In most of A-Level Maths, a curve is written as a single equation linking x and y, such as y = x² + 3x. Parametric equations describe a curve differently: both x and y are written in terms of a third variable, called the parameter and usually labelled t or θ. As t changes, the point (x, y) moves, tracing out the curve. This is a natural way to describe motion — t can represent time — and it also lets us describe curves, such as circles and ellipses, that are awkward to write as y = f(x).
Parametric equations appear in the Pure content of every A-Level Maths specification, including Edexcel, AQA, OCR and OCR MEI. Questions usually combine several skills: converting to Cartesian form, differentiation, integration, trigonometric identities and sometimes modelling. This guide covers each part in turn.
Understanding a parametric curve
Take x = t², y = 2t. To understand the curve, build a short table: when t = −2 the point is (4, −4); when t = 0 it is (0, 0); when t = 2 it is (4, 4). Plotting these shows a parabola lying on its side. The parameter does not appear on the graph itself — it simply tells you which point you are at.
Pay close attention to the domain of the parameter. If the question says 0 ≤ t ≤ 3, the curve is only the section traced between those values, not the whole Cartesian curve. Sketches and areas both depend on this, and examiners frequently restrict the domain deliberately.
Converting parametric equations to Cartesian form
The aim is to eliminate the parameter and get a single equation in x and y. There are two main methods, and recognising which one to use is half the skill.
Method 1: rearrange and substitute
If one equation is easy to rearrange for t, do that and substitute into the other. For x = t + 1, y = t² − 3, rearrange to t = x − 1 and substitute: y = (x − 1)² − 3. This works for most algebraic parametric equations.
Method 2: use a trigonometric identity
If the equations involve sin θ and cos θ, rearranging for θ is messy. Instead, isolate sin θ and cos θ and use sin²θ + cos²θ = 1. For x = 3 cos θ, y = 3 sin θ: cos θ = x/3 and sin θ = y/3, so x²/9 + y²/9 = 1, giving x² + y² = 9 — a circle of radius 3 centred at the origin. For x = 2 + 5 cos θ, y = −1 + 5 sin θ, the same method gives (x − 2)² + (y + 1)² = 25, a circle centred at (2, −1).
| If the equations contain… | Try the identity… |
|---|---|
| sin θ and cos θ | sin²θ + cos²θ = 1 |
| sec θ and tan θ | 1 + tan²θ = sec²θ |
| cosec θ and cot θ | 1 + cot²θ = cosec²θ |
| cos 2θ and sin θ (or cos θ) | cos 2θ = 1 − 2sin²θ or 2cos²θ − 1 |
| sin 2θ and sin θ, cos θ | sin 2θ = 2 sin θ cos θ |
Parametric differentiation
You do not need to convert to Cartesian form to find a gradient. Use the chain rule: dy/dx = (dy/dt) ÷ (dx/dt). Differentiate each equation with respect to t separately, then divide.
For x = t², y = 2t: dx/dt = 2t and dy/dt = 2, so dy/dx = 2/(2t) = 1/t. At the point where t = 2, the gradient is 1/2 and the point is (4, 4), so the tangent is y − 4 = ½(x − 4). The normal has gradient −2. This tangent-and-normal structure is one of the most common parametric exam questions.
- Find dx/dt and dy/dt.
- Divide to get dy/dx in terms of t.
- Substitute the given value of t to find the gradient and the coordinates of the point.
- Use y − y₁ = m(x − x₁) for the tangent, or the negative reciprocal gradient for the normal.
If the question gives a point (x, y) rather than a value of t, you must first find t. Solve one of the parametric equations for t and check that the value also satisfies the other equation — sometimes one equation gives two possible values and only one is correct.
Stationary points, horizontal and vertical tangents
A horizontal tangent occurs where dy/dt = 0 (and dx/dt ≠ 0). A vertical tangent occurs where dx/dt = 0. Questions asking where a curve 'meets the x-axis' or 'has a stationary point' are really asking you to solve y = 0 or dy/dt = 0 for t, then find the coordinates.
Area under a parametric curve
The area under a curve is ∫ y dx. For a parametric curve, change the variable: ∫ y dx = ∫ y (dx/dt) dt, with the limits changed from x-values to t-values. This is the step most students get wrong, either by forgetting dx/dt or by keeping x-limits.
For example, find the area under x = t², y = 2t between x = 0 and x = 4, with t ≥ 0. The limits become t = 0 and t = 2. The integral is ∫ 2t × 2t dt = ∫ 4t² dt from 0 to 2, which equals 32/3. Writing the change of limits on a separate line earns a method mark and protects you from the most common error.
Modelling with parametric equations
Recent specifications include modelling questions, where parametric equations describe a real situation: the path of a ball, the shape of a track, or the position of a point on a wheel. The maths is the same, but you need to interpret your answers in context. Typical parts ask for the maximum height (find where dy/dt = 0), the horizontal distance travelled (find t when y = 0, then x), or a criticism of the model, such as ignoring air resistance or treating an object as a particle.
When asked to comment on a model, give a specific, contextual reason. 'The model is unrealistic' earns nothing; 'the model assumes the ball is not affected by air resistance, so the real horizontal distance would be shorter' is the kind of answer mark schemes reward.
A full exam-style worked example
A curve has parametric equations x = 2 sin t, y = cos 2t, for −π/2 ≤ t ≤ π/2. A typical question asks you to find the Cartesian equation, the gradient at a given point, and the area under the curve. Working through it shows how the separate skills join together.
For the Cartesian equation, the presence of cos 2t and sin t points to the identity cos 2t = 1 − 2sin²t. From the first equation, sin t = x/2, so y = 1 − 2(x/2)² = 1 − x²/2. Because sin t runs from −1 to 1 over the given interval, x runs from −2 to 2, so the domain is −2 ≤ x ≤ 2. The curve is a section of a downward parabola.
For the gradient at t = π/6, differentiate: dx/dt = 2 cos t and dy/dt = −2 sin 2t. So dy/dx = −2 sin 2t / (2 cos t) = −sin 2t / cos t, which simplifies using sin 2t = 2 sin t cos t to −2 sin t. At t = π/6 the gradient is −1, and the point is (1, 1/2). The tangent is therefore y − 1/2 = −(x − 1), or y = −x + 3/2. Notice how a trig identity simplified the derivative — examiners often build this in.
For the area between the curve and the x-axis, you could integrate the Cartesian form, but the question may insist on the parametric route. The curve meets the x-axis where 1 − x²/2 = 0, so x = ±√2, which corresponds to t = ±π/4. The area is ∫ cos 2t × 2 cos t dt between −π/4 and π/4. Checking with the Cartesian form, ∫(1 − x²/2) dx from −√2 to √2 gives 4√2/3, and both routes must agree. Using the Cartesian answer as a check is a habit that catches errors in limits or in the dx/dt factor.
Common mistakes in parametric questions
- Dividing dx/dt by dy/dt instead of dy/dt by dx/dt.
- Forgetting to change the limits from x to t in an area integral.
- Missing the dx/dt factor inside the area integral.
- Ignoring the given domain of the parameter when sketching or finding a Cartesian domain.
- Using the wrong trig identity, or not isolating sin θ and cos θ before squaring.
- Giving a gradient in terms of t when the question asks for a numerical value at a point.
How to revise parametric equations
Because parametric questions draw on trigonometry, differentiation and integration, weaknesses in any of those will show up here. A good revision sequence is: Cartesian conversion (both methods) until it is automatic, then parametric differentiation with tangents and normals, then area integrals, and finally full multi-part past paper questions. Aim to complete at least five full parametric questions from past papers before the exam, marking each strictly.
MathVault offers free A-Level practice by topic, which is useful for building each skill separately. MathVault Premium provides full worked solutions and exam-style question sets, so you can follow every step of a multi-part parametric question and see exactly where marks are awarded.
Getting one-to-one help
If parametric questions keep unravelling, it is often because one of the supporting skills — trig identities, the chain rule or changing limits in integration — is not yet secure. In one-to-one lessons I identify which skill is causing the problem and rebuild it, rather than repeating the same question type. I am a QTS fully qualified teacher and Deputy Head of Maths with over seven years of tutoring experience, and I teach A-Level students online and in London.
The short version
- Parametric equations give x and y in terms of a parameter; respect its domain.
- Convert to Cartesian form by substituting, or with sin²θ + cos²θ = 1 and related identities.
- dy/dx = (dy/dt) ÷ (dx/dt); find t first if you are given a point.
- Area = ∫ y (dx/dt) dt with the limits changed to t-values.
- In modelling questions, interpret answers and criticise the model in context.