| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 4 |
| 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 verification question requiring algebraic manipulation. Part (a) involves simple fraction division to get x/y = t, and part (b) requires substituting this relationship back to verify the Cartesian equation. While it involves rational expressions and some algebraic manipulation, it's a standard technique with clear steps and no novel insight required—slightly easier than average for A-level. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x}{y} = t\) | B1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \dfrac{\left(\frac{x}{y}\right)^3}{2\left(\frac{x}{y}\right)+1}\) or \(x = \dfrac{\left(\frac{x}{y}\right)^4}{2\left(\frac{x}{y}\right)+1}\) | M1 | Uses the \(y\) coordinate to obtain \(y\) in terms of \(x\) and \(y\), or uses the \(x\) coordinate to obtain \(x\) in terms of \(y\) and \(x\) |
| \(y = \dfrac{x^3}{2xy^2+y^3} \Rightarrow y(2xy^2+y^3) = x^3\) or \(x = \dfrac{x^4}{2xy^3+y^4} \Rightarrow x(2xy^3+y^4) = x^4\) | M1 | Uses correct algebra to eliminate the fractions |
| \(x^3 - 2xy^3 - y^4 = 0\)* | A1* | Cso |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x}{y} = t$ | B1 | Cao |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \dfrac{\left(\frac{x}{y}\right)^3}{2\left(\frac{x}{y}\right)+1}$ or $x = \dfrac{\left(\frac{x}{y}\right)^4}{2\left(\frac{x}{y}\right)+1}$ | M1 | Uses the $y$ coordinate to obtain $y$ in terms of $x$ and $y$, or uses the $x$ coordinate to obtain $x$ in terms of $y$ and $x$ |
| $y = \dfrac{x^3}{2xy^2+y^3} \Rightarrow y(2xy^2+y^3) = x^3$ or $x = \dfrac{x^4}{2xy^3+y^4} \Rightarrow x(2xy^3+y^4) = x^4$ | M1 | Uses correct algebra to eliminate the fractions |
| $x^3 - 2xy^3 - y^4 = 0$* | A1* | Cso |
---2. The curve $C$ has parametric equations
$$x = \frac { t ^ { 4 } } { 2 t + 1 } \quad y = \frac { t ^ { 3 } } { 2 t + 1 } \quad t > 0$$
\begin{enumerate}[label=(\alph*)]
\item Write down $\frac { x } { y }$ in terms of $t$, giving your answer in simplest form.
\item Hence show that all points on $C$ satisfy the equation
$$x ^ { 3 } - 2 x y ^ { 3 } - y ^ { 4 } = 0$$
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q2 [4]}}