Introduction to Parametric Equations

Builds on: aqa-alevel-maths-pure-2 / parametric-equations — this lesson is the A2-level deep dive.

This lesson introduces parametric equations, a powerful way to describe curves in the coordinate plane. Instead of expressing y directly in terms of x, we express both x and y in terms of a third variable called a parameter (usually t or θ). Parametric equations are an important topic in A-Level Mathematics, appearing in pure mathematics, mechanics (where time is a natural parameter), and further mathematics.

Spec Mapping — AQA 7357 Section C (Coordinate geometry). This lesson covers the introductory parametric-equations content of Section C — definitions, conversion to Cartesian form, and curve recognition — at the depth required for A2-level synoptic questions. Refer to the official AQA specification document for exact wording.

What Are Parametric Equations?

A curve is defined parametrically when both coordinates are given as functions of a parameter:

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

As the parameter t varies, the point (x, y) traces out the curve. Each value of t gives a specific point on the curve.

Example 1: The parametric equations x = 2t, y = t² define a curve. Find some points.

t	x = 2t	y = t²	Point
−2	−4	4	(−4, 4)
−1	−2	1	(−2, 1)
0	0	0	(0, 0)
1	2	1	(2, 1)
2	4	4	(4, 4)

Plotting these points reveals a parabola.

Converting from Parametric to Cartesian

To find the Cartesian equation (an equation involving only x and y), eliminate the parameter t.

Method 1: Direct substitution

If you can express t in terms of x (or y) from one equation, substitute into the other.

Example 2: Convert x = 2t, y = t² to a Cartesian equation.

\begin{aligned} \text{From } x = 2t &: t = x/2 \\ \text{Substitute into } y = t^{2} &: y = (x/2)^{2} = x^{2}/4 \end{aligned}

Cartesian equation: y = x²/4. This is a parabola.

Method 2: Using trigonometric identities

When the parameter is θ and the equations involve sin θ and cos θ, use the identity sin²θ + cos²θ = 1.

Example 3: Convert x = 3cos θ, y = 3sin θ to a Cartesian equation.

\begin{aligned} \frac{x}{3} = \cos\theta \quad &\text{and} \quad \frac{y}{3} = \sin\theta \\[6pt] \cos^{2}\theta + \sin^{2}\theta &= 1 \\ \left(\frac{x}{3}\right)^{2} + \left(\frac{y}{3}\right)^{2} &= 1 \\ \frac{x^{2}}{9} + \frac{y^{2}}{9} &= 1 \\ x^{2} + y^{2} &= 9 \end{aligned}

This is a circle with centre (0, 0) and radius 3.

Example 4: Convert x = 2 + 5cos θ, y = −1 + 5sin θ to Cartesian form.

\begin{aligned} \frac{x - 2}{5} = \cos\theta \quad &\text{and} \quad \frac{y + 1}{5} = \sin\theta \\[6pt] \cos^{2}\theta + \sin^{2}\theta &= 1 \\ \left(\frac{x - 2}{5}\right)^{2} + \left(\frac{y + 1}{5}\right)^{2} &= 1 \\ (x - 2)^{2} + (y + 1)^{2} &= 25 \end{aligned}

This is a circle with centre (2, −1) and radius 5.

Method 3: Using other identities or algebraic manipulation

Example 5: Convert x = t + 1/t, y = t − 1/t to a Cartesian equation.

\begin{aligned} x + y &= \left(t + \tfrac{1}{t}\right) + \left(t - \tfrac{1}{t}\right) = 2t &&\text{so } t = \frac{x + y}{2} \\[4pt] x - y &= \left(t + \tfrac{1}{t}\right) - \left(t - \tfrac{1}{t}\right) = \tfrac{2}{t} &&\text{so } \tfrac{1}{t} = \frac{x - y}{2} \\[6pt] t \times \tfrac{1}{t} &= 1 \\ \left(\frac{x + y}{2}\right) \times \left(\frac{x - y}{2}\right) &= 1 \\ \frac{(x + y)(x - y)}{4} &= 1 \\ x^{2} - y^{2} &= 4 \end{aligned}

This is a rectangular hyperbola.

Sketching Parametric Curves

To sketch a parametric curve:

Make a table of values for a range of t.
Plot the (x, y) points.
Join them smoothly, noting the direction of travel as t increases (often shown with an arrow).
Note any special features: is the curve closed? Does it self-intersect?

Example 6: Sketch x = cos t, y = sin 2t for 0 ≤ t ≤ 2π.

$t$	$x = \cos t$	$y = \sin 2t$
$0$	$1$	$0$
$\pi/4$	$\sqrt{2}/2 \approx 0.71$	$1$
$\pi/2$	$0$	$0$
$3\pi/4$	$-\sqrt{2}/2 \approx -0.71$	$-1$
$\pi$	$-1$	$0$
$5\pi/4$	$-\sqrt{2}/2$	$1$
$3\pi/2$	$0$	$0$
$7\pi/4$	$\sqrt{2}/2$	$-1$
$2\pi$	$1$	$0$

This traces a figure-of-eight (lemniscate-like) curve. The curve crosses itself at the origin.

Common Parametric Forms

Curve	Parametric Equations	Cartesian Equation
Circle, centre origin, radius r	x = r cos θ, y = r sin θ	x² + y² = r²
Circle, centre (a, b), radius r	x = a + r cos θ, y = b + r sin θ	(x−a)² + (y−b)² = r²
Parabola	x = at², y = 2at	y² = 4ax
Ellipse	x = a cos θ, y = b sin θ	x²/a² + y²/b² = 1

Converting from Cartesian to Parametric

A single Cartesian equation can have many different parametric representations. Common approaches:

For y = f(x): use x = t, y = f(t).
For circles: use cos and sin.
For expressions involving squares: try substitutions like x = t².

Example 7: Write y = x² − 3x + 1 parametrically.

\begin{aligned} &\text{Let } x = t. \text{ Then } y = t^{2} - 3t + 1. \\ &\text{So } x = t, \quad y = t^{2} - 3t + 1. \end{aligned}

Finding Points of Intersection with Axes

x-axis (y = 0)

Set g(t) = 0 and solve for t. Substitute back to find x.

y-axis (x = 0)

Set f(t) = 0 and solve for t. Substitute back to find y.

Example 8: The curve has parametric equations x = t² − 4, y = t³ − t. Find where it crosses the y-axis.

\begin{aligned} \text{Set } x = 0: \quad t^{2} - 4 = 0 \ &\implies\ t = \pm 2 \\[6pt] \text{When } t = 2: \ y = 8 - 2 = 6 \quad &\text{Point: } (0,\ 6) \\ \text{When } t = -2: \ y = -8 + 2 = -6 \quad &\text{Point: } (0,\ -6) \end{aligned}

Specimen Question (Modelled on the AQA Paper Format)

Note: this question is constructed to model AQA Paper 1/2/3 style; it is not a reproduction of any published past paper.

Worked Examination-Style Problem

Problem: A curve is defined parametrically by x = 3t², y = 6t.

(a) Find the Cartesian equation of the curve.

(b) Find the coordinates of the points where the line y = 2x − 6 meets the curve.

Solution:

(a)

\begin{aligned} \text{From } y = 6t &: t = y/6 \\ \text{Substitute} &: x = 3(y/6)^{2} = 3y^{2}/36 = y^{2}/12 \\ &\therefore y^{2} = 12x \end{aligned}

(b) Substitute y = 2x − 6 into y² = 12x:

\begin{aligned} (2x - 6)^{2} &= 12x \\ 4x^{2} - 24x + 36 &= 12x \\ 4x^{2} - 36x + 36 &= 0 \\ x^{2} - 9x + 9 &= 0 \\[4pt] x &= \frac{9 \pm \sqrt{81 - 36}}{2} = \frac{9 \pm \sqrt{45}}{2} = \frac{9 \pm 3\sqrt{5}}{2} \end{aligned}

The corresponding y-values are y = 2x − 6.

Alternatively, using the parameter directly:

\begin{aligned} 6t = 2(3t^{2}) - 6 \\ 6t = 6t^{2} - 6 \\ 6t^{2} - 6t - 6 = 0 \\ t^{2} - t - 1 = 0 \\ t = (1 \pm \sqrt{5}) / 2 \end{aligned}

When $t = \dfrac{1 + \sqrt{5}}{2}$ :

\begin{aligned} x &= 3\left(\frac{1 + \sqrt{5}}{2}\right)^{2} = \frac{3(6 + 2\sqrt{5})}{4} = \frac{18 + 6\sqrt{5}}{4} = \frac{9 + 3\sqrt{5}}{2} \\[4pt] y &= \frac{6(1 + \sqrt{5})}{2} = 3(1 + \sqrt{5}) = 3 + 3\sqrt{5} \end{aligned}

Domain Restrictions

Sometimes the parameter is restricted to a certain interval, which means only part of the Cartesian curve is traced. Always note any parameter restrictions and how they affect the range of x and y.

Example 9: x = 2cos t, y = 2sin t for 0 ≤ t ≤ π.

The Cartesian equation is x² + y² = 4 (a full circle), but since 0 ≤ t ≤ π, we have sin t ≥ 0, so y ≥ 0. Only the upper semicircle is traced.

Synoptic Links

This topic connects to:

Parametric equations (AS-level base) (aqa-alevel-maths-pure-2 / parametric-equations) — the foundational treatment of curves expressed in parameter form.
Parametric differentiation (aqa-alevel-maths-coordinate-geometry-depth / parametric-differentiation) — applying calculus to parametric curves to find gradients of tangents and normals.
Mechanics: projectile motion (aqa-alevel-maths-mechanics / projectile-motion) — projectile equations are naturally parametric in time t.

Summary

Parametric equations express x and y separately as functions of a parameter t (or θ).
To find the Cartesian equation, eliminate the parameter.
Use sin²θ + cos²θ = 1 when the parametric equations involve trigonometric functions.
To sketch, make a table of values and plot the points, noting the direction of travel.
Axis crossings: set x = 0 or y = 0 and solve for the parameter.
Parameter restrictions affect which part of the Cartesian curve is traced.

Exam Tip: When converting parametric to Cartesian form, clearly state how you eliminate the parameter. For trigonometric parametric equations, the identity sin²θ + cos²θ = 1 is almost always the key step. When sketching, always include the direction of increasing parameter — the examiner expects to see arrows on parametric curves.

A-Level Deep Dive: Parametric Equations Introduction

Spec mapping

AQA 7357 specification, Paper 2 — Pure Mathematics, Section C: Coordinate Geometry, Year 2 sub-strand on parametric curves covers parametric equations of curves and conversion between Cartesian and parametric forms; 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 sits squarely in Year 2 Pure content and is examined on Paper 2 alongside other coordinate-geometry items, with synoptic reach into Section H (Differentiation, where $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ underpins parametric calculus), Section I (Integration, where $\int y\,dx = \int y(t)\,\dfrac{dx}{dt}\,dt$ converts parametric area integrals) and Section O (Mechanics — kinematics in two dimensions, where projectile motion is intrinsically parametric in time $t$ ). The AQA formula booklet does not list parametric-to-Cartesian conversion identities — students must recognise the key trigonometric and algebraic eliminations themselves.

Worked example with full mark scheme

Question (8 marks):

A curve $C$ has parametric equations

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

(a) Show that the Cartesian equation of $C$ is $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ . (3)

(b) State the domain and range of the curve in Cartesian form. (2)

Solution with mark scheme:

(a) Step 1 — isolate the trigonometric functions.

From $x = 3\cos t$ , we have $\cos t = \dfrac{x}{3}$ . From $y = 2\sin t$ , we have $\sin t = \dfrac{y}{2}$ .

M1 — rearranging both parametric equations to express $\cos t$ and $\sin t$ in terms of $x$ and $y$ . A common slip is to leave one of these implicit and try to substitute later — examiners reward the explicit isolation step.

Step 2 — apply the Pythagorean identity.

Using $\cos^2 t + \sin^2 t = 1$ :

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

M1 — substituting into the correct identity. The choice of identity is the key reasoning step; $\cos^2 t + \sin^2 t = 1$ is the only Pythagorean identity that eliminates $t$ cleanly here.

Step 3 — present in the printed form.

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

A1 — printed answer obtained. AO2 mark for the deductive chain from parametric form to printed Cartesian equation.

(b) Step 1 — extract domain from parameter range.

Since $\cos t$ ranges over $[-1, 1]$ as $t$ varies, $x = 3\cos t$ ranges over $[-3, 3]$ .

B1 — domain stated as $-3 \leq x \leq 3$ .

Step 2 — extract range from parameter range.

Similarly $y = 2\sin t$ ranges over $[-2, 2]$ .

B1 — range stated as $-2 \leq y \leq 2$ .

$\dfrac{dx}{dt} = -3\sin t$ and $\dfrac{dy}{dt} = 2\cos t$ , so

$\dfrac{dy}{dx} = \dfrac{2\cos t}{-3\sin t} = -\dfrac{2\cos t}{3\sin t}$

M1 — correct application of the parametric chain rule.

Step 2 — set gradient equal to 1 and solve.

$-\dfrac{2\cos t}{3\sin t} = 1 \implies 2\cos t = -3\sin t \implies \tan t = -\dfrac{2}{3}$ .

M1 — equation in $\tan t$ obtained.

Step 3 — back-substitute to coordinates.

Two values of $t$ in $[0, 2\pi)$ satisfy $\tan t = -\tfrac{2}{3}$ , one in the second quadrant and one in the fourth. Using $\sin t = \pm \tfrac{2}{\sqrt{13}}$ and $\cos t = \mp \tfrac{3}{\sqrt{13}}$ , the points are $\left(-\dfrac{9}{\sqrt{13}}, \dfrac{4}{\sqrt{13}}\right)$ and $\left(\dfrac{9}{\sqrt{13}}, -\dfrac{4}{\sqrt{13}}\right)$ .

A1 — both points correct (or rationalised equivalents).

Total: 8 marks (M5 A2 B1 split as shown).

Specimen question modelled on the AQA 7357 Paper 2 format

Question (6 marks): A curve has parametric equations $x = t^2 - 2$ , $y = 2t + 1$ for $t \in \mathbb{R}$ .

(a) Eliminate the parameter $t$ to find the Cartesian equation of the curve. (3)

(b) State the type of conic and identify the vertex. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — making $t$ the subject of $y = 2t + 1$ , giving $t = \dfrac{y - 1}{2}$ .
M1 (AO1.1b) — substituting into $x = t^2 - 2$ to obtain $x = \left(\dfrac{y - 1}{2}\right)^2 - 2$ .
A1 (AO1.1b) — simplified form $x = \dfrac{(y - 1)^2}{4} - 2$ or equivalent $4(x + 2) = (y - 1)^2$ .

(b)

B1 (AO2.4) — identifying the curve as a parabola (sideways-opening, axis horizontal).
M1 (AO2.2a) — recognising vertex as point where $(y - 1)^2 = 0$ , i.e. $y = 1$ .
A1 (AO1.1b) — vertex coordinates $(-2, 1)$ .

Total: 6 marks split AO1 = 4, AO2 = 2. The synoptic AO2 marks reward recognising the standard form of a horizontally-oriented parabola — a connection back to coordinate-geometry conics that pure-procedural candidates often miss.

Synoptic links

Connects to:

Section H — Parametric differentiation: the chain rule $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ unlocks tangent and normal computation directly from parametric form, without first eliminating $t$ . For complex curves this is faster and avoids implicit-differentiation traps.
Section E — Trigonometric identities: elimination of the parameter relies on $\cos^2 t + \sin^2 t = 1$ , $\sec^2 t - \tan^2 t = 1$ , $\cosh^2 t - \sinh^2 t = 1$ (further maths). Recognising which identity matches a given parametric pair is itself a synoptic skill.

Introduction to Parametric Equations

Introduction to Parametric Equations

What Are Parametric Equations?

Converting from Parametric to Cartesian

Sketching Parametric Curves

Common Parametric Forms

Converting from Cartesian to Parametric

Finding Points of Intersection with Axes

x-axis (y = 0)

y-axis (x = 0)

Specimen Question (Modelled on the AQA Paper Format)

Worked Examination-Style Problem

Domain Restrictions

Synoptic Links

Summary

A-Level Deep Dive: Parametric Equations Introduction

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 2 format

Synoptic links

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