| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 12 |
| 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 standard C4 parametric equations question with routine techniques: substituting coordinates to find parameter value, finding gradient via dy/dx = (dy/dθ)/(dx/dθ) and equation of normal, and eliminating parameter using cos²θ + sin²θ = 1 to get an ellipse equation. All steps are textbook exercises requiring no novel insight, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(x = 1 \therefore -1 + 4 \cos \theta = 1, \cos \theta = \frac{1}{2}, \theta = \frac{\pi}{3}, \frac{5\pi}{3}\) | M1 | |
| \(y > 0 \therefore \sin \theta > 0 \therefore \theta = \frac{\pi}{3}\) | A1 | |
| (b) \(\frac{dx}{d\theta} = -4\sin\theta, \quad \frac{dy}{d\theta} = 2\sqrt{2}\cos\theta\) | M1 | |
| \(\therefore \frac{dy}{dx} = \frac{2\sqrt{2}\cos\theta}{-4\sin\theta}\) | M1 A1 | |
| at \(P\), grad \(= \frac{2\sqrt{2} \times \frac{1}{2}}{4 \times \frac{\sqrt{3}}{2}} = \frac{\sqrt{2}}{2\sqrt{3}}\) | M1 | |
| grad of normal \(= \frac{2\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{6}\) | A1 | |
| \(\therefore y - \sqrt{6} = \sqrt{6}(x-1)\) | M1 | |
| \(y = \sqrt{6}x, \quad\) when \(x = 0, y = 0 \therefore\) passes through origin | A1 | |
| (c) \(\cos\theta = \frac{x+1}{4}, \sin\theta = \frac{y}{2\sqrt{2}}\) | M1 | |
| \(\therefore \frac{(x+1)^2}{16} + \frac{y^2}{8} = 1\) | M1 A1 | (12) |
**(a)** $x = 1 \therefore -1 + 4 \cos \theta = 1, \cos \theta = \frac{1}{2}, \theta = \frac{\pi}{3}, \frac{5\pi}{3}$ | M1 |
$y > 0 \therefore \sin \theta > 0 \therefore \theta = \frac{\pi}{3}$ | A1 |
**(b)** $\frac{dx}{d\theta} = -4\sin\theta, \quad \frac{dy}{d\theta} = 2\sqrt{2}\cos\theta$ | M1 |
$\therefore \frac{dy}{dx} = \frac{2\sqrt{2}\cos\theta}{-4\sin\theta}$ | M1 A1 |
at $P$, grad $= \frac{2\sqrt{2} \times \frac{1}{2}}{4 \times \frac{\sqrt{3}}{2}} = \frac{\sqrt{2}}{2\sqrt{3}}$ | M1 |
grad of normal $= \frac{2\sqrt{3}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \sqrt{6}$ | A1 |
$\therefore y - \sqrt{6} = \sqrt{6}(x-1)$ | M1 |
$y = \sqrt{6}x, \quad$ when $x = 0, y = 0 \therefore$ passes through origin | A1 |
**(c)** $\cos\theta = \frac{x+1}{4}, \sin\theta = \frac{y}{2\sqrt{2}}$ | M1 |
$\therefore \frac{(x+1)^2}{16} + \frac{y^2}{8} = 1$ | M1 A1 | **(12)**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b71c9832-e502-4a25-85fb-a49c03ea9209-12_495_784_246_461}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the curve with parametric equations
$$x = - 1 + 4 \cos \theta , \quad y = 2 \sqrt { 2 } \sin \theta , \quad 0 \leq \theta < 2 \pi$$
The point $P$ on the curve has coordinates $( 1 , \sqrt { 6 } )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\theta$ at $P$.
\item Show that the normal to the curve at $P$ passes through the origin.
\item Find a cartesian equation for the curve.\\
7. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [12]}}