| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (sin/cos identities) |
| Difficulty | Moderate -0.3 This is a straightforward parametric-to-Cartesian conversion using the double angle formula cos(2t) = 2cos²(t) - 1, leading to a simple quadratic relationship. The sketch requires recognizing the parabola shape and domain restrictions, making it slightly easier than average but still requiring proper technique. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 3\cos t \Rightarrow \cos t = \frac{x}{3}\) | M1 | Using \(\cos 2t = 2\cos^2 t - 1\) |
| \(y = \cos 2t = 2\cos^2 t - 1 = 2\left(\frac{x}{3}\right)^2 - 1\) | ||
| \(y = \frac{2x^2}{9} - 1\) | A1 | Correct Cartesian equation |
| Answer | Marks | Guidance |
|---|---|---|
| Parabola shape, vertex at \((0, -1)\) | B1 | Correct shape |
| Endpoints at \((-3, 1)\) and \((3, 1)\) when \(t=0\) gives \((3,1)\), \(t=\pi\) gives \((-3,1)\) | B1 | Correct endpoints/range indicated |
# Question 2:
## Part (a)
| $x = 3\cos t \Rightarrow \cos t = \frac{x}{3}$ | M1 | Using $\cos 2t = 2\cos^2 t - 1$ |
| $y = \cos 2t = 2\cos^2 t - 1 = 2\left(\frac{x}{3}\right)^2 - 1$ | | |
| $y = \frac{2x^2}{9} - 1$ | A1 | Correct Cartesian equation |
## Part (b)
| Parabola shape, vertex at $(0, -1)$ | B1 | Correct shape |
| Endpoints at $(-3, 1)$ and $(3, 1)$ when $t=0$ gives $(3,1)$, $t=\pi$ gives $(-3,1)$ | B1 | Correct endpoints/range indicated |
---2. The curve $C$ is described by the parametric equations
$$x = 3 \cos t , \quad y = \cos 2 t , \quad 0 \leq t \leq \pi .$$
\begin{enumerate}[label=(\alph*)]
\item Find a cartesian equation of the curve $C$.
\item Draw a sketch of the curve $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q2 [4]}}