Why Branches Matter
A parametric curve is given by coordinate formulas involving a parameter, such as \(x=x(t)\) and \(y=y(t)\), together with an allowed set of parameter values. Eliminating the parameter means finding a relationship between \(x\) and \(y\) that no longer contains \(t\).
That relationship may describe more points than the parametrization actually traces. For instance, squaring an equation can erase a sign condition, and an unrestricted Cartesian equation may include several branches even when the parameter interval traces only one. To describe the traced curve accurately, keep track of both the algebra and the parameter restrictions.
What It Means to Eliminate a Parameter
Suppose \(x=x(t)\) and \(y=y(t)\) are defined for \(t\) in a specified domain \(D\). The parametrization traces the set of points \((x(t),y(t))\) as \(t\) ranges through \(D\). To eliminate \(t\), use one or both coordinate equations to obtain a relationship involving only \(x\) and \(y\).
A Cartesian equation is useful, but it is not automatically a complete description of that set. A complete description must include every traced point and exclude every point the parametrization does not reach. Restrictions such as \(t\geq 0\), \(t\in[a,b]\), or a sign condition on one coordinate can be essential.
Worked Example: A Restricted Parabola
Worked Example: A Restricted Parabola
Let \(x=t^2\) and \(y=t\), where \(0\leq t\leq 2\). Eliminate \(t\) and describe exactly the part of the curve that is traced.
Since \(y=t\), substitute \(t=y\) into \(x=t^2\):
The parameter restriction \(0\leq t\leq2\), together with \(y=t\), gives \(0\leq y\leq2\). Therefore the exact traced set is
Equivalently, \(0\leq x\leq4\) and \(y=\sqrt{x}\). The equation \(x=y^2\) by itself also includes points with \(y<0\), which this parameter interval never reaches.
To check the description, take any \(t\) with \(0\leq t\leq2\). Then \(y=t\) lies in \([0,2]\) and \(x=t^2=y^2\), so every parametrized point satisfies the description. Conversely, if \(x=y^2\) and \(0\leq y\leq2\), choose \(t=y\). This \(t\) is allowed and gives \(x=t^2\), \(y=t\). Thus the description includes exactly the traced points.
Parameter Restrictions Become Coordinate Restrictions
The first example used a coordinate equal to the parameter, making the restriction easy to transfer. In other parametrizations, a restricted interval may select a particular part of a familiar implicit curve. Trigonometric parametrizations provide a clear example: the same circle equation can represent a full circle, a semicircle, or a smaller arc, depending on the interval for \(t\).
Worked Example: The Upper Half of an Ellipse
Let \(x=2\cos t\) and \(y=\sin t\), where \(0\leq t\leq\pi\). Eliminate \(t\) and identify the traced part of the ellipse.
Divide the first equation by \(2\), square both coordinate expressions, and use \(\sin^2t+\cos^2t=1\):
Thus the Cartesian equation is \(\frac{x^2}{4}+y^2=1\). But because \(0\leq t\leq\pi\), \(\sin t\geq0\), so \(y\geq0\). The exact traced part is
The ellipse equation also implies \(-2\leq x\leq2\), since \(x^2/4\leq1\). To verify that no upper-half points are missing, suppose \(\frac{x^2}{4}+y^2=1\) and \(y\geq0\). Since \(-2\leq x\leq2\), there is a \(t\in[0,\pi]\) with \(\cos t=x/2\). For that \(t\), \(\sin t=\sqrt{1-x^2/4}\). The ellipse equation and \(y\geq0\) give \(y=\sqrt{1-x^2/4}\), so \(y=\sin t\). Hence every point satisfying the stated conditions is reached.
At the endpoints, \(t=0\) gives \((2,0)\), and \(t=\pi\) gives \((-2,0)\). The interval traces the upper half, not the lower half or the full ellipse.
When Algebra Can Add a Branch
Squaring both sides of an equation can remove information about sign. For example, if \(y=A\), then \(y^2=A^2\) follows. The reverse implication is not always true: \(y^2=A^2\) allows either \(y=A\) or \(y=-A\). If the original parametrization imposes a sign, preserve it after squaring.
Worked Example: A Sign Condition Selects the Branch
Let \(x=t^2-1\) and \(y=t(t^2-1)\), where \(t\geq0\). Eliminate \(t\) and describe the exact branch.
From \(x=t^2-1\), obtain \(t^2=x+1\), so \(x\geq-1\). Also, \(y=tx\). Squaring this equation and substituting \(t^2=x+1\) gives
This squared equation does not by itself retain the sign of \(t\). The original relation \(y=tx\) and \(t\geq0\) give \(xy=t x^2\geq0\). Therefore a Cartesian description is
The same branch can be written as the explicit equation \(y=x\sqrt{x+1}\), with \(x\geq-1\). Indeed, from \(t\geq0\) and \(t^2=x+1\), we have \(t=\sqrt{x+1}\), and hence \(y=tx=x\sqrt{x+1}\).
To check the converse carefully, suppose \(x\geq-1\), \(y^2=x^2(x+1)\), and \(xy\geq0\). Set \(t=\sqrt{x+1}\), which is real and nonnegative. Then \(x=t^2-1\). If \(x\neq0\), the equation for \(y^2\) gives \(|y|=|x|t\); the condition \(xy\geq0\) selects the sign of \(y\) to match the sign of \(x\), so \(y=xt\). If \(x=0\), the equation gives \(y^2=0\), hence \(y=0=xt\). In all cases, \(y=t(t^2-1)\), so the point is reached by an allowed parameter.
A Reliable Branch-Preserving Method
A useful way to organize elimination is to distinguish what follows from the parametrization from what is needed to recover an allowed parameter. The forward algebra produces necessary conditions. The reverse check determines whether those conditions are sufficient.
Write down the restriction on \(t\) before eliminating it. It may determine the signs or coordinate ranges needed later.
Substitute one coordinate expression into the other, or use an identity to obtain an equation involving \(x\) and \(y\).
If you square, take a square root, or divide by an expression that might be zero, record the sign condition or excluded case that the operation affects.
Starting with a point satisfying the proposed Cartesian description, recover a parameter and check that it belongs to the allowed domain and gives both coordinates.
Why the Converse Check Matters
Substitution usually proves that every point traced by the parametrization satisfies the equation you derived. In set language, it shows that the traced set is contained in the proposed Cartesian set. But it does not automatically show that the two sets are equal. The Cartesian equation may include extra points, as happens when a sign restriction is lost.
The converse check addresses the missing direction: take a point satisfying the Cartesian conditions and determine whether an allowed parameter produces it. If the parameter cannot be recovered, or the recovered value lies outside its permitted interval, the Cartesian description is too broad.
In practice, proving these directions often takes only a few lines. In the ellipse example, \(t\in[0,\pi]\) follows from choosing \(t=\arccos(x/2)\), while \(y\geq0\) ensures that the sine coordinate has the required value. In the third example, \(t=\sqrt{x+1}\) recovers the parameter, and the sign condition makes the original equation \(y=tx\) hold rather than just its square.
Common Pitfall: Keeping Only the Implicit Equation
For \(x=2\cos t\), \(y=\sin t\), \(0\leq t\leq\pi\), the calculation
is correct, but presenting that equation as the full answer is incomplete. For example, \((0,-1)\) satisfies it because \(0^2/4+(-1)^2=1\). Yet \(y=\sin t\geq0\) throughout the allowed interval, so \((0,-1)\) is not traced. The correction is to include \(y\geq0\).
Whenever an algebraic step can merge positive and negative possibilities, pause before discarding sign information. The goal is not just to obtain an equation that the curve satisfies; it is to characterize precisely the points the given parametrization reaches.
Check Your Understanding
For each question, pay attention to whether the Cartesian equation alone identifies the traced set.
- For \(x=t^2\), \(y=t\), and \(-1\leq t\leq1\), what equation and coordinate restriction describe the traced curve?
- Why does \(\frac{x^2}{4}+y^2=1\) fail to describe only the curve traced when \(x=2\cos t\), \(y=\sin t\), and \(0\leq t\leq\pi\)?
- When an equation is squared during elimination, what information might be lost?
- For \(x=t^2-1\), \(y=t(t^2-1)\), \(t\geq0\), explain why the condition \(xy\geq0\) helps preserve the correct branch.
- What two directions must be checked to show that a Cartesian description is exactly the parametrized curve?
