Parametric Equations

In many situations it is convenient to express the x- and y-coordinates of a curve separately in terms of a third variable called a parameter (usually t or θ). This lesson introduces parametric equations and their graphical interpretation, as required by the Edexcel 9MA0 specification.

What Are Parametric Equations?

Instead of writing y as a function of x directly (the Cartesian equation), we write:

x = f(t), y = g(t)

As the parameter t varies, the point (x, y) traces out a curve.

Example: x = 2t, y = t² defines a parabola. When t = 0: (0, 0). When t = 1: (2, 1). When t = −1: (−2, 1). When t = 2: (4, 4).

Why Use Parametric Equations?

Some curves cannot be written as y = f(x) because they fail the vertical line test (e.g. circles).
Parametric equations can describe the direction and speed of motion along a curve.
Many physical applications (projectile motion, circular motion) naturally use a parameter for time.

Common Parametric Forms

Circle

A circle with centre (a, b) and radius r:

x = a + r cos(θ), y = b + r sin(θ)

As θ varies from 0 to 2π, the point traces the full circle.

Example: x = 3 cos(θ), y = 3 sin(θ) is a circle of radius 3 centred at the origin.

Parabola

The parabola y² = 4ax can be parametrised as:

x = at², y = 2at

Example: For y² = 8x (where a = 2): x = 2t², y = 4t.

When t = 1: (2, 4). When t = −1: (2, −4). When t = 0: (0, 0).

Trigonometric Parametrisation

Many curves use trigonometric parametrisation:

x = a cos(t), y = b sin(t) gives an ellipse.

Sketching Parametric Curves

To sketch a parametric curve:

Create a table of values for the parameter t.
Calculate the corresponding x and y values.
Plot the points and join them in order of increasing t.
Indicate the direction of travel with arrows.

Example: Sketch x = t + 1, y = t² − 1 for −3 ≤ t ≤ 3.

t	x = t + 1	y = t² − 1
−3	−2	8
−2	−1	3
−1	0	0
0	1	−1
1	2	0
2	3	3
3	4	8

The curve is a parabola passing through (0, 0) and (2, 0) with vertex at (1, −1).

Finding Points from Parameters

Given specific parameter values, substitute into both x and y equations.

Example: A curve is defined by x = 3t − 1, y = t² + 2t. Find the coordinates when t = 2.

x = 3(2) − 1 = 5, y = 4 + 4 = 8. The point is (5, 8).

Restricting the Parameter

The domain of the parameter determines which portion of the curve is traced. For instance:

x = cos(t), y = sin(t), 0 ≤ t ≤ π traces only the upper semicircle.
x = cos(t), y = sin(t), 0 ≤ t ≤ 2π traces the full circle.

Determining Direction of Travel

As t increases, the point moves along the curve in a specific direction. This is important in mechanics and can be shown with arrows on a sketch.

Example: For x = cos(t), y = sin(t), as t increases from 0, the point moves anticlockwise around the circle starting from (1, 0).

Exam Tip: When sketching parametric curves, always include arrows showing the direction as the parameter increases. Mark specific parameter values at key points (such as t = 0, t = π/2, etc.) to demonstrate your understanding.

Key Terms

Term	Definition
Parameter	A third variable (usually t or θ) in terms of which x and y are expressed
Parametric equations	A pair of equations x = f(t), y = g(t) defining a curve
Cartesian equation	An equation relating x and y directly, with no parameter
Trace	The path followed by the point (x, y) as the parameter varies

Summary

Parametric equations express x and y separately as functions of a parameter.
They are useful for curves that cannot be written as y = f(x), and for describing motion.
Standard forms exist for circles (using cos and sin), parabolas (using t²) and ellipses.
To sketch, create a table of values and plot in order, marking the direction of travel.
The domain of the parameter controls which part of the curve is traced.

A-Level Deep Dive: Parametric Equations

Spec mapping

Edexcel 9MA0 specification, Year 2 Pure section 8 — Parametric equations covers parametric equations in modelling in a variety of contexts; understand and use the parametric equations of curves and conversion between Cartesian and parametric forms (refer to the official specification document for exact wording). This sub-strand sits squarely on Paper 2 (Pure Mathematics 2) but its tendrils reach across the rest of A-Level. Section 9 (Differentiation) introduces parametric differentiation $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , used to find tangents and normals to parametrically-defined curves. Section 10 (Integration) extends to areas under parametric curves via $\int y \, \dfrac{dx}{dt}\,dt$ . Conics — circles, ellipses and parabolas — are routinely expressed in parametric form, and the technique reappears in 9MA0-03 Mechanics when projectile-motion trajectories are written as $x = (u\cos\alpha)t,\ y = (u\sin\alpha)t - \tfrac{1}{2}gt^2$ . Section 11 (Vectors) also uses parametric thinking: a position vector $\mathbf{r}(t) = \mathbf{a} + t\mathbf{d}$ is exactly a parametric equation in disguise. The Edexcel formula booklet does not give the parametric differentiation formula or the trig identity $\cos^2 t + \sin^2 t = 1$ in this section — both must be memorised and recalled at speed.

Worked example with full mark scheme

Question (8 marks):

A curve $C$ is defined by the parametric equations

$x = 2\cos t + 1, \qquad y = 3\sin t - 2, \qquad 0 \leq t < 2\pi.$

(a) Find a Cartesian equation of $C$ in the form $\dfrac{(x - a)^2}{p^2} + \dfrac{(y - b)^2}{q^2} = 1$ , stating the values of $a, b, p, q$ . (5)

(b) State the range of $x$ and the range of $y$ for points on $C$ , and identify the curve geometrically. (3)

Solution with mark scheme:

(a) Step 1 — isolate the trigonometric terms.

From $x = 2\cos t + 1$ : $\cos t = \dfrac{x - 1}{2}$ .

From $y = 3\sin t - 2$ : $\sin t = \dfrac{y + 2}{3}$ .

M1 — rearranging both parametric equations to isolate $\cos t$ and $\sin t$ . A common slip here is to forget the $+1$ shift on $x$ or the $-2$ shift on $y$ — the constant must move with the trig term, not be absorbed into it. Writing $\cos t = \dfrac{x}{2}$ (dropping the $-1$ ) loses both the M1 and every subsequent A mark.

A1 — both isolations correct, including the sign of the shift on $y$ (the $-2$ in the original $y$ -equation becomes $+2$ when transposed).

Step 2 — apply the Pythagorean identity $\cos^2 t + \sin^2 t = 1$ .

$\left(\dfrac{x - 1}{2}\right)^2 + \left(\dfrac{y + 2}{3}\right)^2 = 1$

M1 — substitution into the identity. This step is the engine of the elimination: candidates who try to eliminate $t$ by, say, taking inverse trig functions and equating arguments invariably hit an impasse. The identity-based elimination is the standard examiner-rewarded route.

A1 — correct substitution with both fractions squared.

Step 3 — state the canonical form.

$\dfrac{(x - 1)^2}{4} + \dfrac{(y + 2)^2}{9} = 1$

So $a = 1$ , $b = -2$ , $p = 2$ , $q = 3$ .

A1 — values of $a, b, p, q$ all correct, with the $b = -2$ recognised as negative (because the curve is centred at $(1, -2)$ , and the form $(y - b)^2$ requires $b$ to carry the sign).

(b) Step 1 — bound the parametric expressions.

Since $-1 \leq \cos t \leq 1$ :

$-1 \leq \dfrac{x - 1}{2} \leq 1 \implies -2 \leq x - 1 \leq 2 \implies -1 \leq x \leq 3.$

M1 — using the bounds of $\cos t$ to bound $x$ . The mark is awarded for showing the bounding inequality, not just stating the answer.

Since $-1 \leq \sin t \leq 1$ :

$-1 \leq \dfrac{y + 2}{3} \leq 1 \implies -3 \leq y + 2 \leq 3 \implies -5 \leq y \leq 1.$

A1 — both ranges correct: $x \in [-1, 3]$ and $y \in [-5, 1]$ .

Step 2 — geometric identification.

The Cartesian equation $\dfrac{(x-1)^2}{4} + \dfrac{(y+2)^2}{9} = 1$ is an ellipse with centre $(1, -2)$ , semi-major axis $3$ (vertical, since $q > p$ ) and semi-minor axis $2$ (horizontal). As $t$ varies over $[0, 2\pi)$ , the point traces the entire ellipse exactly once.

A1 — geometric description: ellipse, with centre and semi-axes correctly identified, plus the observation that the full curve is traced once.

Total: 8 marks (M3 A5, split as shown).

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Question (6 marks): A curve $C$ has parametric equations

$x = t^2 - 4, \qquad y = 2t + 1, \qquad t \in \mathbb{R}.$

(a) Find a Cartesian equation for $C$ , giving your answer in the form $x = f(y)$ . (4)

(b) State the coordinates of the point on $C$ where the curve crosses the $x$ -axis. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — making $t$ the subject of the simpler equation: $y = 2t + 1 \implies t = \dfrac{y - 1}{2}$ .
M1 (AO1.1b) — substituting into $x = t^2 - 4$ : $x = \left(\dfrac{y - 1}{2}\right)^2 - 4$ .
A1 (AO1.1b) — expanding correctly: $x = \dfrac{(y - 1)^2}{4} - 4$ or equivalently $x = \dfrac{(y-1)^2 - 16}{4}$ .
A1 (AO2.5) — fully simplified form $x = \dfrac{y^2 - 2y - 15}{4}$ or factorised $x = \dfrac{(y - 5)(y + 3)}{4}$ , presented in the form $x = f(y)$ as requested.

(b)

M1 (AO1.1b) — recognising that on the $x$ -axis $y = 0$ , then either substituting into $y = 2t + 1$ to get $t = -\tfrac{1}{2}$ , or substituting $y = 0$ into the Cartesian equation.
A1 (AO1.1b) — correct point: $x = (-\tfrac{1}{2})^2 - 4 = -\tfrac{15}{4}$ , giving $\left(-\tfrac{15}{4}, 0\right)$ .

Total: 6 marks split AO1 = 5, AO2 = 1. This is an AO1-dominated procedural question, characteristic of the first parametric question on Paper 2: technique is rewarded over insight, and the AO2 mark is reserved for the final fully-simplified or factorised presentation. Typical examiner expectation is that the algebraic elimination route (rearrange one equation, substitute into the other) is the canonical method when one of the parametric equations is linear in $t$ ; the trig identity is reserved for trigonometric parametrisations.

Synoptic links

Connects to:

Section 9 — Parametric differentiation: the chain rule gives $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ , valid wherever $\dfrac{dx}{dt} \neq 0$ . For $x = 2\cos t + 1, y = 3\sin t - 2$ : $\dfrac{dx}{dt} = -2\sin t$ , $\dfrac{dy}{dt} = 3\cos t$ , so $\dfrac{dy}{dx} = -\dfrac{3\cos t}{2\sin t} = -\dfrac{3}{2}\cot t$ . Tangent and normal questions on Paper 2 routinely require this chain, often without first eliminating the parameter.
Section 5 — Trigonometric identities: the elimination technique $\cos^2 t + \sin^2 t = 1$ is the pivotal identity, but harder questions exploit $\sec^2 t - \tan^2 t = 1$ , $\cosh^2 t - \sinh^2 t = 1$ (further maths), and double-angle identities $\sin 2t = 2\sin t\cos t$ . A parametric curve given as $x = \cos 2t, y = \sin t$ requires $\cos 2t = 1 - 2\sin^2 t$ to eliminate, producing the parabola $x = 1 - 2y^2$ .
Section 10 — Integration of parametric curves: the area under a parametric curve between $t = a$ and $t = b$ is $\int_a^b y \, \dfrac{dx}{dt}\,dt$ . For an ellipse $x = a\cos t, y = b\sin t$ , the enclosed area $\pi a b$ falls out of $\int_0^{2\pi} (b\sin t)(-a\sin t)\,dt = -ab\int_0^{2\pi} \sin^2 t\,dt$ (with appropriate sign convention).
Conic sections: circles $(x = a + r\cos t, y = b + r\sin t)$ , ellipses $(x = a\cos t, y = b\sin t)$ , parabolas $(x = at^2, y = 2at)$ and hyperbolas $(x = a\sec t, y = b\tan t)$ all admit clean parametrisations. Recognising the parametric signature of each conic — trig pair for circle/ellipse, polynomial pair for parabola, sec/tan pair for hyperbola — is examiner-rewarded pattern recognition.
9MA0-03 Mechanics — projectile motion: the position of a projectile at time $t$ is $x = (u\cos\alpha)t,\ y = (u\sin\alpha)t - \tfrac{1}{2}gt^2$ , a parametric pair with $t$ as time. Eliminating $t$ gives the Cartesian trajectory $y = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha}$ — a parabola. The elimination technique is identical to Pure section 8.

Mark-scheme literacy

Parametric equations questions on 9MA0 Paper 2 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Eliminating the parameter, rearranging parametric equations, applying the Pythagorean identity, finding points by substitution, sketching with correct table-of-values method
AO2 (reasoning / interpretation)	20–30%	Identifying the resulting Cartesian curve geometrically (ellipse, parabola, etc.), stating domain and range correctly, justifying choice of elimination method, recognising orientation
AO3 (problem-solving)	5–15%	Modelling-context questions where parametric equations describe a real situation (motion, curve design); choosing a strategy when neither elimination route is direct

Examiner-rewarded phrasing: "since $\cos^2 t + \sin^2 t = 1$ for all $t$ "; "the parameter $t$ ranges over $[0, 2\pi)$ , so the entire ellipse is traced once"; "the Cartesian equation does not capture the direction of traversal, which is information held by the parametric form". Phrases that lose marks: omitting the domain of the parameter when it is restricted ("for $0 \leq t \leq \pi$ "); writing $y = \sqrt{...}$ when both branches are needed (or vice versa); leaving the answer as an implicit relation when the question demands $x = f(y)$ or $y = f(x)$ .

A specific Edexcel pattern: questions that say "find the Cartesian equation of $C$ , stating the values of any constants" require the candidate to declare each constant explicitly — answers buried inside an unsimplified fraction often lose the final A1.

Grade-band model answers

3-mark question

Question: A curve has parametric equations $x = t^2, y = 2t$ for $-2 \leq t \leq 2$ . Find the coordinates of the point on the curve when $t = -\tfrac{3}{2}$ , and state whether this point lies on the upper or lower branch of the curve.

Grade C response (~210 words):

Substitute $t = -\tfrac{3}{2}$ into both equations:

$x = (-\tfrac{3}{2})^2 = \tfrac{9}{4}$ and $y = 2 \times -\tfrac{3}{2} = -3$ .

The point is $(\tfrac{9}{4}, -3)$ .

Since $y = -3 < 0$ , the point lies on the lower branch.

Parametric Equations

Parametric Equations

What Are Parametric Equations?

Why Use Parametric Equations?

Common Parametric Forms

Circle

Parabola

Trigonometric Parametrisation

Sketching Parametric Curves

Finding Points from Parameters

Restricting the Parameter

Determining Direction of Travel

Key Terms

Summary

A-Level Deep Dive: Parametric Equations

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the Edexcel 9MA0 Paper 2 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

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