| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2018 |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (sin/cos identities) |
| Difficulty | Standard +0.3 This is a straightforward parametric-to-Cartesian conversion using standard trigonometric identities. Part (a) requires expanding cos(t + π/6) using the compound angle formula, then simple algebraic manipulation. Part (b) involves eliminating the parameter using sin²t + cos²t = 1, which is a standard technique. The question is slightly easier than average as it provides the target forms and requires only routine application of well-practiced methods with no novel insight needed. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right)\) | M1 | \(\cos\left(t + \frac{\pi}{6}\right) \to \cos t\cos\left(\frac{\pi}{6}\right) \pm \sin t\sin\left(\frac{\pi}{6}\right)\) |
| \(\{x + y\} = 4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right) + 2\sin t\) | dM1 | Adds expanded \(x\) (in terms of \(t\)) to \(2\sin t\) |
| \(= 4\left(\frac{\sqrt{3}}{2}\cos t - \frac{1}{2}\sin t\right) + 2\sin t\) | — | — |
| \(= 2\sqrt{3}\cos t\) | A1* | cso; evidence of \(\cos\left(\frac{\pi}{6}\right)\) and \(\sin\left(\frac{\pi}{6}\right)\) evaluated, proof correct with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left(\frac{x+y}{2\sqrt{3}}\right)^2 + \left(\frac{y}{2}\right)^2 = 1\) | M1 | Applies \(\cos^2 t + \sin^2 t = 1\) to achieve equation containing only \(x\)'s and \(y\)'s |
| \(\Rightarrow \frac{(x+y)^2}{12} + \frac{y^2}{4} = 1\) | — | — |
| \(\Rightarrow (x+y)^2 + 3y^2 = 12\) | A1 | \(\{a = 3,\ b = 12\}\) |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right)$ | M1 | $\cos\left(t + \frac{\pi}{6}\right) \to \cos t\cos\left(\frac{\pi}{6}\right) \pm \sin t\sin\left(\frac{\pi}{6}\right)$ |
| $\{x + y\} = 4\left(\cos t\cos\left(\frac{\pi}{6}\right) - \sin t\sin\left(\frac{\pi}{6}\right)\right) + 2\sin t$ | dM1 | Adds expanded $x$ (in terms of $t$) to $2\sin t$ |
| $= 4\left(\frac{\sqrt{3}}{2}\cos t - \frac{1}{2}\sin t\right) + 2\sin t$ | — | — |
| $= 2\sqrt{3}\cos t$ | A1* | cso; evidence of $\cos\left(\frac{\pi}{6}\right)$ and $\sin\left(\frac{\pi}{6}\right)$ evaluated, proof correct with no errors |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{x+y}{2\sqrt{3}}\right)^2 + \left(\frac{y}{2}\right)^2 = 1$ | M1 | Applies $\cos^2 t + \sin^2 t = 1$ to achieve equation containing **only** $x$'s and $y$'s |
| $\Rightarrow \frac{(x+y)^2}{12} + \frac{y^2}{4} = 1$ | — | — |
| $\Rightarrow (x+y)^2 + 3y^2 = 12$ | A1 | $\{a = 3,\ b = 12\}$ |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4de08317-5fb9-4789-8d57-ccf463224c78-21_664_1244_301_351}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve $C$ with parametric equations
$$x = 4 \cos \left( t + \frac { \pi } { 6 } \right) \quad y = 2 \sin t \quad 0 \leqslant t \leqslant 2 \pi$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$x + y = 2 \sqrt { 3 } \cos t$$
\item Show that a cartesian equation of $C$ is
$$( x + y ) ^ { 2 } + a y ^ { 2 } = b$$
where $a$ and $b$ are integers to be found.\\
\includegraphics[max width=\textwidth, alt={}, center]{4de08317-5fb9-4789-8d57-ccf463224c78-22_2673_1948_107_118}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2018 Q7 [5]}}