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Sketching parametric curves

Sketch parametric curves using domains, intercepts, symmetry, direction and limiting behaviour. Explore a self-intersecting loop and separate rational branches, with original worked practice.

Before you startParametric coordinates, elimination, graph shapes and limits

01 / Build a sketch from justified features

A table supports a sketch; the rules explain its shape.

Begin with the valid parameter domain. Find intercepts and endpoints, check symmetry, then use a table in increasing parameter order to connect the features. Eliminate the parameter when it helps identify the underlying curve.

Domain → key points → symmetry → tracing order → limiting behaviour

A loop or an undefined parameter can make a simple left-to-right graphing approach fail. Follow the parameter itself in the model.

Follow the parameterExplore
Trace a loop or two branchesFor x equals t squared minus one and y equals t cubed minus t, increasing t from minus two to two traces a loop through the origin.xy0−136−6

x = t² − 1; y = t³ − t; −2 ≤ t ≤ 2.

t = −2 gives (3, −6).

The origin occurs at t = −1 and t = 1. The path between them is a loop.

Grey shows the displayed domain; blue follows the parameter up to your selection. The reciprocal graph continues beyond the frame near its pole. Separate branches are never joined across an undefined parameter.

02 / Find intercepts from the appropriate coordinate rule

Solve x = 0 and y = 0 separately, then pair the coordinates.

For x = t² − 1 and y = t³ − t, with t real, find the axis crossings.Worked example

x = 0 gives t = −1 or t = 1

Both give y = 0, so the y-axis is crossed at one point: (0, 0).

y = t(t − 1)(t + 1) = 0

The parameter values are −1, 0 and 1.

At t = 0, x = −1

The x-axis crossings are (−1, 0) and (0, 0).

Three parameter values give only two distinct x-axis points

Repeated parameter roots must be converted to positions before counting points.

An intercept parameter must also belong to the stated domain. Algebraic roots outside it do not contribute.

03 / Include values between repeated points

A sparse table can hide a loop.

For the same curve, use values between t = −1 and t = 1 as well as the integer values.

t = −2 → (3, −6)
t = −1 → (0, 0)
t = −1/2 → (−3/4, 3/8)
t = 0 → (−1, 0)
t = 1/2 → (−3/4, −3/8)
t = 1 → (0, 0)
t = 2 → (3, 6)

As t increases, the curve approaches the origin along the lower-right branch, travels around the upper and then lower sides of a loop to the left, returns to the origin, and leaves along the upper-right branch.

Watch the loop traced in parameter order

Pause, replay or seek freely. The notes explain the same idea and stay in view.

04 / Test symmetry using the parameter rules

Compare t and −t when the domain permits both.

Show that x = t² − 1, y = t³ − t has x-axis symmetry for all real t.Worked example

x(−t) = t² − 1 = x(t)

The horizontal coordinate is unchanged.

y(−t) = −t³ + t = −y(t)

The vertical coordinate changes sign.

Every (x, y) has a matching (x, −y)

Both parameter values are allowed.

This conclusion can fail after restricting the domain. For example, keeping only t ≥ 0 removes the paired negative parameters, so symmetry of the formulas alone no longer proves symmetry of the restricted locus.

Same x, opposite y → x-axis reflection.
Opposite x, same y → y-axis reflection.
Opposite x and y → half-turn about the origin.

05 / Justify the loop algebraically

Repeated positions do not automatically mean the whole path is retraced.

Eliminate t from x = t² − 1 and y = t³ − t.Worked example

y = t(t² − 1) = tx

Factor the y-rule.

t² = x + 1

Use the x-rule.

y² = x²(x + 1), x ≥ −1

Square y = tx and substitute.

For −1 < x < 0, there are two opposite non-zero y-values

These form the two sides of the loop. They meet at (−1, 0) and (0, 0).

The origin is a self-intersection: t = −1 and t = 1 reach it by different parts of the path. The segment between those parameter values is not retraced; it goes around a loop. For x > 0, two branches extend to the right.

06 / Show only the permitted part of a curve

The same formulas can describe a complete curve, an arc or a segment.

Restrict the loop model to different parameter intervals.Worked example

−1 ≤ t ≤ 1

Only the closed loop is traced.

0 ≤ t ≤ 1

Only the lower half of the loop, from (−1, 0) to (0, 0), is traced.

−2 ≤ t ≤ 2

The loop and finite pieces of both right-hand branches are included.

−1 < t < 1

The origin is excluded, but (−1, 0) remains included at t = 0.

Parameter endpoints and geometric endpoints are not interchangeable. Two ends of a parameter interval may land on the same point.

07 / Split a curve at undefined parameters

Never draw a connecting stroke through a pole.

Sketch x = 1/(t − 1), y = t + 2 for all real t except 1.Worked example

t ≠ 1

The x-rule has a zero denominator there.

t = y − 2, so x = 1/(y − 3)

Substitute using the linear y-rule.

y = 3 + 1/x, x ≠ 0

This is a translated reciprocal curve.

t < 1 gives x < 0, y < 3; t > 1 gives x > 0, y > 3

The two pieces lie on separate branches.

A plotting program that connects adjacent sample points across t = 1 can create a false line between the branches. Keep those pieces separate.

08 / Translate parameter limits into coordinate limits

A pole in a parameter rule is not automatically a vertical graph asymptote.

Find the asymptotes of x = 1/(t − 1), y = t + 2.Worked example

As t → 1⁻, x → −∞ and y → 3⁻

This approaches the horizontal line y = 3 on the left branch.

As t → 1⁺, x → +∞ and y → 3⁺

The right branch approaches the same horizontal line.

As t → +∞, x → 0⁺ and y → +∞

This approaches the vertical line x = 0.

As t → −∞, x → 0⁻ and y → −∞

This approaches x = 0 from the other side.

Neither asymptote is part of this locus: x can never equal zero, and y = 3 would require the excluded parameter t = 1.

09 / Mark the direction on each valid branch

Increasing t can move x backwards.

On the reciprocal example, as t increases below 1, x decreases from near 0⁻ towards −∞ while y increases towards 3. Above 1, x decreases from +∞ towards 0⁺ while y increases without bound.

Use finite points to check the arrows.Worked example

t = 0 gives (−1, 2); t = 1/2 gives (−2, 5/2)

The left branch moves up and left.

t = 2 gives (1, 4); t = 3 gives (1/2, 5)

The right branch also moves up and left.

If t represents time, there is no defined position at t = 1 in this model. Do not describe the branches as a continuous physical jump through infinity.

10 / Audit a finished sketch

Check what the drawing claims, not just its general appearance.

  • Are all axis crossings labelled with coordinates rather than parameters?
  • Are endpoint symbols consistent with the valid parameter domain?
  • Does each arrow follow increasing parameter values?
  • Have loops and repeated points been distinguished from retracing?
  • Are separate branches separated at every pole?
  • Do asymptotes come from actual coordinate limits?

If a feature is inferred from a few samples only, treat it as a conjecture until the algebra or further analysis supports it.

11 / Your turn

Use the parameter to justify the sketch features.

01 · Intercepts

x = t² − 4, y = t, with t real. Find all axis crossings.

Hint

Set each coordinate to zero separately.

Worked solution

x = 0 at t = ±2 gives (0, −2), (0, 2). y = 0 at t = 0 gives (−4, 0).

02 · Symmetry

For the preceding curve, compare t with −t.

Hint

The square stays the same; t changes sign.

Worked solution

x(−t) = x(t), y(−t) = −y(t), so the locus has x-axis symmetry.

03 · Parameter direction

x = t² − 4, y = t for −2 ≤ t ≤ 2. Describe the path.

Hint

Track the points at −2, 0 and 2.

Worked solution

From (0, −2) it moves up and left to (−4, 0), then up and right to (0, 2). It traces x = y² − 4 with −2 ≤ y ≤ 2.

04 · Distinct points

For x = t² − 1, y = t³ − t, how many distinct x-axis crossings are there?

Hint

Convert t = −1, 0, 1 into coordinate pairs.

Worked solution

Two: (0, 0) and (−1, 0). The origin is reached twice.

05 · A hidden loop

Why is the table containing only t = −1 and t = 1 insufficient for the loop example?

Hint

Both give the origin.

Worked solution

It hides the non-constant path between them. For instance t = 0 gives (−1, 0), and t = ±1/2 give opposite non-zero y-values. The point travels around a loop.

06 · Restricted loop

Describe the loop example on −1 ≤ t ≤ 0.

Hint

Here t is non-positive and x = t² − 1 is non-positive.

Worked solution

It traces the upper half of the loop from (0, 0) to (−1, 0), since y = tx ≥ 0. Both endpoints are included.

07 · Open parameter endpoints

For the loop example on −1 < t < 1, is the origin included? Is (−1, 0) included?

Hint

Find every parameter for each point.

Worked solution

The origin needs t = ±1, so it is excluded. (−1, 0) is included at t = 0.

08 · Reciprocal elimination

x = 1/(t + 2), y = t − 1, t ≠ −2. Eliminate t and identify the asymptotes.

Hint

t = y + 1.

Worked solution

y = 1/x − 3, x ≠ 0. Asymptotes x = 0 and y = −3.

09 · Parameter versus graph limit

In the preceding example, what happens as t → −2⁺?

Hint

t + 2 approaches zero through positive values.

Worked solution

x → +∞ and y → −3⁺. This is approach to the horizontal asymptote y = −3, not to x = −2.

10 · Branch separation

Why must you not connect sample points on opposite sides of t = −2 in that example?

Hint

The original x-coordinate is undefined at −2.

Worked solution

The two samples belong to separate branches; the curve does not contain the straight connecting segment. Split the plot at the pole.

11 · A domain breaks symmetry

x = t², y = t, but only 0 ≤ t ≤ 2. Is the locus symmetric about the x-axis?

Hint

Negative t-values are not permitted.

Worked solution

No. The locus is y = √x on 0 ≤ x ≤ 4; reflected negative-y points are missing except the origin.

12 · Retrace or loop?

Compare x = t², y = 2t² for −1 ≤ t ≤ 1 with the loop model.

Hint

Eliminate t and follow from −1 through 0 to 1.

Worked solution

The first follows y = 2x from (1, 2) to (0, 0) and back along the same segment. That is retracing, not a loop enclosing a region.

13 · Boundaries

x = t − 1, y = t² for 0 < t ≤ 2. Give the endpoints and their inclusion.

Hint

Substitute the limiting parameter values.

Worked solution

(−1, 0) is excluded and (1, 4) is included. The locus is y = (x + 1)² with −1 < x ≤ 1.

14 · Spot the false point

A sketch of x = 1/(t − 1), y = t + 2 includes (0, 3). Explain why it is wrong.

Hint

Check both original coordinate rules.

Worked solution

x = 0 is impossible for a finite valid t. y = 3 would require t = 1, where x is undefined. (0, 3) is the intersection of the asymptotes, not a point on the curve.

12 / Recap

A useful sketch explains the path as well as the locus.

  • Pair coordinates using one valid parameter.
  • Use enough key points to reveal loops and repeated visits.
  • Check symmetry against the actual domain.
  • Separate branches at poles and derive asymptotes from coordinate limits.
  • Label endpoints and arrows accurately.

Section 1 of 12 · Build a sketch from justified features