| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find normal equation |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: differentiation using the chain rule (dy/dx = (dy/dt)/(dx/dt)), finding a normal line equation, and eliminating the parameter. All steps are routine C4 material with no conceptual challenges, making it slightly easier than average but not trivial due to the multi-part nature and algebraic manipulation required. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(y=\frac{2}{x-3}-1\) and differentiate | M1 | Differentiate expression in \(y\) and \(x\) |
| \(\frac{dy}{dx}=\frac{-2}{(x-3)^2}\) | A1 | Correct |
| \(x=5\) | ||
| \(\frac{dy}{dx}=\frac{-2}{(5-3)^2}\) | m1 | Find and therefore use \(x\) (and \(y\)) |
| \(\frac{dy}{dx}=-\frac{1}{2}\) | A1 |
# Question 2:
## Part 2(a)
$y=\frac{2}{x-3}-1$ and differentiate | M1 | Differentiate expression in $y$ and $x$
$\frac{dy}{dx}=\frac{-2}{(x-3)^2}$ | A1 | Correct
$x=5$ | |
$\frac{dy}{dx}=\frac{-2}{(5-3)^2}$ | m1 | Find and therefore use $x$ (and $y$)
$\frac{dy}{dx}=-\frac{1}{2}$ | A1 |2 A curve is defined, for $t \neq 0$, by the parametric equations
$$x = 4 t + 3 , \quad y = \frac { 1 } { 2 t } - 1$$
At the point $P$ on the curve, $t = \frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve at the point $P$.
\item Find an equation of the normal to the curve at the point $P$.
\item Find a cartesian equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q2 [10]}}