Why Eliminating a Parameter Needs Care
A parametrized curve gives its coordinates in terms of a variable \(t\). Removing \(t\) can reveal a familiar equation in \(x\) and \(y\), making the curve easier to describe. But the equation alone may include points that the original parameter was not allowed to produce.
The issue is that algebraic elimination often forgets information such as \(t\ge 0\), an interval restriction, or the sign of a square root. A relation between \(x\) and \(y\) is therefore only part of the description: the original parameter domain must also be considered.
Parameter Domain and Curve Trace
Suppose a curve is given by \(x=x(t)\) and \(y=y(t)\), with \(t\) restricted to a domain \(D\). Its trace is the set of points in the plane produced by the allowed parameter values:
Eliminating the parameter means using the coordinate formulas to obtain one or more conditions involving \(x\) and \(y\), without \(t\). For example, solving one coordinate equation for \(t\) and substituting into the other can give such a condition.
Every point of the trace must satisfy a correctly derived coordinate relation. The reverse need not be true: a point might satisfy the relation but correspond only to a parameter value outside \(D\). An exact description must therefore meet two tests: every traced point satisfies the description, and every point satisfying the description is produced by some \(t\in D\).
Worked Example: A Restricted Line Segment
Worked Example: A Restricted Line Segment
Eliminate \(t\) from \(x=2t+1\), \(y=t-3\), where \(0\le t\le 2\), and state the exact trace.
First solve the \(x\)-equation for \(t\):
Substitute this expression into \(y=t-3\):
Now use the parameter restriction. Since \(0\le t\le 2\), the formula \(x=2t+1\) gives
Thus the exact trace is the line segment described by \(y=(x-7)/2\) and \(1\le x\le 5\). Its endpoint values are \(y=(1-7)/2=-3\) and \(y=(5-7)/2=-1\).
To check the description in the reverse direction, take any \(x\) with \(1\le x\le 5\) and set \(t=(x-1)/2\). Then \(0\le t\le 2\), \(2t+1=x\), and \(t-3=(x-7)/2=y\). So every point satisfying the line equation and coordinate restriction is produced by an allowed parameter. For a direct check, \(t=1\) gives \((x,y)=(3,-2)\), and \((3-7)/2=-2\).
Quadratic Equations and Branches
When a coordinate depends on \(t^2\), solving for \(t\) can introduce a choice of sign. For instance, \(t^2=x\) gives \(t=\sqrt{x}\) or \(t=-\sqrt{x}\) when \(x>0\). If the original parameter is restricted to \(t\ge 0\), only the nonnegative choice is permitted. Squaring an equation can hide this choice, so the resulting relation must be paired with the appropriate branch condition.
Worked Example: One Branch of a Parabola
Eliminate \(t\) from \(x=t^2\), \(y=t\), with \(t\ge 0\), and describe the exact trace.
The second equation gives \(t=y\). Substituting into the first gives
The parameter restriction \(t\ge 0\) becomes \(y\ge 0\). Therefore the exact trace is
To verify that this description is sufficient as well as necessary, let a point satisfy \(x=y^2\) and \(y\ge 0\). Choose \(t=y\). Then \(t\ge 0\), \(y=t\), and \(x=y^2=t^2\), as required. For example, \(t=2\) produces \((x,y)=(4,2)\); indeed \(4=2^2\) and \(2\ge0\). The point \((4,-2)\) also satisfies \(x=y^2\), but it is not in the trace because its \(y\)-coordinate is negative.
Worked Example: A Restricted Cusp
Eliminate \(t\) from \(x=t^2\), \(y=t^3\), with \(t\ge 0\), and retain the correct branch.
Squaring \(y=t^3\) gives \(y^2=t^6\), while cubing \(x=t^2\) gives \(x^3=t^6\). Hence
Because \(x=t^2\), we also know \(x\ge0\); in this equation that condition follows already from \(x^3=y^2\ge0\). The sign information not captured by squaring \(y=t^3\) is \(y\ge0\), since \(t\ge0\). Thus an exact description is
For the converse, suppose \(y^2=x^3\) and \(y\ge0\). Then \(x\ge0\). Set \(t=\sqrt{x}\), so \(t\ge0\) and \(t^2=x\). Also, \(y^2=x^3=(\sqrt{x})^6=t^6\). Since \(y\ge0\) and \(t^3\ge0\), equality of their squares gives \(y=t^3\). Thus the point is produced by an allowed parameter.
As a numerical check, \(t=2\) gives \(x=4\), \(y=8\), and \(y^2=64=x^3\), with \(y\ge0\). The relation \(y^2=x^3\) by itself also admits \((4,-8)\), but that point requires \(t=-2\), which is outside the domain.
Trigonometric Elimination and Coordinate Ranges
Trigonometric elimination often uses an identity such as \(\cos^2 t+\sin^2 t=1\). The identity gives a curve equation, but a restricted interval for \(t\) can still select only part of that curve. The signs and ranges of the trigonometric functions on the allowed interval supply the needed coordinate restrictions.
Worked Example: A Quarter of an Ellipse
Eliminate \(t\) from \(x=2\cos t\), \(y=3\sin t\), where \(0\le t\le \pi/2\), and describe the exact trace.
Divide the coordinate equations by their scale factors:
Square both equations and add them. Using \(\cos^2 t+\sin^2 t=1\) gives
On \(0\le t\le\pi/2\), both \(\cos t\) and \(\sin t\) are nonnegative, so \(x\ge0\) and \(y\ge0\). The exact trace is therefore described by
For the converse, take a point satisfying these conditions. Since \(0\le x/2\le1\), there is an angle \(t\in[0,\pi/2]\) with \(\cos t=x/2\). Its sine is nonnegative, and the ellipse equation gives \((y/3)^2=1-(x/2)^2=\sin^2 t\). Since \(y/3\ge0\), it follows that \(\sin t=y/3\). The original coordinate equations are then satisfied. As a check, \(t=\pi/4\) gives \(x=\sqrt{2}\) and \(y=3\sqrt{2}/2\); the two terms in the ellipse equation are \(1/2\) and \(1/2\), summing to \(1\).
Why the Coordinate Restrictions Matter
The examples illustrate a general logical distinction. A calculation that produces \(F(x,y)=0\) establishes that every traced point lies on that equation. It does not, by itself, establish that every point on the equation comes from an allowed parameter.
Use the coordinate formulas to remove \(t\), keeping track of operations such as squaring that may discard sign information.
Convert restrictions on \(t\) into restrictions on \(x\), \(y\), or both, using the original formulas.
Confirm that traced points satisfy the proposed description and that any point satisfying it can be obtained from an allowed parameter.
This two-direction check is a useful justification for an eliminated description. The forward direction comes from substitution into the coordinate equations. The reverse direction usually requires constructing a suitable parameter and verifying both coordinates and its domain restriction.
Common Mistake: Forgetting the Sign
For \(x=t^2\), \(y=t\), \(t\ge0\), a tempting calculation is \(x=y^2\), followed by the claim that the entire parabola \(x=y^2\) is the trace. That is incorrect: the equation includes points with \(y<0\), which would require \(t=y<0\).
The correction is to carry the original restriction through the elimination. Since \(y=t\) and \(t\ge0\), we must also impose \(y\ge0\). The exact description is \(x=y^2\) together with \(y\ge0\), not the full parabola. A reliable check is to take a point on the proposed equation and ask whether an allowed parameter actually produces it.
When an Implicit Equation Is Not the Whole Answer
An equation alone can describe a full curve even when the parametrization traces only a portion of it. An interval restriction may produce a segment, an arc, or one branch. Sometimes the restriction becomes a simple coordinate inequality, as in the examples. In other cases, one can state the exact trace as the implicit equation together with coordinate bounds derived from the parameter interval.
For a restricted linear parameter, coordinate bounds often give endpoints of a segment. For quadratic formulas, sign restrictions distinguish branches. For trigonometric formulas, interval restrictions can determine signs or coordinate ranges. These are not extra details: they are part of identifying which points the parametrization actually produces.
Check Your Understanding
For each question, distinguish the coordinate equation from any restrictions needed to describe the exact trace.
- For \(x=3t-2\), \(y=2t+1\), with \(0\le t\le1\), eliminate \(t\) and find the range of \(x\).
- For \(x=t^2\), \(y=t\), with \(t\le0\), what coordinate inequality selects the correct branch?
- For \(x=\cos t\), \(y=\sin t\), with \(0\le t\le\pi\), what equation and coordinate restriction describe the trace?
- Why does the equation \(y^2=x^3\) alone fail to describe the trace of \(x=t^2,\ y=t^3,\ t\ge0\)?
- When checking an eliminated description, what must be shown in the reverse direction?
