Precalculus / Unit 9: Conic Sections and Parametric Equations
9.6 Parametric Equations
Notes
Parametric equations give $x$ and $y$ separately in terms of a parameter $t$.
- Plug in a value of $t$ to find a point.
- To eliminate the parameter, solve one equation for $t$ and substitute, or use $\cos^2 t + \sin^2 t = 1$.
- Increasing $t$ shows the direction the curve is traced.
- Example: $x = t + 1$, $y = t^2$ gives $t = x - 1$ and $y = (x - 1)^2$.
Practice
For $x = 3t$ and $y = t^2 - 1$, what point corresponds to $t = 2$?
$x = 3(2) = 6$ and $y = 4 - 1 = 3$.
Eliminate the parameter: $x = 4\cos t$, $y = 3\sin t$.
$\displaystyle \cos t = \frac{x}{4}$ and $\displaystyle \sin t = \frac{y}{3}$, and $\cos^2 t + \sin^2 t = 1$.
Eliminate the parameter: $x = t + 1$, $y = 2t - 3$.
$t = x - 1$, so $y = 2(x - 1) - 3 = 2x - 5$.
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