| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Moderate -0.3 This is a straightforward parametric question requiring standard elimination (part b) and a simple distance calculation using Pythagoras (part a). The algebra is routine with no conceptual challenges—slightly easier than average due to the clean parameters and direct approach needed. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.10g Problem solving with vectors: in geometry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 4t^2,\ y = 8t\); \(PQ = 4 + 4t^2\) | B1 | |
| \(PR^2 = (4t^2 - 4)^2 + (8t)^2\) | M1 | Use of distance formula |
| \(= 16t^4 + 32t^2 + 16 = (4 + 4t^2)^2 = PQ^2\), so equidistant | M1, A1 [4] | Must show some algebraic manipulation; correct expression for \(PR\) or \(PR^2\); conclusion comparing \(PQ\) and \(PR\) or \(PQ^2\) and \(PR^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitute \(t = \frac{y}{8}\) to obtain \(x = 4\left(\frac{y}{8}\right)^2\) | M1 | Rearranged equation must be used; allow use of \(t = \frac{\sqrt{x}}{2}\) oe for M1A0 |
| \(y^2 = 16x\) | A1 [2] | Allow any equivalent form including \(y = \pm 4\sqrt{x}\) |
## Question 12:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 4t^2,\ y = 8t$; $PQ = 4 + 4t^2$ | B1 | |
| $PR^2 = (4t^2 - 4)^2 + (8t)^2$ | M1 | Use of distance formula |
| $= 16t^4 + 32t^2 + 16 = (4 + 4t^2)^2 = PQ^2$, so equidistant | M1, A1 [4] | Must show some algebraic manipulation; correct expression for $PR$ or $PR^2$; conclusion comparing $PQ$ and $PR$ or $PQ^2$ and $PR^2$ |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitute $t = \frac{y}{8}$ to obtain $x = 4\left(\frac{y}{8}\right)^2$ | M1 | Rearranged equation must be used; allow use of $t = \frac{\sqrt{x}}{2}$ oe for M1A0 |
| $y^2 = 16x$ | A1 [2] | Allow any equivalent form including $y = \pm 4\sqrt{x}$ |12 Fig. 12 shows a curve C with parametric equations $x = 4 t ^ { 2 } , y = 4 t$. The point P , with parameter $t$, is a general point on the curve. Q is the point on the line $x + 4 = 0$ such that PQ is parallel to the $x$-axis. R is the point $( 4,0 )$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{59e924e6-8fa9-4035-9173-705fce487bd9-6_766_584_413_255}
\captionsetup{labelformat=empty}
\caption{Fig. 12}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show algebraically that P is equidistant from Q and R .
\item Find a cartesian equation of C .
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2019 Q12 [6]}}