Repeated Points and Self-Intersections

When Does a Parametric Curve Return to the Same Point?

A parametrization assigns a point in the plane to each parameter value. Write it as \(x=x(t)\), \(y=y(t)\), or equivalently as the position \(\mathbf{r}(t)=(x(t),y(t))\). As \(t\) varies over its stated domain, these points form the curve's trace.

Different parameter values can sometimes produce the same point. This may happen at an isolated point, at the endpoints of a periodic parametrization, or throughout a retraced portion of the curve. To distinguish these situations, we must compare both coordinates and keep track of which parameter values are distinct.

Definition: A point is a repeated point of a parametrized curve if there are distinct parameter values \(t_1\) and \(t_2\) in the domain such that \(x(t_1)=x(t_2)\) and \(y(t_1)=y(t_2)\). Equivalently, \(\mathbf{r}(t_1)=\mathbf{r}(t_2)\) with \(t_1\ne t_2\). A self-intersection is commonly used to describe a point in the trace reached at distinct parameter values; the repetition itself does not imply that the curve crosses in any particular way.

The definition is about equality of points, not equality of just one coordinate. Two parameter values may give the same \(x\)-coordinate while giving different \(y\)-coordinates, in which case the curve has not returned to the same point.

A Coordinate Test

To locate repeated points, take two parameter values \(u\) and \(v\), require \(u\ne v\), and solve the simultaneous equations \(x(u)=x(v)\) and \(y(u)=y(v)\). The restriction \(u\ne v\) matters: \(u=v\) always makes both equations true, but says nothing about repetition.

Key takeaway: A pair \((u,v)\) identifies a repeated point only if \(u\) and \(v\) are distinct, lie in the stated domain, and make both coordinates equal. The repeated point is then \((x(u),y(u))\).

Often, subtracting the coordinate expressions reveals a factor of \(u-v\). Since we are looking for distinct values, we can exclude the solution \(u=v\) and analyze the remaining factors. For trigonometric expressions, identities and periodicity serve the same purpose. The exact parameter domain must be included in either approach: changing the domain can add or remove valid pairs.

Write the domain.
Record the permitted parameter values, including whether endpoints are included.

Compare the first coordinates.
Solve \(x(u)=x(v)\), keeping \(u\ne v\) in view.

Compare the second coordinates.
Substitute the possible pairs into \(y(u)=y(v)\) and retain only pairs that satisfy it.

Verify the point.
Evaluate both coordinates at each retained parameter value and state the common point.

Why Comparing Both Coordinates Is Complete

The test is not just a convenient procedure; it follows directly from equality of ordered pairs. Two points in the plane are equal exactly when their first coordinates are equal and their second coordinates are equal.

Justification: Suppose \(u\) and \(v\) are parameters in the domain. By the definition of the parametrization, the corresponding points are \((x(u),y(u))\) and \((x(v),y(v))\). These are equal if and only if their first entries agree and their second entries agree. Thus

\mathbf{r}(u)=\mathbf{r}(v) \quad\Longleftrightarrow\quad x(u)=x(v)\ \text{and}\ y(u)=y(v).

Consequently, any distinct pair satisfying both coordinate equations really does give a repeated point, and every repeated point must arise from such a pair. Solving only one equation can produce candidates, but it cannot establish a repeated point.

Worked Examples

Worked Example: An Isolated Repeated Point on a Polynomial Curve

Consider \(x(t)=t^2-1\), \(y(t)=t^3-t\) for \(-2\le t\le2\). Find all distinct parameter values that produce the same point.

Let \(u\ne v\). Equality of the first coordinates gives

u^2-1=v^2-1 \quad\Longrightarrow\quad (u-v)(u+v)=0.

Since \(u\ne v\), the factor \(u-v\) is not zero, so \(u+v=0\), or \(v=-u\). Now impose equality of the second coordinates:

$$ u^3-u=v^3-v=(-u)^3-(-u)=-u^3+u. $$

Moving all terms to one side gives \(2u^3-2u=0\), hence \(2u(u^2-1)=0\). The possibilities are \(u=0\), \(u=1\), and \(u=-1\). If \(u=0\), then \(v=-u=0\), contradicting \(u\ne v\). The remaining distinct pair is \(\{u,v\}=\{1,-1\}\), and both values lie in the domain.

At \(t=1\), the point is \((1-1,1-1)=(0,0)\). At \(t=-1\), it is \((1-1,-1+1)=(0,0)\). Thus the repeated point is \((0,0)\), reached at \(t=1\) and \(t=-1\). Substitution verifies both coordinates.

Worked Example: Repeated Points in a Trigonometric Parametrization

Consider \(x(t)=\sin t\), \(y(t)=\sin(2t)\) on \(0\le t\le2\pi\). Find all distinct parameter values that produce the same point.

For real \(u\) and \(v\), equality \(\sin u=\sin v\) occurs when \(u=v+2k\pi\) or \(u=\pi-v+2k\pi\), where \(k\) is an integer. In the first case, the domain restriction \(0\le u,v\le2\pi\) allows either \(u=v\) or the endpoint pair \(\{u,v\}=\{0,2\pi\}\). The endpoint pair also satisfies \(\sin(2u)=\sin(2v)=0\), giving the point \((0,0)\).

For a distinct pair from the second case, write \(u+v=\pi+2k\pi\). Since both parameters are between \(0\) and \(2\pi\), their sum is between \(0\) and \(4\pi\); the possible sums are \(\pi\) and \(3\pi\). If \(v=\pi-u+2k\pi\), then

\sin(2v)=\sin(2\pi-2u+4k\pi)=-\sin(2u).

Equality of the second coordinates therefore requires \(\sin(2u)=-\sin(2u)\), so \(\sin(2u)=0\). Within the domain, this gives \(u=n\pi/2\) for an integer \(n\). For \(u+v=\pi\), the possible pairs are \((0,\pi)\), \((\pi/2,\pi/2)\), and \((\pi,0)\); the distinct pair is \(\{0,\pi\}\). For \(u+v=3\pi\), the possible pairs are \((\pi,2\pi)\), \((3\pi/2,3\pi/2)\), and \((2\pi,\pi)\); the distinct pair is \(\{\pi,2\pi\}\).

At \(t=0,\pi,2\pi\), both coordinates equal zero. The distinct pairs found above, together with the endpoint pair, show that the point \((0,0)\) is reached at all three parameter values. The cases considered exhaust the sine-equality possibilities, so there are no other repeated points on this domain.

Worked Example: Retracing a Polynomially Parametrized Arc

Consider \(x(t)=t^2\), \(y(t)=t^4\) for \(-2\le t\le2\). For every \(t\ne0\), compare \(t\) with \(-t\). Both are in the domain, and

x(-t)=(-t)^2=t^2=x(t), \qquad y(-t)=(-t)^4=t^4=y(t).

Thus every nonzero parameter value in the domain has a distinct partner producing the same point. For example, \(t=1\) and \(t=-1\) both give \((1,1)\), since \(1^2=1\), \((-1)^2=1\), \(1^4=1\), and \((-1)^4=1\). At \(t=0\), the partner \(-t\) is still \(0\), so this particular comparison does not give a distinct parameter pair there.

Here the repetition is not confined to one isolated point: the parametrization traverses the same traced portion for positive and negative parameter values. Indeed, \(y=x^2\) and \(0\le x\le4\) describe the trace. This example shows why it is useful to distinguish a repeated point from a single isolated self-intersection.

Periodic Duplication and the Role of the Domain

Trigonometric parametrizations often repeat because sine and cosine are periodic. For example, a coordinate pair may recur when the parameter increases by a full period. This is a repeated point according to the definition whenever both parameter values belong to the domain and are distinct. If the domain includes both endpoints of a full period, the starting and ending points may therefore be counted as repeated.

That fact alone does not establish that the trace has an additional isolated self-intersection. A parametrization may simply close up at its endpoints, as the trigonometric example above also visits \((0,0)\) at the interior value \(t=\pi\). Conversely, an algebraic relation such as \(x(-t)=x(t)\) and \(y(-t)=y(t)\) may show that an entire portion is retraced. State what the parameters do, rather than assuming every repeated point has the same geometric cause.

Domain check: A parameter pair is relevant only when both values are in the domain. For example, two values that differ by a period represent the same trigonometric point, but that repetition is part of the given parametrization only if both values are allowed.

Common Mistake: Checking Only One Coordinate

A frequent error is to find two parameter values with equal \(x\)-coordinates and call the result a self-intersection without checking \(y\). For instance, use \(x(t)=t^2-1\), \(y(t)=t^3-t\) on \([-2,2]\). Since \(x(2)=2^2-1=3\) and \(x(-2)=(-2)^2-1=3\), one might incorrectly conclude that \(t=2\) and \(t=-2\) give the same point.

But the second coordinates are \(y(2)=2^3-2=6\) and \(y(-2)=(-2)^3-(-2)=-8+2=-6\). The points are \((3,6)\) and \((3,-6)\), not the same point. The correction is to solve both coordinate equations and then substitute every surviving parameter value back into both coordinates.

Why This Test Is Useful

Identifying repeated points prevents a parameter description from being confused with the geometric trace it produces. A single geometric point can correspond to several parameter values, while distinct points can share one coordinate. Pairwise coordinate comparison gives a direct way to separate these cases and works for both polynomial and trigonometric expressions.

It also makes conclusions precise. Report the parameter pair or pairs, the common point, and the domain on which the comparison was made. If a point is visited more than twice, list all the distinct parameter values found. If the equations show that an interval of parameter values is retraced, describe that repetition rather than reporting only one sample pair.

Check Your Understanding

Use the two-coordinate test in each question, and respect the stated domain.

For \(x(t)=t^2\), \(y(t)=t^3-t\) on \([-2,2]\), find any distinct parameter values that give the same point.
Why is \(x(u)=x(v)\) alone insufficient to establish that two parameters give the same point?
For \(x(t)=\cos t\), \(y(t)=\sin t\) on \([0,2\pi]\), identify a distinct parameter pair giving the same point and state that point.
What distinguishes an isolated repeated point from the retracing in \(x(t)=t^2,\ y(t)=t^4\)?
Why must the parameter domain be checked after solving the coordinate equations?