| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Tangent/normal intersection problems |
| Difficulty | Standard +0.3 This is a straightforward parametric differentiation question requiring standard techniques: finding coordinates at a given parameter, computing dy/dx using the chain rule, writing the tangent equation, and finding intercepts to calculate a triangle area. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| P is \((\sqrt{2}, \frac{\sqrt{2}}{2})\) | B1 | oe |
| \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}\) | M1 | |
| \(= \frac{\cos\theta}{-2\sin\theta}\) | A1 | |
| When \(\theta = \frac{\pi}{4}\), \(\frac{dy}{dx} = -\frac{1}{2}\) | A1 | |
| Equation of tangent is \((y - \frac{\sqrt{2}}{2}) = -\frac{1}{2}(x - \sqrt{2})\) | B1 | |
| \(\Rightarrow y = -\frac{1}{2}x + \frac{1}{2}\sqrt{2} + \frac{1}{2}\sqrt{2}\) | E1 | AG |
| \(\Rightarrow x + 2y = 2\sqrt{2}\) | ||
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x = 0\), \(y = \sqrt{2}\) so A is \((0, \sqrt{2})\) | B1 | |
| When \(y = 0\), \(x = 2\sqrt{2}\) so B is \((2\sqrt{2}, 0)\) | B1 | |
| Area of triangle \(= \frac{1}{2}\sqrt{2} \times 2\sqrt{2} = 2\) units² | B1 | |
| [3] |
## Question 11:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| P is $(\sqrt{2}, \frac{\sqrt{2}}{2})$ | B1 | oe |
| $\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta}$ | M1 | |
| $= \frac{\cos\theta}{-2\sin\theta}$ | A1 | |
| When $\theta = \frac{\pi}{4}$, $\frac{dy}{dx} = -\frac{1}{2}$ | A1 | |
| Equation of tangent is $(y - \frac{\sqrt{2}}{2}) = -\frac{1}{2}(x - \sqrt{2})$ | B1 | |
| $\Rightarrow y = -\frac{1}{2}x + \frac{1}{2}\sqrt{2} + \frac{1}{2}\sqrt{2}$ | E1 | AG |
| $\Rightarrow x + 2y = 2\sqrt{2}$ | | |
| **[6]** | | |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x = 0$, $y = \sqrt{2}$ so A is $(0, \sqrt{2})$ | B1 | |
| When $y = 0$, $x = 2\sqrt{2}$ so B is $(2\sqrt{2}, 0)$ | B1 | |
| Area of triangle $= \frac{1}{2}\sqrt{2} \times 2\sqrt{2} = 2$ units² | B1 | |
| **[3]** | | |
---11 Fig. 11 shows the curve with parametric equations
$$x = 2 \cos \theta , y = \sin \theta , 0 \leq \theta \leq 2 \pi .$$
The point P has parameter $\frac { 1 } { 4 } \pi$. The tangent at P to the curve meets the axes at A and B .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ff44367e-c992-4e79-b255-5a04e0b8e21e-10_668_1075_543_255}
\captionsetup{labelformat=empty}
\caption{Fig. 11}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the line AB is $x + 2 y = 2 \sqrt { 2 }$.
\item Determine the area of the triangle AOB .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 Q11 [9]}}