Eliminating the Parameter, precalculus lesson

A table of points from $x = f(t)$ and $y = g(t)$ often looks like a familiar shape, a line, a parabola or a circle. To be sure, get rid of $t$ and see which equation in $x$ and $y$ the points satisfy. The catch is that the new equation usually says more than the parametric one about where the curve is, and less about how it is traced, so a careful answer keeps a short record next to it.

Key terms

Rectangular equation

Definition A rectangular equation of a parametric curve $x = f(t)$ , $y = g(t)$ , $t \in I$ , is an equation in $x$ and $y$ alone that is satisfied by every point $(f(t), g(t))$ with $t \in I$ . The curve is always contained in the graph of such an equation. It equals the whole graph only when every point of the graph is reached by some $t$ in $I$ . A rectangular equation is also called a Cartesian equation.

In plain words The rectangular equation names the familiar curve the point travels on. It does not promise that the point visits all of it.

Example For $x = 3 - 2t$ , $y = 1 + 4t$ with $t$ any real number, solving the first equation gives $t = \frac{3 - x}{2}$ , and then $y = 1 + 2(3 - x) = 7 - 2x$ . Every point of this line is reached, because any $x$ comes from $t = \frac{3 - x}{2}$ . At $t = 1$ , for instance, the point is $(1, 5)$ , and indeed $5 = 7 - 2 \cdot 1$ .

Elimination by substitution

Definition Elimination by substitution means solving one of the equations $x = f(t)$ or $y = g(t)$ for $t$ in terms of $x$ or $y$ and substituting the result into the other equation. This is valid on any part of the parameter interval where the solved equation can be inverted, that is, where each value of $x$ (or $y$ ) comes from only one $t$ .

In plain words Pick the equation that is easier to solve for $t$ , usually a linear one, solve it, and plug $t$ into the other equation.

Example For $x = 2t + 1$ , $y = t^2 - 3$ with $t$ any real number, the first equation gives

t = \frac{x - 1}{2}, \qquad y = \frac{(x - 1)^2}{4} - 3.

This is a parabola with vertex $(1, -3)$ . Since $x = 2t + 1$ takes every real value, the whole parabola is traced, from left to right as $t$ increases.

Elimination by an identity

Definition When $t$ appears inside trigonometric functions, eliminate it with an identity that holds for every $t$ in the parameter interval, such as

\begin{gathered} \cos^2 t + \sin^2 t = 1, \\[0.4em] 1 + \tan^2 t = \sec^2 t, \\[0.4em] \cos 2t = 1 - 2\sin^2 t. \end{gathered}

First solve each coordinate equation for the trigonometric expression it contains, then substitute into the identity. The identity $1 + \tan^2 t = \sec^2 t$ needs $\cos t \ne 0$ .

In plain words Solving for $t$ itself would need inverse trigonometric functions. It is easier to find an identity that links $x$ and $y$ directly.

Example For $x = 3 + 2\cos t$ , $y = -1 + 2\sin t$ with $0 \le t < 2\pi$ , solve for $\cos t = \frac{x - 3}{2}$ and $\sin t = \frac{y + 1}{2}$ . The Pythagorean identity gives

\left(\frac{x - 3}{2}\right)^2 + \left(\frac{y + 1}{2}\right)^2 = 1,

(x - 3)^2 + (y + 1)^2 = 4.

The curve is the circle with center $(3, -1)$ and radius $2$ , traced once counterclockwise from $(5, -1)$ .

Inherited restrictions

Definition The inherited restrictions of a parametric curve are the conditions on $x$ and $y$ that come from the parametrization, namely that $x$ can only take the values of $f(t)$ and $y$ only the values of $g(t)$ for $t$ in the parameter interval. A complete answer states the rectangular equation together with these restrictions. They are necessary but not always sufficient, so check which part of the graph is actually reached by following the point as $t$ runs through $I$ .

In plain words The rectangular equation forgets where the parameter was allowed to go. Write the ranges of $x$ and $y$ next to it to rule out points the path can never reach.

Example For $x = \sqrt{t}$ , $y = 4 - t$ with $t \ge 0$ , squaring the first equation gives $t = x^2$ , and then $y = 4 - x^2$ . The graph of $y = 4 - x^2$ is a whole parabola, but $x = \sqrt{t}$ is never negative. The curve is

y = 4 - x^2, \qquad x \ge 0,

the right half of the parabola, starting at $(0, 4)$ when $t = 0$ and moving right and down.

Extraneous branches

Definition An extraneous branch is a piece of the graph of the rectangular equation that the parametric curve never reaches. Extraneous branches often appear after squaring or after using an identity, because those steps can produce an equation that holds on a larger set than the curve. The inherited restrictions remove many of them, and following the point as $t$ runs through the parameter interval shows exactly which part is reached.

In plain words Some pieces of the graph are imposters. They satisfy the equation, but no value of $t$ ever lands the point there.

Example For $x = \sec t$ , $y = \tan t$ with $-\frac{\pi}{2} < t < \frac{\pi}{2}$ , the identity $1 + \tan^2 t = \sec^2 t$ gives

x^2 - y^2 = 1,

a hyperbola with two branches. On this interval $\cos t > 0$ , so $x = \sec t$ is positive and the point never reaches the left branch, where $x \le -1$ . Also $y = \tan t$ takes every real value, and for each one the point has $x = \sec t = \sqrt{1 + y^2}$ , the point of the right branch at that height. So the curve is the whole right branch and nothing else, and the left branch is extraneous.

Orientation information lost

Definition A rectangular equation describes only a set of points. It does not record the orientation, the initial and terminal points, the tracing speed, or how many times each part of the curve is traced. A full description of a parametric curve keeps this information next to the rectangular equation, for example as endpoints and arrows on a sketch.

In plain words Once $t$ is gone, so is the clock. The equation tells you where the path is, but not which way the point moved along it or how often.

Example For $x = \sin t$ , $y = \cos 2t$ with $0 \le t \le 2\pi$ , the identity $\cos 2t = 1 - 2\sin^2 t$ gives $y = 1 - 2x^2$ with $-1 \le x \le 1$ , an arc of a parabola. The table shows what that equation hides.

$t$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$
$(x, y)$	$(0, 1)$	$(1, -1)$	$(0, 1)$	$(-1, -1)$	$(0, 1)$

The point swings from the top of the arc to its right end, back to the top, out to the left end and back again. Every point of the arc except its two ends is passed at least twice.

Check yourself

Eliminate the parameter from $x = t - 2$ , $y = \sqrt{t}$ with $t \ge 0$ , and state the restrictions.

Show the answer

The first equation gives $t = x + 2$ , so $y = \sqrt{x + 2}$ with $x \ge -2$ and $y \ge 0$ . This is the upper half of the parabola $y^2 = x + 2$ , traced from $(-2, 0)$ upward and to the right. Writing only $y^2 = x + 2$ would add the lower half as an extraneous branch.

Eliminate the parameter from $x = 2\tan t$ , $y = 3\sec t$ with $-\frac{\pi}{2} < t < \frac{\pi}{2}$ , and say which part of the graph is traced.

Show the answer

With $\tan t = \frac{x}{2}$ and $\sec t = \frac{y}{3}$ , the identity $1 + \tan^2 t = \sec^2 t$ gives

\frac{y^2}{9} - \frac{x^2}{4} = 1.

On this interval $\sec t \ge 1$ , so $y \ge 3$ . Only the upper branch is traced, from left to right as $t$ increases, and the lower branch is extraneous.

The curves $x = \cos t$ , $y = \sin t$ for $0 \le t \le \pi$ and $x = t$ , $y = \sqrt{1 - t^2}$ for $-1 \le t \le 1$ have the same rectangular equation and restrictions. What does the elimination hide?

Show the answer

Both give $x^2 + y^2 = 1$ with $y \ge 0$ , the upper half of the unit circle. The first starts at $(1, 0)$ and runs counterclockwise to $(-1, 0)$ . The second starts at $(-1, 0)$ and runs clockwise to $(1, 0)$ . The elimination hides these opposite orientations.

Eliminating the Parameter

Key terms

Rectangular equation

Elimination by substitution

Elimination by an identity

Inherited restrictions

Extraneous branches

Orientation information lost

Check yourself

How this lesson connects

Builds on

Where this leads in Calculus