Edexcel A level Maths (9MA0) · Pure mathematics › Coordinate geometry
Practise Parametric equations. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
The curve \(C\) has parametric equations \(x = t - 3, \quad y = t^{2} - 2t + 1, \quad t \in \mathbb{R}\).
(a) Find the coordinates of the point where \(C\) crosses the \(y\)-axis.[2]
(b) Find a Cartesian equation of \(C\), giving your answer in the form \(y = ax^2 + bx + c\), where \(a\), \(b\) and \(c\) are constants to be found.[3]
Show the answer and mark scheme
(a)Answer: \((0, 4)\)
M1 for setting \(t - 3 = 0\) to obtain \(t = 3\)
A1 for \((0, 4)\)
Worked solution: \(t - 3 = 0 \Rightarrow t = 3\), so \(y = 4\). The point is \((0, 4)\).
(b)Answer: \(y = x^2 + 4x + 4\)
M1 for rearranging to \(t = x + 3\)
M1 for substituting their \(t\) into \(y = t^{2} - 2t + 1\)
The curve \(C\) has parametric equations \(x = 2t + 4, \quad y = 3t^2 + 4, \quad -2 \le t \le 2\). The curve \(C\) can be written in the form \(y = f(x)\).
(a) Find an expression for \(f(x)\).[3]
(b) State the domain of \(f\).[1]
(c) Find the range of \(f\).[2]
Show the answer and mark scheme
(a)Answer: \(f(x) = \frac{3}{4}(x - 4)^2 + 4\)
M1 for rearranging to \(t = \frac{x - 4}{2}\)
M1 for substituting their \(t\) into \(y = 3t^2 + 4\)
Worked solution: At \(t = -2\), \(x = 0\) and at \(t = 2\), \(x = 8\), so \(0 \le x \le 8\).
(c)Answer: \(4 \le f(x) \le 16\)
M1 for considering the turning point at \(t = 0\) (\(y = 4\)) and the end point with the larger value of \(t^2\)
A1 for \(4 \le f(x) \le 16\)
Worked solution: \(y = 3t^2 + 4\) has its least value \(4\) at \(t = 0\), which lies in the interval. The greatest value occurs where \(t^2\) is largest, i.e. \(t = 2\), giving \(y = 16\). So \(4 \le f(x) \le 16\).
Question 3Hard11 marks
The curve \(C\) has parametric equations \[x = t^2, \quad y = t^3\]The point \(P\) on \(C\) has parameter \(p\), where \(p \ne 0\).
(a) Show that the tangent to \(C\) at \(P\) has equation \(2y = 3px - p^3\).[3]
(b) The tangent at \(P\) meets \(C\) again at the point \(Q\). Show that \(Q\) has parameter \(-\frac{p}{2}\).[4]
(c) Find the exact values of \(p\) for which the tangent at \(P\) is also the normal to \(C\) at \(Q\).[4]
M1 for substituting \(x = t^2\), \(y = t^3\) into the tangent: \(2t^3 - 3pt^2 + p^3 = 0\)
M1 for recognising that \(t = p\) is a (repeated) root, e.g. by the factor theorem
M1 for factorising: \((t - p)^2(2t + p) = 0\)
A1* for \(t = -\frac{p}{2}\), with the comment that it differs from \(p\) since \(p \ne 0\)
Worked solution: A point with parameter \(t\) lies on the tangent when \(2t^3 = 3pt^2 - p^3\), i.e. \(2t^3 - 3pt^2 + p^3 = 0\). \(t = p\) is a root (the point \(P\)), and it is a repeated root because the line is a tangent there. Dividing by \((t - p)^2 = t^2 - 2pt + p^2\): \(2t^3 - 3pt^2 + p^3 = (t - p)^2(2t + p)\). So the other intersection has \(t = -\frac{p}{2}\), which is different from \(p\) as \(p \ne 0\).
(c)Answer: \(p = \pm\dfrac{2\sqrt{2}}{3}\)
M1 for the gradient at \(Q\): \(\frac{3}{2} \times \left(-\frac{p}{2}\right) = -\frac{3p}{4}\)
M1 for perpendicular gradients: \(\frac{3p}{2} \times \left(-\frac{3p}{4}\right) = -1\)
A1 for \(p^2 = \frac{8}{9}\)
A1 for \(p = \pm\frac{2\sqrt{2}}{3}\)
Worked solution: The gradient of \(C\) at \(Q\) (parameter \(-\frac{p}{2}\)) is \(\frac{3}{2}\left(-\frac{p}{2}\right) = -\frac{3p}{4}\). The line through \(Q\) is the normal there if it is perpendicular to the tangent at \(Q\): \(\dfrac{3p}{2} \times \left(-\dfrac{3p}{4}\right) = -1\), so \(\dfrac{9p^2}{8} = 1\). \(p^2 = \frac{8}{9}\), \(p = \pm\dfrac{2\sqrt{2}}{3}\).