| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2007 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Tangent/normal meets curve again |
| Difficulty | Standard +0.3 This is a standard C4 parametric differentiation question with routine techniques. Parts (i)-(iii) involve straightforward differentiation and algebraic manipulation (6 marks total). Part (iv) requires iterating the process twice but follows the same method established earlier. All steps are predictable for a well-prepared student, 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 |
|---|---|---|
| (i) Use \(\frac{dy}{dx} = -\frac{dx/dt}{dy/dt}\) and \(-\frac{1}{m}\) for grad of normal | M1 | or change to cartesian, diff & use \(-\frac{1}{m}\) |
| \(= -p\) | A1 | 2 marks |
| (ii) Use correct formula to find gradient of line | M1 | |
| Obtain \(\frac{2}{p+q}\) | A1 | 2 marks |
| (iii) State \(-p = \frac{-2}{p+q}\) | M1 | Or find eqn normal at P & subst \(\{2q^2, 4q\}\) |
| Simplify to \(p^2 + pq + 2 = 0\) | A1 | 2 marks |
| (iv) \((8,8) \to t\) or \(p\) or \(q = 2\) only | B1 | No possibility of \(-2\) |
| Subst \(p = 2\) in eqn (iii) to find \(q_1\) | M1 | Or eqn normal, solve simult with cartes/param |
| Subst \(p = q_1\) in eqn (iii) to find \(q_2\) | M1 | Ditto |
| \(q_2 = \frac{11}{8} \to (\frac{242}{64}, \frac{44}{8})\) | A1 | 4 marks |
(i) Use $\frac{dy}{dx} = -\frac{dx/dt}{dy/dt}$ and $-\frac{1}{m}$ for grad of normal | M1 | or change to cartesian, diff & use $-\frac{1}{m}$
$= -p$ | A1 | 2 marks | AG WWW; Not $-t$
(ii) Use correct formula to find gradient of line | M1 |
Obtain $\frac{2}{p+q}$ | A1 | 2 marks | AG WWW; Minimum of denom $= 2(p - q)(p + q)$
(iii) State $-p = \frac{-2}{p+q}$ | M1 | Or find eqn normal at P & subst $\{2q^2, 4q\}$
Simplify to $p^2 + pq + 2 = 0$ | A1 | 2 marks | AG WWW; With sufficient evidence
(iv) $(8,8) \to t$ or $p$ or $q = 2$ only | B1 | No possibility of $-2$
Subst $p = 2$ in eqn (iii) to find $q_1$ | M1 | Or eqn normal, solve simult with cartes/param
Subst $p = q_1$ in eqn (iii) to find $q_2$ | M1 | Ditto
$q_2 = \frac{11}{8} \to (\frac{242}{64}, \frac{44}{8})$ | A1 | 4 marks | No follow-through; accept $(26.9, 14.7)$ |
---The parametric equations of a curve are $x = 2t^2$, $y = 4t$. Two points on the curve are $P(2p^2, 4p)$ and $Q(2q^2, 4q)$.
\begin{enumerate}[label=(\roman*)]
\item Show that the gradient of the normal to the curve at $P$ is $-p$. [2]
\item Show that the gradient of the chord joining the points $P$ and $Q$ is $\frac{2}{p + q}$. [2]
\item The chord $PQ$ is the normal to the curve at $P$. Show that $p^2 + pq + 2 = 0$. [2]
\item The normal at the point $R(8, 8)$ meets the curve again at $S$. The normal at $S$ meets the curve again at $T$. Find the coordinates of $T$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C4 2007 Q8 [10]}}