| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring elimination of the parameter to find a Cartesian equation (part a) and then solving simultaneous linear equations (part b). The algebra is routine: from the given equations, students can easily express t in terms of y and substitute into x, or add the equations directly. Both parts use standard A-level techniques with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| E.g. \(x=\frac{t-1}{2t+1}\Rightarrow t=\frac{x+1}{1-2x}\) or \(y=\frac{6}{2t+1}\Rightarrow t=\frac{6-y}{2y}\) | M1 | Attempt to get \(t\) in terms of \(x\) or \(y\), or eliminate \(t\); allow \(\frac{x}{y}=\frac{t-1}{6}\Rightarrow t=\frac{6x+y}{y}\) |
| E.g. \(y=\frac{6}{2\times\left(\frac{x+1}{1-2x}\right)+1}\) or \(x=\frac{\frac{6-y}{2y}-1}{2\times\frac{6-y}{2y}+1}\) | A1 | Forms correct equation linking \(x\) and \(y\) only |
| E.g. \(y=\frac{6}{\frac{2(x+1)}{1-2x}+1} \Rightarrow y=\frac{6(1-2x)}{2(x+1)+1(1-2x)}\) | dM1 | Depends on first M; attempts to simplify fraction reaching linear form in \(x\) and \(y\); allow slips but \(ax+by=c\) must be achieved |
| E.g. \(y=\frac{6(1-2x)}{3}\), \(y=2(1-2x)\), so linear | A1* | Achieves \(y=2(1-2x)\) o.e. and states linear/hence a line etc. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y=2(1-2x)\) and \(y=x+12 \Rightarrow 2(1-2x)=x+12 \Rightarrow x=...\) | M1 | Solves their \(y=2(1-2x)\) (may not be linear) with \(y=x+12\) |
| \(x=-2\) | A1 cao | Ignore references to \(y\) coordinate; do not accept \(-\frac{10}{5}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dx}{dt}=\frac{(2t+1)-2(t-1)}{(2t+1)^2}\) and \(\frac{dy}{dt}=\frac{-12}{(2t+1)^2}\) | M1, A1 | Attempts \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) using appropriate rule for at least one; both correct |
| \(\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}\) | dM1 | Depends on first M; leads to constant |
| \(\frac{dy}{dx}=\frac{-12}{(2t+1)^2}\div\frac{3}{(2t+1)^2}=-4\) (constant), hence linear | A1* | Achieves \(\frac{dy}{dx}=-4\) and makes suitable conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x=\frac{t-1}{2t+1}\Rightarrow x=A-\frac{B}{2t+1}\); \(x=\frac{1}{2}-\frac{3}{2(2t+1)}\) | M1; A1 | Attempts to write \(x\) in terms of just \(2t+1\) |
| \(x=\frac{1}{2}-\frac{3}{2(6/y)}\) | dM1 | Uses \(y=\frac{6}{2t+1}\) to form equation linking \(x\) and \(y\) |
| \(x=\frac{1}{2}-\frac{1}{4}y\), hence linear | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(ax+by=\frac{at-a+6b}{2t+1}=\frac{k(2t+1)}{2t+1}\Rightarrow a=2k, 6b-a=k\) | M1 | |
| \(a=12b-2a\Rightarrow a=4b\) | A1 | |
| E.g. \(4x+y=\frac{2(2t+1)}{2t+1}=...\) | dM1 | |
| \(4x+y=2\) (oe), hence linear | A1* |
# Question 2:
## Part (a)
| Working | Mark | Guidance |
|---------|------|----------|
| E.g. $x=\frac{t-1}{2t+1}\Rightarrow t=\frac{x+1}{1-2x}$ or $y=\frac{6}{2t+1}\Rightarrow t=\frac{6-y}{2y}$ | M1 | Attempt to get $t$ in terms of $x$ or $y$, or eliminate $t$; allow $\frac{x}{y}=\frac{t-1}{6}\Rightarrow t=\frac{6x+y}{y}$ |
| E.g. $y=\frac{6}{2\times\left(\frac{x+1}{1-2x}\right)+1}$ or $x=\frac{\frac{6-y}{2y}-1}{2\times\frac{6-y}{2y}+1}$ | A1 | Forms correct equation linking $x$ and $y$ only |
| E.g. $y=\frac{6}{\frac{2(x+1)}{1-2x}+1} \Rightarrow y=\frac{6(1-2x)}{2(x+1)+1(1-2x)}$ | dM1 | Depends on first M; attempts to simplify fraction reaching linear form in $x$ and $y$; allow slips but $ax+by=c$ must be achieved |
| E.g. $y=\frac{6(1-2x)}{3}$, $y=2(1-2x)$, so linear | A1* | Achieves $y=2(1-2x)$ o.e. and states linear/hence a line etc. |
## Part (b)
| Working | Mark | Guidance |
|---------|------|----------|
| $y=2(1-2x)$ and $y=x+12 \Rightarrow 2(1-2x)=x+12 \Rightarrow x=...$ | M1 | Solves their $y=2(1-2x)$ (may not be linear) with $y=x+12$ |
| $x=-2$ | A1 cao | Ignore references to $y$ coordinate; do not accept $-\frac{10}{5}$ |
## Part (a) Alt 1 (via differentiation)
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dx}{dt}=\frac{(2t+1)-2(t-1)}{(2t+1)^2}$ and $\frac{dy}{dt}=\frac{-12}{(2t+1)^2}$ | M1, A1 | Attempts $\frac{dx}{dt}$ and $\frac{dy}{dt}$ using appropriate rule for at least one; both correct |
| $\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}$ | dM1 | Depends on first M; leads to constant |
| $\frac{dy}{dx}=\frac{-12}{(2t+1)^2}\div\frac{3}{(2t+1)^2}=-4$ (constant), hence linear | A1* | Achieves $\frac{dy}{dx}=-4$ and makes suitable conclusion |
## Part (a) Alt 2 (via division)
| Working | Mark | Guidance |
|---------|------|----------|
| $x=\frac{t-1}{2t+1}\Rightarrow x=A-\frac{B}{2t+1}$; $x=\frac{1}{2}-\frac{3}{2(2t+1)}$ | M1; A1 | Attempts to write $x$ in terms of just $2t+1$ |
| $x=\frac{1}{2}-\frac{3}{2(6/y)}$ | dM1 | Uses $y=\frac{6}{2t+1}$ to form equation linking $x$ and $y$ |
| $x=\frac{1}{2}-\frac{1}{4}y$, hence linear | A1* | |
## Part (a) Alt 3
| Working | Mark | Guidance |
|---------|------|----------|
| $ax+by=\frac{at-a+6b}{2t+1}=\frac{k(2t+1)}{2t+1}\Rightarrow a=2k, 6b-a=k$ | M1 | |
| $a=12b-2a\Rightarrow a=4b$ | A1 | |
| E.g. $4x+y=\frac{2(2t+1)}{2t+1}=...$ | dM1 | |
| $4x+y=2$ (oe), hence linear | A1* | |\begin{enumerate}
\item A set of points $P ( x , y )$ is defined by the parametric equations
\end{enumerate}
$$x = \frac { t - 1 } { 2 t + 1 } \quad y = \frac { 6 } { 2 t + 1 } \quad t \neq - \frac { 1 } { 2 }$$
(a) Show that all points $P ( x , y )$ lie on a straight line.\\
(b) Hence or otherwise, find the $x$ coordinate of the point of intersection of this line and the line with equation $y = x + 12$
\hfill \mbox{\textit{Edexcel P4 2023 Q2 [6]}}