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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover 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 are Stage 2 content, so they are A Level only. Learn the methods first from the study guide and revision notes. The wider course is on the OCR A Level Mathematics hub, and the printable checklist lists every skill. Marks are given in bold brackets. Use unrounded values until your final line, and give decimals to 3 significant figures unless told otherwise.

Questions

1. A curve has parametric equations x = 4t − 3, y = 2t² + t.

(a) Find the coordinates of the points where t = −1 and t = 2. [2] (b) At which point does the curve cut the y-axis? [2]

2. A curve has parametric equations x = 2t + 3, y = 1/t, for t > 0. Find a Cartesian equation of the curve and state the set of possible values of x. [3]

3. A curve has parametric equations x = 2 sin θ − 1, y = 3 cos θ + 2, for 0 ≤ θ < 2π.

(a) Find a Cartesian equation of the curve. [3] (b) State the greatest and least values of x. [1]

4. A curve has parametric equations x = 4 sin θ, y = 3 − 2 cos 2θ, for 0 ≤ θ ≤ 2π. Show that y = 1 + x²/4, and state the range of values of x. [4]

5.

(a) Write the circle x² + y² + 8x − 6y + 9 = 0 in parametric form. [3] (b) Write down parametric equations for the line segment from A(−2, 5) to B(4, −1), and use them to show that the segment lies on the line x + y = 3. [2]

6. A curve has parametric equations x = t² − 1, y = t³ − t.

(a) Find the coordinates of the points on the curve for t = −2, −1, 0, 1 and 2. [2] (b) Show that y² = x²(x + 1). [2] (c) Describe the shape of the curve for −1 ≤ x ≤ 0, and say what happens at the origin. [1]

7. A stone is thrown from the top of a cliff. Its position t seconds later is modelled by x = 8t, y = 10 + 12t − 5t², where x is the horizontal distance in metres from the cliff edge and y is the height in metres above the sea.

(a) Find the greatest height of the stone above the sea. [3] (b) Find the horizontal distance from the cliff edge at which the stone enters the sea. [3]

8. A drone flies so that its position t seconds after take-off is x = 30 cos(πt/20), y = 20 sin(πt/20), where x and y are in metres.

(a) Find a Cartesian equation of the drone’s path. [2] (b) Find the first time at which x = 15. [2] (c) Find the exact value of y at that time. [2]

9. A curve has parametric equations x = 3t + 2, y = t² − 6t. The line y = x − 10 intersects the curve in two places, called P and Q.

(a) Work out the coordinates of both points. [4] (b) Find the exact length of PQ. [2]

10. Particle P moves so that its position at time t seconds (t ≥ 0) is (3t, t²). Particle Q moves so that its position at time t seconds (t ≥ 0) is (15 − 3t, t + 1). Lengths are in metres.

(a) Find a Cartesian equation of the path of each particle. [3] (b) Show that the two paths cross at only one point, and find it. [2] (c) Show that P and Q do not collide. [2]

11. A football leaves the ground at t = 0 and is modelled by x = 15t, y = 10t − 5t², with t in seconds, where x is the horizontal distance in metres and y is the height in metres. A crossbar 2.4 m high is 24 m from the kick, in the line of the ball’s flight.

(a) Find a Cartesian equation of the path. [3] (b) Decide whether the ball passes above the crossbar. [3] (c) For how long is the ball higher than 4 m? [3] (d) State one way in which the model is unrealistic. [1]

Answers

1. (a) t = −1: x = −4 − 3 = −7, y = 2 − 1 = 1, giving (−7, 1). [1] t = 2: x = 5, y = 8 + 2 = 10, giving (5, 10). [1] (b) On the y-axis x = 0, so 4t − 3 = 0 and t = 3/4. [1] y = 2(9/16) + 3/4 = 15/8, so the point is (0, 15/8). [1] Examiner insight: give both coordinates of every point, and keep the exact fraction instead of a rounded decimal unless the question asks for one.

2. From x = 2t + 3, t = (x − 3)/2. [1] Then y = 1/t = 2/(x − 3), so y = 2/(x − 3). [1] Since t > 0, x = 2t + 3 > 3, so x > 3. [1] Examiner insight: the equation alone is not the full answer when the question asks for the values of x. Link the stated range of t to x explicitly.

3. (a) sin θ = (x + 1)/2 and cos θ = (y − 2)/3. [1] Use sin²θ + cos²θ = 1. [1] ((x + 1)/2)² + ((y − 2)/3)² = 1. [1] (b) sin θ ranges from −1 to 1, so x runs from −3 to 1: greatest 1, least −3. [1] Examiner insight: writing the identity you are using makes the method clear to the reader. Mark the centre’s coordinates carefully, since (x + 1) means the centre has x-coordinate −1.

4. cos 2θ = 1 − 2sin²θ. [1] So y = 3 − 2(1 − 2sin²θ) = 1 + 4sin²θ. [1] From x = 4 sin θ, sin θ = x/4, so y = 1 + 4(x²/16) = 1 + x²/4. [1] Since −1 ≤ sin θ ≤ 1, −4 ≤ x ≤ 4. [1] Examiner insight: in a “show that” question every line of algebra must appear, with the substitution stated. The final range is easy to forget, so write it as a separate line.

5. (a) Complete the square: (x + 4)² − 16 + (y − 3)² − 9 + 9 = 0, so (x + 4)² + (y − 3)² = 16. [1] Centre (−4, 3), radius 4. [1] x = −4 + 4 cos θ, y = 3 + 4 sin θ, 0 ≤ θ < 2π. [1] (b) x = −2 + 6t, y = 5 − 6t, 0 ≤ t ≤ 1. [1] Then x + y = (−2 + 6t) + (5 − 6t) = 3, as required. [1] Examiner insight: the parametric form of a circle is not unique, but the centre and radius must match the completed square. Check by substituting back.

6. (a) t = −2: (3, −6); t = −1: (0, 0); t = 0: (−1, 0); t = 1: (0, 0); t = 2: (3, 6). [1] (any three correct) [1] (all five correct) (b) y = t(t² − 1) = t·x, and t² = x + 1. [1] So y² = t²x² = x²(x + 1). [1] (c) For −1 ≤ x ≤ 0 the curve forms a loop between (−1, 0) and (0, 0). The curve crosses itself at the origin, because t = −1 and t = 1 both give (0, 0). [1] Examiner insight: a table of values with both t and the matching points shows why a curve crosses itself. Equal points from different parameter values are the evidence.

7. (a) The greatest value of y is at the turning point of the quadratic 10 + 12t − 5t². [1] This is at t = 12/10 = 1.2 (halfway between the roots of 12t − 5t² = 0). [1] y = 10 + 14.4 − 7.2 = 17.2 m. [1] (b) The stone enters the sea when y = 0: 5t² − 12t − 10 = 0. [1] t = (12 + √344)/10 = 3.0547…, rejecting the negative root. [1] x = 8 × 3.0547… = 24.4 m (3 s.f.). [1] Examiner insight: reject the negative root because the model begins at t = 0, and say so. Use the unrounded t when you calculate x.

8. (a) cos(πt/20) = x/30 and sin(πt/20) = y/20. [1] x²/900 + y²/400 = 1. [1] (b) 30 cos(πt/20) = 15, so cos(πt/20) = 1/2. [1] The first solution is πt/20 = π/3, so t = 20/3 s (6.67 s). [1] (c) y = 20 sin(π/3) = 20 × √3/2. [1] y = 10√3. [1] Examiner insight: “exact” means leave √3 in the answer. Choose the smallest positive angle for “first time” and work in radians, since the model uses π.

9. (a) Put x and y into y = x − 10: t² − 6t = (3t + 2) − 10. [1] t² − 9t + 8 = 0, so (t − 1)(t − 8) = 0. [1] t = 1 gives (5, −5). [1] t = 8 gives (26, 16). [1] (b) PQ² = (26 − 5)² + (16 + 5)² = 882. [1] PQ = 21√2. [1] Examiner insight: find t first, then both coordinates for each value. Check each point on the line, since 5 − 10 = −5 and 26 − 10 = 16.

10. (a) For P, t = x/3, so y = x²/9. [1] For Q, t = (15 − x)/3, so y = (15 − x)/3 + 1 = 6 − x/3. [1] A statement of each domain, x ≥ 0 for P and x ≤ 15 for Q. [1] (b) Set x²/9 = 6 − x/3, so x² + 3x − 54 = 0. [1] (x + 9)(x − 6) = 0. x = −9 is impossible for P as x ≥ 0, so the only crossing is (6, 4). [1] (c) P is at (6, 4) when 3t = 6, so t = 2. [1] Q is at (6, 4) when 15 − 3t = 6, so t = 3. The times differ, so they do not collide. [1] Examiner insight: “crossing paths” and “colliding” are different conclusions. Compare the times on the same clock, and give the time for each particle.

11. (a) t = x/15. [1] y = 10(x/15) − 5(x²/225). [1] y = 2x/3 − x²/45. [1] (b) x = 24 gives t = 8/5. [1] y = 16 − 5(64/25) = 16 − 12.8 = 3.2. [1] Since 3.2 > 2.4, the ball passes above the crossbar. [1] (c) Solve 10t − 5t² > 4, so 5t² − 10t + 4 < 0. [1] t = 1 ± √(1/5), about 0.553 and 1.447. [1] The duration is 2/√5 = 0.894 s (3 s.f.). [1] (d) For example, air resistance is ignored, or the ball is treated as a particle. [1] Examiner insight: in a barrier question, state the comparison and the conclusion in words. Give a limitation that is specific to the model, not just “it is not realistic”.

Where marks are usually lost

• Omitting the range of x or y after eliminating a trigonometric parameter.
• Using a root outside the stated range of the parameter, such as a negative time.
• Rounding t before using it to find x or y.
• Writing a circle’s or ellipse’s centre with the wrong sign.
• Not writing down the identity used in a “show that” question.
• Giving the time of a collision from one particle only.
• Mixing degrees and radians when the model has π in it.
• Stopping at the Cartesian equation when the question asks for a restriction.
• Giving a bare number with no unit in a modelling question.

Next steps

• Revisit the revision notes for the method boxes.
• Reread the study guide for the full worked examples.
• Check progress on the OCR A Level Mathematics hub and the printable checklist.
• Take the free 10-minute diagnostics for wider Pure Mathematics.
• Book a free trial class.

Official syllabus

These questions 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.