| Exam Board | OCR MEI |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2008 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent parallel to axis condition |
| Difficulty | Challenging +1.2 This is a structured Further Maths parametric equations question with clear scaffolding through parts (i)-(v). While it requires differentiation of parametric equations and algebraic manipulation, each part guides students systematically. The self-intersection and stationary points are standard FP2 techniques, making this moderately above average difficulty but not requiring exceptional insight. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Three sketches for \(\lambda=-1, 0, 1\) showing appropriate features | B1B1B1 | Two different features (cusp, loop, asymptote) correctly identified |
| 5 |
| Answer | Marks |
|---|---|
| \(x=1\) | B1 |
| 1 |
| Answer | Marks |
|---|---|
| Intersects itself when \(y=0\) | M1 |
| \(t=(\pm)\sqrt{\lambda}\) | A1 |
| \(\begin{pmatrix}\frac{\lambda}{1+\lambda},0\end{pmatrix}\) | A1 |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dt}=3t^2-\lambda=0\) | M1 | |
| \(t=\pm\sqrt{\frac{\lambda}{3}}\) | A1 ag | |
| \(x=\frac{t^3}{1+t^2}=\frac{\lambda}{3+\lambda}\) | A1 ag | One value sufficient |
| \(y=\pm\left(\frac{\lambda}{3}\right)^{\frac{3}{2}}-\lambda\left(\frac{\lambda}{3}\right)^{\frac{1}{2}} = \pm\lambda^{\frac{3}{2}}\left(\frac{1}{3\sqrt{3}}-\frac{1}{\sqrt{3}}\right)=\pm\lambda^{\frac{3}{2}}\left(-\frac{2}{3\sqrt{3}}\right) = \pm\sqrt{\frac{4\lambda^3}{27}}\) | M1, A1 ag | One value sufficient |
| 4 |
| Answer | Marks |
|---|---|
| From asymptote, \(a=8\) | B1 |
| From intersection point, \(\frac{d\lambda}{1+\lambda}=2\) | M1 |
| \(\lambda=\frac{1}{3}\) | A1 |
| From maximum point, \(b\sqrt{\frac{4\lambda^3}{27}}=2\) | M1 |
| \(b=27\) | A1 |
| 5 |
## 5(i)
| Three sketches for $\lambda=-1, 0, 1$ showing appropriate features | B1B1B1 | Two different features (cusp, loop, asymptote) correctly identified |
|---|---|---|
| | 5 | |
## 5(ii)
| $x=1$ | B1 | |
|---|---|---|
| | 1 | |
## 5(iii)
| Intersects itself when $y=0$ | M1 | |
|---|---|---|
| $t=(\pm)\sqrt{\lambda}$ | A1 | |
|---|---|---|
| $\begin{pmatrix}\frac{\lambda}{1+\lambda},0\end{pmatrix}$ | A1 | |
|---|---|---|
| | 3 | |
## 5(iv)
| $\frac{dy}{dt}=3t^2-\lambda=0$ | M1 | |
|---|---|---|
| $t=\pm\sqrt{\frac{\lambda}{3}}$ | A1 ag | |
|---|---|---|
| $x=\frac{t^3}{1+t^2}=\frac{\lambda}{3+\lambda}$ | A1 ag | One value sufficient |
|---|---|---|
| $y=\pm\left(\frac{\lambda}{3}\right)^{\frac{3}{2}}-\lambda\left(\frac{\lambda}{3}\right)^{\frac{1}{2}} = \pm\lambda^{\frac{3}{2}}\left(\frac{1}{3\sqrt{3}}-\frac{1}{\sqrt{3}}\right)=\pm\lambda^{\frac{3}{2}}\left(-\frac{2}{3\sqrt{3}}\right) = \pm\sqrt{\frac{4\lambda^3}{27}}$ | M1, A1 ag | One value sufficient |
|---|---|---|
| | 4 | |
## 5(v)
| From asymptote, $a=8$ | B1 | |
|---|---|---|
| From intersection point, $\frac{d\lambda}{1+\lambda}=2$ | M1 | |
|---|---|---|
| $\lambda=\frac{1}{3}$ | A1 | |
|---|---|---|
| From maximum point, $b\sqrt{\frac{4\lambda^3}{27}}=2$ | M1 | |
|---|---|---|
| $b=27$ | A1 | |
|---|---|---|
| | 5 | |5 A curve has parametric equations $x = \frac { t ^ { 2 } } { 1 + t ^ { 2 } } , y = t ^ { 3 } - \lambda t$, where $\lambda$ is a constant.\\
(i) Use your calculator to obtain a sketch of the curve in each of the cases
$$\lambda = - 1 , \quad \lambda = 0 \quad \text { and } \quad \lambda = 1 .$$
Name any special features of these curves.\\
(ii) By considering the value of $x$ when $t$ is large, write down the equation of the asymptote.
For the remainder of this question, assume that $\lambda$ is positive.\\
(iii) Find, in terms of $\lambda$, the coordinates of the point where the curve intersects itself.\\
(iv) Show that the two points on the curve where the tangent is parallel to the $x$-axis have coordinates
$$\left( \frac { \lambda } { 3 + \lambda } , \pm \sqrt { \frac { 4 \lambda ^ { 3 } } { 27 } } \right)$$
Fig. 5 shows a curve which intersects itself at the point ( 2,0 ) and has asymptote $x = 8$. The stationary points A and B have $y$-coordinates 2 and - 2 .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{43b4c7ed-3556-4d87-8aef-0111fe747982-4_791_609_1482_769}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(v) For the curve sketched in Fig. 5, find parametric equations of the form $x = \frac { a t ^ { 2 } } { 1 + t ^ { 2 } } , y = b \left( t ^ { 3 } - \lambda t \right)$, where $a , \lambda$ and $b$ are to be determined.
\hfill \mbox{\textit{OCR MEI FP2 2008 Q5 [18]}}