Parametric equations are great for motion, but to recognize the shape of the path it helps to have one equation in $x$ and $y$. Removing $t$ is called eliminating the parameter.
Key idea
Either solve one equation for $t$ and substitute into the other, or find an identity that links the two, such as $\cos^2 t + \sin^2 t = 1$ or $\sec^2 t - \tan^2 t = 1$. Then check which $x$- and $y$-values the original equations can actually reach.
Why it works
Every point on the path satisfies both equations for the same $t$. If $t$ can be written in terms of $x$, putting that into the $y$-equation gives a relation that every point of the path satisfies. With trig functions, the Pythagorean identity plays the same role without solving for $t$.
Worked examples
Worked example: Solve for $t$
Eliminate the parameter: $x = 2t - 1$, $y = 3t + 2$.
- From $x$: $t = \dfrac{x + 1}{2}$.
- Substitute: $y = 3 \cdot \dfrac{x + 1}{2} + 2 = \dfrac{3}{2}x + \dfrac{3}{2} + 2$.
Answer: $y = \frac{3}{2}x + \frac{7}{2}$, a line
Worked example: Use an identity
Eliminate the parameter: $x = 3\cos t$, $y = 2\sin t$.
- Isolate the trig functions: $\cos t = \frac{x}{3}$ and $\sin t = \frac{y}{2}$.
- Use $\cos^2 t + \sin^2 t = 1$: $\left(\frac{x}{3}\right)^2 + \left(\frac{y}{2}\right)^2 = 1$.
Answer: $\frac{x^2}{9} + \frac{y^2}{4} = 1$, an ellipse
Worked example: Watch the restrictions
Eliminate the parameter: $x = \sqrt{t}$, $y = t + 2$.
- From $x$: $t = x^2$, so $y = x^2 + 2$.
- But $x = \sqrt{t}$ is never negative, so only the right half of that parabola is traced.
Answer: $y = x^2 + 2$ with $x \ge 0$
Common mistake
Forgetting the restrictions. For $x = t^2$, $y = t^4$ you get $y = x^2$, but $x = t^2$ can never be negative, so the path is only the right half of the parabola. Check the range of $x(t)$ and $y(t)$ before you sketch.