2}=1\\ \frac{{x}^{2}}{25}+\frac{{y}^{2}}{4}=1\end{gathered}[/latex]
Analysis of the Solution
In Figure 5, the data from the parametric equations and the rectangular equation are plotted together. The parametric equations are plotted in blue; the graph for the rectangular equation is drawn on top of the parametric in a dashed style colored red. Clearly, both forms produce the same graph.
