OCR A Level Mathematics A: Parametric equations of curves and their use in modelling (H240) – Revision Notes

These notes condense parametric equations of curves for OCR A Level Mathematics A (H240), specification Version 3.1 (August 2026), content statements 1.03g and 1.03h. Both statements are in the Stage 2 column, so they are A Level only. For full explanations and longer worked examples, read the study guide first. Differentiating parametric equations is statement 1.07s and is not covered here. Find the rest of the course on the OCR A Level Mathematics hub and tick off skills on the printable checklist.

Key terms

• Parameter: the extra variable (t or θ) that gives x and y separately.
• Parametric equations: x = f(t) and y = g(t). Each value of the parameter gives one point.
• Direction: the curve is traced in the direction of increasing parameter. A Cartesian equation has no direction.
• Range of the parameter: decides which part of the curve exists, and can make some Cartesian points unreachable.

Conversion toolkit

Situation	Move
x or y is linear in t	Make t the subject of that equation and substitute into the other
Both x and y contain sin θ and cos θ	Rearrange to sin θ and cos θ alone, then use sin²θ + cos²θ ≡ 1
One equation has a double angle	Use cos 2θ ≡ 1 − 2sin²θ ≡ 2cos²θ − 1 or sin 2θ ≡ 2 sin θ cos θ
tan and sec appear	Use sec²θ ≡ 1 + tan²θ
Sum or difference of sin and cos	Square it, then use sin²θ + cos²θ ≡ 1 and sin 2θ
Cartesian y = f(x) to parametric	Let x = t, so y = f(t)

Standard parametric forms

Curve	Cartesian	Parametric
Circle of radius r about (a, b)	(x − a)² + (y − b)² = r²	y = b + r sin θ with x = a + r cos θ
Ellipse, centre (a, b)	((x − a)/p)² + ((y − b)/q)² = 1	x = a + p cos θ, y = b + q sin θ
Line segment from A to B	straight line between the points	x = x_A + t(x_B − x_A), y = y_A + t(y_B − y_A), 0 ≤ t ≤ 1

Parametric forms are not unique. Many correct answers exist for the same curve.

Method: parametric to Cartesian

1. Look at which equation is simplest to rearrange.
2. Eliminate the parameter, by substitution or by an identity.
3. Simplify to a relation between x and y.
4. Work out the range of x (and y) that the parameter allows.
5. State the final answer with its restriction.

Reminder. x = 2 − 5t and y = t² + 1 give t = (2 − x)/5 and y = (2 − x)²/25 + 1. If t ≥ 0 is stated, then x ≤ 2.

Reminder with restriction. x = 3 − 2cos θ means −1 ≤ cos θ ≤ 1, so 1 ≤ x ≤ 5.

Method: Cartesian to parametric

1. For a circle or ellipse, complete the square to find the centre and radii.
2. Match each squared bracket to cos²θ and sin²θ.
3. Write x and y with a range 0 ≤ θ < 2π, unless told otherwise.
4. Check by substituting back into the Cartesian equation.

Reminder. x² + y² + 10x − 2y + 10 = 0 gives (x + 5)² + (y − 1)² = 16. So x = −5 + 4 cos θ, y = 1 + 4 sin θ.

Method: sketching

1. Table of values, including both ends of the parameter range.
2. Axis crossings: y-axis when x(t) = 0, x-axis when y(t) = 0.
3. Cartesian form for the overall shape.
4. Greatest and least x and y.
5. Arrow for the direction of increasing parameter.

If a curve crosses itself, two different values of t give the same point. Equal x and y at two parameter values means a self-intersection.

Reminder. For x = t² − 2, y = t³ − 2t, both t = √2 and t = −√2 give (0, 0), so the curve passes through the origin twice. The value t = 0 gives (−2, 0).

Modelling checklist

1. Identify what the parameter means. Usually it is time.
2. Substitute a time into both equations to get a position.
3. For greatest height of a projectile, find the turning point of the y quadratic in t (a quadratic ax² + bx + c has its turning point at x = −b/(2a)).
4. For landing, solve y = 0 and reject roots outside the range of t.
5. For range, substitute the landing time into x.
6. To test a barrier at a given horizontal distance, find t from the x-equation, then check y.
7. For a time spent above or below a height, solve an inequality in t.
8. Use exact or unrounded values until the last line.
9. Comment on the model: air resistance ignored, particle model, valid only for the stated range of t.

Reminder: barrier test. A ball has x = 10t, y = 12t − 5t². A wall 3 m high stands 15 m away.

x = 15  gives  t = 1.5
y = 12(1.5) - 5(1.5)² = 18 - 11.25 = 6.75
6.75 > 3, so the ball clears the wall

Reminder: time above a height. For the same ball, y ≥ 6 means 5t² − 12t + 6 ≤ 0. Solve the quadratic equation and take the interval between the roots. The roots are t = (6 ± √6)/5, about 0.710 and 1.690, so the duration is 2√6/5, about 0.980 s.

Paths versus objects

To find where two paths cross, eliminate the parameter in each and solve the Cartesian equations together. To find whether two objects meet, the parameters must also be equal at that point, because the objects must be there at the same time. Two paths can cross with no collision.

Must-know distinctions

• Cartesian equation vs parametric curve. The Cartesian equation may describe more points than the parametric curve does.
• Circle vs ellipse. A circle has the same coefficient on cos θ and sin θ. An ellipse has different coefficients.
• Centre sign. (x − 3)² gives centre x-coordinate +3, so x = 3 + r cos θ.
• Degrees vs radians. If the parameter appears as πt/15 or similar, work in radians.
• Path vs motion. The path is the shape. The parametric form also tells you when each point is reached.

Quick self-test

1. Find a Cartesian equation of x = 2t − 5, y = 4 − t.
2. Find a Cartesian equation of x = t², y = t³.
3. Find a Cartesian equation of x = 3 + 2 sin θ, y = 4 cos θ − 1.
4. Find a Cartesian equation of x = sin θ, y = 4 + cos 2θ, and state the range of x.
5. Write (x − 1)² + (y + 3)² = 9 in parametric form.
6. Give parametric equations for the segment from (0, 4) to (6, −2).
7. A ball has x = 5t, y = 20t − 5t² for t ≥ 0. Find the greatest height and the range.
8. A seat has x = 8 sin(πt/6), y = 10 − 8 cos(πt/6). Find its greatest height and the time for one turn.
9. Does the point (7, −3) lie on x = 2t + 1, y = t² − 4t?
10. State the range of x for x = 3 − 2 cos θ.

Answers

1. t = 4 − y, so x = 2(4 − y) − 5 = 3 − 2y. y = (3 − x)/2.
2. Square y: y² = t⁶ = (t²)³. y² = x³, with x ≥ 0.
3. ((x − 3)/2)² + ((y + 1)/4)² = 1, an ellipse.
4. cos 2θ = 1 − 2sin²θ, so y = 4 + 1 − 2x² = 5 − 2x², for −1 ≤ x ≤ 1.
5. x = 1 + 3 cos θ, y = −3 + 3 sin θ, 0 ≤ θ < 2π.
6. x = 6t, y = 4 − 6t, 0 ≤ t ≤ 1. Other correct forms exist.
7. Turning point at t = 2, so y = 40 − 20 = 20. Landing: 5t(4 − t) = 0, t = 4, so x = 20.
8. Greatest height at cos = −1, so y = 10 + 8 = 18, reached at t = 6. One turn is 2π ÷ (π/6) = 12 s.
9. x = 7 gives t = 3. Then y = 9 − 12 = −3. Yes, at t = 3.
10. −1 ≤ cos θ ≤ 1, so 1 ≤ x ≤ 5.

Where marks are usually lost

• Leaving out the range of x or y after eliminating a trigonometric parameter.
• Forgetting a point that is missing because the parameter can never reach it.
• Giving a circle’s centre with the wrong sign.
• Using a single coefficient for an ellipse that needs two.
• Not writing down the identity used, so the step looks unsupported. In a “show that” question, show each stage.
• Treating “paths cross” as “objects meet”.
• Rounding t early and carrying a rounded value into x or y.
• Keeping a negative time or a time past landing.
• Mixing degrees and radians in a model with π in it.
• Finishing a modelling question with a bare number and no units or context.

Next steps

Try the practice questions under timed conditions, and use the free 10-minute diagnostics to check wider Pure Mathematics. The course-level Pure Mathematics revision notes put this unit in context.

Official syllabus

These notes follow the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. The content statements used are 1.03g and 1.03h.