Parametric equations - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 17 exam-style questions on Parametric equations for Edexcel A-Level Maths (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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3–9 marks each
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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

The curve \(C\) has parametric equations
\[x = 5 - 2t, \quad y = 3t + 1, \quad t \in \mathbb{R}\]
By eliminating \(t\), show that \(C\) is a straight line.

Find the gradient of this line and the coordinates of the point where it crosses the \(y\)-axis.

Model answer

\(t = \dfrac{5 - x}{2}\)
\(y = 3\left(\dfrac{5 - x}{2}\right) + 1 \Rightarrow 2y = 15 - 3x + 2 \Rightarrow 3x + 2y = 17\)
This is a linear equation, so \(C\) is a straight line.
\(y = -\dfrac{3}{2}x + \dfrac{17}{2}\): gradient \(-\dfrac{3}{2}\), crossing the \(y\)-axis at \(\left(0, \dfrac{17}{2}\right)\).

Explanation

When both \(x\) and \(y\) are linear in \(t\), eliminating \(t\) always gives a linear equation, here \(3x + 2y = 17\), so the curve is a straight line. Writing it as \(y = -\dfrac{3}{2}x + \dfrac{17}{2}\) gives the gradient and the \(y\)-intercept; the gradient is also the change in \(y\) (3) divided by the change in \(x\) (\(-2\)) per unit of \(t\). A common error is a sign slip giving a gradient of \(\dfrac{3}{2}\).

Mark scheme

1 mark
M1 Makes \(t\) the subject of one equation and substitutes into the other: \(t = \dfrac{5 - x}{2}\), \(y = \dfrac{3(5 - x)}{2} + 1\), leading to a linear equation such as \(3x + 2y = 17\). An answer with no working scores 0 marks for this mark
An answer of gradient \(-\dfrac{3}{2}\) and \(\left(0, \dfrac{17}{2}\right)\) with no working scores 2 marks (the two B marks); the M mark needs the elimination shown.
1 mark
B1 Gradient \(-\dfrac{3}{2}\)
From \(y = -\dfrac{3}{2}x + \dfrac{17}{2}\), or from \(\dfrac{\text{change in } y}{\text{change in } x} = \dfrac{3}{-2}\) per unit of \(t\). Accept \(-1.5\).
1 mark
B1 \(\left(0, \dfrac{17}{2}\right)\)
Accept \((0, 8.5)\). The \(y\) intercept 8.5 without coordinates is condoned.

Question 2

5 marks

In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.

The curve \(C\) has parametric equations
\[x = t^3, \quad y = t^2 + 2t, \quad t \geqslant 0\]
The line with equation \(y = x + 2\) meets \(C\) at the points \(P\) and \(Q\).

Show that the values of \(t\) at \(P\) and \(Q\) satisfy
\[t^3 - t^2 - 2t + 2 = 0\]
Hence find the exact coordinates of \(P\) and of \(Q\).

Model answer

Substituting into \(y = x + 2\): \(t^2 + 2t = t^3 + 2\)
\(t^3 - t^2 - 2t + 2 = 0\)
\(t = 1\): \(1 - 1 - 2 + 2 = 0\), so \((t - 1)\) is a factor
\(t^3 - t^2 - 2t + 2 = t^2(t - 1) - 2(t - 1) = (t - 1)(t^2 - 2)\)
So \(t = 1\) or \(t = \pm\sqrt{2}\); \(t = -\sqrt{2}\) is rejected because \(t \geqslant 0\)
\(t = 1\): \(x = 1\), \(y = 1 + 2 = 3\), giving \((1, 3)\)
\(t = \sqrt{2}\): \(x = (\sqrt{2})^3 = 2\sqrt{2}\), \(y = 2 + 2\sqrt{2}\), giving \(\left(2\sqrt{2}, 2 + 2\sqrt{2}\right)\)
So \(P\) and \(Q\) are \((1, 3)\) and \(\left(2\sqrt{2}, 2 + 2\sqrt{2}\right)\).

Explanation

Substituting \(x = t^3\) and \(y = t^2 + 2t\) into the equation of the line gives a cubic in \(t\), which the factor theorem (\(t = 1\)) or grouping factorises as \((t - 1)(t^2 - 2)\). The domain \(t \geqslant 0\) rules out \(t = -\sqrt{2}\), so there are only two points, each found by substituting its value of \(t\) back into both parametric equations. Common errors are keeping \(t = -\sqrt{2}\), giving decimal coordinates and finding only \(x\) or only \(y\).

Mark scheme

1 mark
M1 Substitutes the parametric equations into \(y = x + 2\): \(t^2 + 2t = t^3 + 2\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: the question demands working, so the coordinates \((1, 3)\) and \(\left(2\sqrt{2}, 2 + 2\sqrt{2}\right)\) with no working score 0 marks. Substituting into \(y = x + 2\) the wrong way round (e.g. \(t^3 = t^2 + 2t + 2\)) is M0.
1 mark
A1* \(t^3 - t^2 - 2t + 2 = 0\)
cso. Given answer: the substitution and a correct rearrangement must be seen.
1 mark
M1 Factorises the cubic, e.g. shows that \(t = 1\) is a root (\(1 - 1 - 2 + 2 = 0\)) and divides by \((t - 1)\), or groups the terms: \(t^2(t - 1) - 2(t - 1) = (t - 1)(t^2 - 2) = 0\)
Roots stated directly from a calculator, with no factor theorem, division or factorisation seen, score M0 (the question demands working).
1 mark
A1 \(t = 1\) and \(t = \sqrt{2}\), rejecting \(t = -\sqrt{2}\) because \(t \geqslant 0\)
Both values and the rejection with a reason are needed. Keeping \(t = -\sqrt{2}\) is A0.
1 mark
A1 \((1, 3)\) and \(\left(2\sqrt{2}, 2 + 2\sqrt{2}\right)\)
Depends on the second M mark. Both points, exact, and no others; accept \(\sqrt{8}\) for \(2\sqrt{2}\). Decimals only, e.g. \((2.83, 4.83)\), score A0. Including the point \(\left(-2\sqrt{2}, 2 - 2\sqrt{2}\right)\) from \(t = -\sqrt{2}\) is A0.

Question 3

7 marks

A dinghy sails round a buoy during a race. With \(x\) metres and \(y\) metres measured east and north of a fixed point, the buoy is at \(B(40, 25)\). The path of the dinghy is a semicircle of radius 15 m with centre \(B\): the dinghy starts due south of \(B\) when \(t = 0\), moves anticlockwise at constant speed, and reaches the point due north of \(B\) when \(t = 30\), where \(t\) is the time in seconds.

The position of the dinghy is modelled by the parametric equations
\[x = p + 15\sin(kt), \quad y = q - 15\cos(kt), \quad 0 \leqslant t \leqslant 30\]
where \(p\), \(q\) and \(k\) are positive constants.

Find the values of \(p\), \(q\) and \(k\).

Find the exact coordinates of the dinghy when \(t = 10\).

Find the values of \(t\) at which the dinghy is 7.5 m further east than the buoy.

Model answer

When \(t = 0\): \((p, q - 15)\) is due south of \(B(40, 25)\), so \(p = 40\), \(q = 25\).
Half a revolution (\(\pi\) radians) takes 30 s: \(30k = \pi \Rightarrow k = \dfrac{\pi}{30}\)
\(t = 10\): \(x = 40 + 15\sin\dfrac{\pi}{3} = 40 + \dfrac{15\sqrt{3}}{2}\), \(y = 25 - 15\cos\dfrac{\pi}{3} = 25 - \dfrac{15}{2} = \dfrac{35}{2}\)
Position \(\left(40 + \dfrac{15\sqrt{3}}{2}, \dfrac{35}{2}\right)\)
7.5 m east of the buoy: \(40 + 15\sin\dfrac{\pi t}{30} = 47.5 \Rightarrow \sin\dfrac{\pi t}{30} = \dfrac{1}{2}\)
\(\dfrac{\pi t}{30} = \dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\) (for \(0 \leqslant \dfrac{\pi t}{30} \leqslant \pi\)), so \(t = 5\) or \(t = 25\)

Explanation

The parametric equations describe a circle of radius 15 about \((p, q)\), so \(p\) and \(q\) are the coordinates of the buoy; the angle \(kt\) increases by \(\pi\) in 30 seconds, giving \(k = \dfrac{\pi}{30}\). Substituting \(t = 10\) uses the exact values of \(\sin\dfrac{\pi}{3}\) and \(\cos\dfrac{\pi}{3}\), and \(\sin\dfrac{\pi t}{30} = \dfrac{1}{2}\) has two solutions in the half-revolution. Common errors are using \(2\pi\) for a half-revolution and working in degrees.

Mark scheme

1 mark
B1 \(p = 40\) and \(q = 25\) (the coordinates of the centre \(B\)); correct answers to the whole question with no working score full marks
Correct answers of \(p = 40\), \(q = 25\), \(k = \dfrac{\pi}{30}\), \(\left(40 + \dfrac{15\sqrt{3}}{2}, \dfrac{35}{2}\right)\) and \(t = 5\), \(t = 25\) with no working score 7 marks. Both \(p\) and \(q\) are needed for this mark.
1 mark
M1 Uses the fact that half a revolution takes 30 seconds: \(30k = \pi\)
Using \(30k = 2\pi\) or \(30k = 180\) is M0.
1 mark
A1 \(k = \dfrac{\pi}{30}\)
Accept awrt 0.105, i.e. [0.1045, 0.1055), only if \(\dfrac{\pi}{30}\) is seen.
1 mark
M1 Substitutes \(t = 10\) into both equations with their \(p\), \(q\) and \(k\): \(x = 40 + 15\sin\dfrac{\pi}{3}\), \(y = 25 - 15\cos\dfrac{\pi}{3}\)
The angle must be in radians if \(k\) is in radians.
1 mark
A1 \(\left(40 + \dfrac{15\sqrt{3}}{2}, \dfrac{35}{2}\right)\)
Exact values required: accept \((40 + 7.5\sqrt{3}, 17.5)\). Decimals only, e.g. \((52.99, 17.5)\), are A0.
1 mark
M1 Sets \(x = 47.5\): \(\sin\left(\dfrac{\pi t}{30}\right) = \dfrac{1}{2}\), and finds one value of \(t\)
Follow through their \(k\).
1 mark
A1 \(t = 5\) and \(t = 25\)
Both needed and no others in \(0 \leqslant t \leqslant 30\).

Question 4

8 marks

The curve \(C\) has parametric equations
\[x = t - 1, \quad y = t + \dfrac{4}{t}, \quad t > 0\]
A sketch of \(C\) is to be drawn.

Find a Cartesian equation of \(C\) in the form \(y = \mathrm{f}(x)\), stating the domain of \(\mathrm{f}\).

Find the coordinates of the point where \(C\) meets the \(y\)-axis, and explain why \(C\) does not meet the \(x\)-axis.

Show that \(y - 4 = \dfrac{(t - 2)^2}{t}\), and hence find the coordinates of the minimum point of \(C\).

Write down the equation of the vertical asymptote to \(C\).

Model answer

\(x = t - 1 \Rightarrow t = x + 1\), so \(y = x + 1 + \dfrac{4}{x + 1}\)
Since \(t > 0\), the domain of \(\mathrm{f}\) is \(x > -1\).
\(y\)-axis: \(x = 0 \Rightarrow t = 1 \Rightarrow y = 1 + 4 = 5\), giving \((0, 5)\)
For \(t > 0\), \(t\) and \(\dfrac{4}{t}\) are both positive, so \(y > 0\): \(C\) never meets the \(x\)-axis.
\(y - 4 = t + \dfrac{4}{t} - 4 = \dfrac{t^2 - 4t + 4}{t} = \dfrac{(t - 2)^2}{t}\)
For \(t > 0\), \(\dfrac{(t - 2)^2}{t} \geqslant 0\), with equality only when \(t = 2\), so \(y \geqslant 4\) and the least value \(y = 4\) occurs when \(t = 2\), \(x = 1\).
The minimum point is \((1, 4)\).
As \(t \to 0\) from above, \(x \to -1\) and \(y \to \infty\), so the vertical asymptote is \(x = -1\).

Explanation

Eliminating \(t = x + 1\) gives \(y = x + 1 + \dfrac{4}{x + 1}\), and \(t > 0\) gives the domain \(x > -1\), with the vertical asymptote \(x = -1\) as \(t \to 0\). Both terms of \(y = t + \dfrac{4}{t}\) are positive, so the curve stays above the \(x\)-axis, and writing \(y - 4\) as \(\dfrac{(t - 2)^2}{t}\) shows that \(y \geqslant 4\), with the minimum \((1, 4)\) at \(t = 2\). Common errors are giving the domain as \(x ≠ -1\), stating the minimum without a reason, and writing the asymptote as \(y = -1\).

Mark scheme

1 mark
M1 Substitutes \(t = x + 1\) into the equation for \(y\); or a correct Cartesian equation with no working
Correct answers (\(y = x + 1 + \dfrac{4}{x + 1}\), \(x > -1\), \((0, 5)\), minimum \((1, 4)\), \(x = -1\)) with no working score 5 marks: M1 A1 for the equation, B1 for the domain, B1 for \((0, 5)\) and B1 for the asymptote. The explanation needs a reason, and the minimum must be shown from the given identity, so those 3 marks need working.
1 mark
A1 \(y = x + 1 + \dfrac{4}{x + 1}\)
Any equivalent form, e.g. \(y = \dfrac{x^2 + 2x + 5}{x + 1}\) or \(y = \dfrac{(x + 1)^2 + 4}{x + 1}\).
1 mark
B1 Domain \(x > -1\)
From \(t > 0\). Accept \(\{x : x > -1\}\) or \((-1, \infty)\). \(x ≠ -1\) or \(x \geqslant -1\) is B0.
1 mark
B1 \((0, 5)\)
At \(t = 1\).
1 mark
B1 Explains that \(y = t + \dfrac{4}{t} > 0\) for every \(t > 0\) (both terms are positive), so \(y\) is never 0 and \(C\) does not meet the \(x\)-axis
Accept the argument from the next part: \(y \geqslant 4\). Accept \(t^2 + 4 = 0\) having no real solutions (from \(y = 0\)). "The curve does not reach the \(x\)-axis" with no reason is B0.
1 mark
M1 Shows that \(y - 4 = t + \dfrac{4}{t} - 4 = \dfrac{t^2 - 4t + 4}{t} = \dfrac{(t - 2)^2}{t}\)
Given result: the combination into a single fraction and the factorisation must be seen.
1 mark
A1 Since \(\dfrac{(t - 2)^2}{t} \geqslant 0\) for \(t > 0\), with equality only when \(t = 2\), the minimum point is \((1, 4)\)
The reason (a square divided by a positive number is never negative, and is zero at \(t = 2\)) and the coordinates \((1, 4)\) are both needed. Using calculus instead of the given identity scores M0 A0 here, as the question says "hence".
1 mark
B1 \(x = -1\)
As \(t \to 0\) (from above), \(x \to -1\) and \(y \to \infty\). Must be an equation; \(y = -1\) is B0.

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