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Eliminating the parameter

Convert parametric equations to Cartesian form using substitution, powers, exponentials and logarithms. Preserve the correct branch, coordinate ranges, endpoints and excluded points.

Before you startParametric coordinate pairs, rearranging equations, indices and logarithms

01 / Remove the parameter, keep the restrictions

A Cartesian equation describes a relation between x and y.

Eliminating a parameter means combining the two coordinate rules so that the final equation uses only x and y. The equation is only part of the answer: also state which points are actually reached by the permitted parameter values.

Coordinate rules → Cartesian relation + exact restrictions

Explore three examples. Change the domain and see why an unrestricted equation can include too much of a curve.

Keep the right part of the curveExplore

Coordinate rulesParameter domainTest parameter t

y = (x − 1)²; −1 ≤ x ≤ 3.

y-range: 0 ≤ y ≤ 4.

t = −2 gives (−1, 4), an included point.

Blue is the permitted locus. Filled endpoints are included; hollow circles are excluded. The gold point appears only for an allowed parameter. The rational rule always excludes t = 1, even after cancellation.

02 / Rearrange the easier rule first

Substitution works especially well when one coordinate is linear.

Eliminate t from x = 2t − 3 and y = t² + 2, where −1 ≤ t ≤ 2.Worked example

t = (x + 3)/2

Rearrange the linear rule.

y = (x + 3)²/4 + 2

Substitute the entire expression for t into the y-rule.

−5 ≤ x ≤ 1

The increasing x-rule sends the endpoint parameters to −5 and 1.

2 ≤ y ≤ 6

The minimum occurs at t = 0 inside the interval. Endpoint values alone would miss it.

Use brackets when squaring a substituted expression. You may leave a compact factorised form; expanding is not always helpful.

03 / Squaring can hide a branch restriction

An equation can be necessary without describing exactly the same locus.

Eliminate t from x = t² and y = t, where −2 ≤ t ≤ 0.Worked example

x = y²

The second coordinate is exactly t.

−2 ≤ y ≤ 0

Keep the original parameter restriction.

Equivalently, y = −√x with 0 ≤ x ≤ 4

The negative square-root branch is selected.

Writing only y² = x would also allow y > 0

Those points are not reached here.

Watch a domain select one branch

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

04 / Use root rules with their original domains

The square-root symbol denotes the non-negative root.

Given x = √(t + 2) and y = t − 1 for −2 ≤ t < 7, find the Cartesian form.Worked example

t = x² − 2

Square the x-rule, retaining x ≥ 0.

y = x² − 3

Substitute for t.

0 ≤ x < 3

The lower endpoint is attained; the upper is approached.

−3 ≤ y < 6

The y-rule increases throughout the parameter interval.

If no parameter domain is supplied, the root still requires t ≥ −2. A restriction supplied by the formula is just as important as one stated separately.

05 / Cancellation does not fill a hole

Check every original denominator before simplifying.

Eliminate t from x = t and y = (t² − 1)/(t − 1), with −2 ≤ t ≤ 2.Worked example

t ≠ 1

The original y-rule is undefined there.

y = t + 1 for t ≠ 1

Cancel the factor only on the existing domain.

y = x + 1; −2 ≤ x ≤ 2, x ≠ 1

The line segment has a hole at (1, 2).

−1 ≤ y ≤ 3, y ≠ 2

The missing y-value has no other parameter producing it.

For x = 1/(t − 2), y = 3t + 1 with t ≠ 2, rearrange to t = 2 + 1/x. Hence y = 7 + 3/x with x ≠ 0. The original excluded parameter corresponds to a vertical asymptote, while x = 0 is never attained.

06 / Eliminate exponential parameters by using powers

Positive exponentials carry a coordinate restriction.

For x = 2eᵗ and y = 3e²ᵗ − 1, with t any real number, eliminate t.Worked example

eᵗ = x/2

In particular x > 0.

e²ᵗ = (eᵗ)² = x²/4

Use the index law.

y = 3x²/4 − 1, x > 0

Only the positive-x half of the parabola is reached.

y > −1

The limiting height −1 is approached, never attained.

If instead t ≤ 0, then 0 < x ≤ 2 and −1 < y ≤ 2. The limit as t tends to −∞ is not an included endpoint.

07 / Use logarithm laws on their valid domains

A logarithm input must be positive.

Given x = ln t and y = ln(4t²), with t > 0, eliminate t.Worked example

y = ln 4 + 2 ln t

The positive t-domain makes this law valid.

y = 2x + ln 4

Replace ln t by x.

x can be any real number

ln t ranges over all real numbers when t > 0.

Another common pattern is x = eᵗ + 1, y = t². Then t = ln(x − 1), so y = [ln(x − 1)]² with x > 1. Its minimum y = 0 occurs at x = 2; there is no upper bound.

08 / Find actual coordinate ranges

Transform the domain through each rule, including any internal turning point.

Find the coordinate ranges of x = t² − 1, y = 2t + 3 for −2 < t ≤ 1.Worked example

−1 ≤ x < 3

The minimum at t = 0 is included; t = −2 is excluded, so x = 3 is not attained.

−1 < y ≤ 5

The linear y-rule is strictly increasing.

x = (y − 3)²/4 − 1, −1 < y ≤ 5

Using a y-restriction preserves which branch portions occur.

Separate x- and y-ranges alone may not identify the correct portion of an implicit curve. Retain a branch/sign condition when needed. Check whether another allowed parameter produces the same point before marking it excluded.

09 / An identity can still miss a point

A rational parameterisation need not cover a whole circle.

For real t, consider x = (1 − t²)/(1 + t²) and y = 2t/(1 + t²). The denominator is always positive.

Eliminate t without solving either rule for it.Worked example

x² + y² = [(1 − t²)² + 4t²]/(1 + t²)²

Square the coordinate rules and add.

(1 − t²)² + 4t² = (1 + t²)²

Expand or recognise the identity.

x² + y² = 1, with (−1, 0) excluded

x = −1 would require 1 − t² = −1 − t², which is impossible.

Every other point of the circle is reached

For x ≠ −1, take t = y/(1 + x) and substitute using x² + y² = 1.

As |t| grows, the point approaches (−1, 0). A limiting point is not necessarily an actual point of a parameterised curve.

10 / Translate a bounded parameter into a line segment

Elimination gives the line; endpoints determine the segment.

For x = 1 + 2t and y = 4 − t, where 0 ≤ t ≤ 3, find the segment and its length.Worked example

t = (x − 1)/2, so y = (9 − x)/2

This is the Cartesian line equation.

1 ≤ x ≤ 7

The curve is only a segment.

Endpoints: (1, 4) and (7, 1)

Substitute t = 0 and t = 3.

Someone eliminates t from x = √t, y = t for t ≥ 0 and writes “y = x² for every real x”. What is wrong?

Hint

The root rule has a sign restriction.

Worked solution

x = √t requires x ≥ 0. Only the right half of y = x² is traced. Squaring the rule lost that restriction.

12 / Recap

A complete answer preserves the original point set.

Rearrange the easier coordinate rule, or use an identity.
Retain root signs, logarithm domains and original denominator exclusions.
Find coordinate ranges using endpoints and internal extrema.
Check whether squaring or cancellation introduced extra points.
Remember that a Cartesian equation loses tracing order and timing.

Section 1 of 12 · Remove the parameter, keep the restrictions