Eliminating Parameters in Parametric Curves

# Eliminating Parameters in Parametric Curves

We are now going to look at taking a parametric curve $C$ defined by the parametric equations $x = f(t)$ and $y = g(t)$ and eliminate the parameter $t$ so that we write $C$ in terms of $x$ and $y$ only.

 Note: While eliminating the parameter may be nice in verifying a type of curve, often times parameter eliminating will result in more complex equations.

## Example 1

Eliminate the parameter $t$ of the parametric curve $C$ defined by the equations $x = 3t + 1$ and $y = t^2$.

We will first solve for $t$ by isolating it from the $x$-equation to get:

(1)
\begin{align} x = 3t + 1 \\ 3t = x - 1 \\ t = \frac{x - 1}{3} \end{align}

Now we will substitute this value of $t$ into the $y$-equation as follows:

(2)
\begin{align} y = t^2 \\ y = \left ( \frac{x - 1}{3} \right )^2 \end{align}

We have thus eliminated the parameter $t$ and have written $C$ in terms of $x$ and $y$ only.

## Example 2

Eliminate the parameter $t$ of the parametric curve $C$ defined by the equations $x = 3 \cos t$ and $y = 4 \sin t$.

First let's isolate $t$ from the $x$-equation:

(3)
\begin{align} x = 3 \cos t \cos t = \frac{x}{3} \\ t = \cos ^{-1} \left ( \frac{x}{3} \right ) \end{align}

Substituting this value of $t$ into the $y$-equation we get that:

(4)
\begin{align} y = 4 \sin \left (\cos ^{-1} \left ( \frac{x}{3} \right ) \right) \end{align}

Now recall the following trigonometric identity $\sin ^2 a + \cos ^2 a = 1$. Suppose that we let $a = \cos ^{-1} x$, and then:

(5)
\begin{align} \sin ^2 (\cos ^{-1} x) + \cos ^2 (\cos ^{-1} x) = 1 \\ \sin ^2 (\cos ^{-1} x) + \cos (\cos ^{-1} x)\cos (\cos ^{-1} x) = 1 \\ \sin ^2 (\cos ^{-1} x) + x^2 = 1 \\ \sin ^2 (\cos ^{-1} x) = 1 - x^2 \\ \sin (\cos^{-1} x) = \sqrt{1 - x^2} \end{align}

It follows that:

(6)
\begin{align} y = 4 \sin \left (\cos ^{-1} \left ( \frac{x}{3} \right )^2 \right) \\ y = 4 \sqrt{1 - \left ( \frac{x}{3} \right )^2} \\ y = 4 \sqrt{\frac{9}{9} - \frac{x^2}{9}} \\ y = \frac{4}{3} \sqrt{9 - x^2} \\ \frac{3}{4} y = \sqrt{9 - x^2} \\ \frac{9}{16} y^2 = 9 - x^2 \\ x^2 + \frac{9}{16} y^2 = 9 \\ 16x^2 + 9y^2 = 144 \\ 16x^2 + 9y^2 = 12^2 \end{align}

We have thus eliminated the parameter $t$.