| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question with routine techniques: substituting a parameter value, finding derivatives for tangent equations, and eliminating the parameter. Part (b) requires algebraic manipulation but follows directly from adding/subtracting the given equations. Slightly above average due to the multi-step nature and the need to recognize x±y as useful combinations, but all techniques are textbook standard. |
| 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 |
|---|---|
| \(t = \frac{1}{2}\), \(x = 2 \times \frac{1}{2} + \frac{1}{\left(\frac{1}{2}\right)^2}\), \(y = 2 \times \frac{1}{2} - \frac{1}{\left(\frac{1}{2}\right)^2}\) | M1 |
| \(x = 5\), \(y = -3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dt} = 2 + 2t^{-3}\), \(\frac{dx}{dt} = 2 - 2t^{-3}\) | M1 A1 | 2 and \(d\left(\frac{1}{t^2}\right)\) attempted in both derivatives |
| \(t = \frac{1}{2}\): \(\frac{dy}{dx} = \frac{2 + \frac{2}{8}}{2 - \frac{2}{8}} = \frac{-9}{7}\) | M1 A1 | use chain rule; expressions can be in terms of \(t\) or evaluated. CAO or any equivalent fraction (not decimals) |
| \(y + 3 = -\frac{9}{7}(x - 5)\) | B1 F | ft on \(x, y\) and gradient; if \(y = mx + c\) used, \(c\) must be found correctly and the equation must be re-written |
| Answer | Marks | Guidance |
|---|---|---|
| either \(x - y = \frac{2}{t^2}\) or both of \(x - y = 4t\) and \(x + y = \frac{2}{t^2}\) | M1 | |
| \(\frac{2}{(x-y)} = \left(\frac{x+y}{4}\right)^2\) | M1 | eliminate \(t\) |
| \(32 = (x-y)(x+y)^2\) | A1 | or \((x-y)(x+y)^2 = \frac{2}{t^2} \times(4t)^2 = 32\); \(k = 32\) alone, no marks |
**5(a)(i)**
| $t = \frac{1}{2}$, $x = 2 \times \frac{1}{2} + \frac{1}{\left(\frac{1}{2}\right)^2}$, $y = 2 \times \frac{1}{2} - \frac{1}{\left(\frac{1}{2}\right)^2}$ | M1 | |
| $x = 5$, $y = -3$ | A1 | |
**5(a)(ii)**
| $\frac{dv}{dt} = 2 + 2t^{-3}$, $\frac{dx}{dt} = 2 - 2t^{-3}$ | M1 A1 | 2 and $d\left(\frac{1}{t^2}\right)$ attempted in both derivatives |
| $t = \frac{1}{2}$: $\frac{dy}{dx} = \frac{2 + \frac{2}{8}}{2 - \frac{2}{8}} = \frac{-9}{7}$ | M1 A1 | use chain rule; expressions can be in terms of $t$ or evaluated. CAO or any equivalent fraction (not decimals) |
| $y + 3 = -\frac{9}{7}(x - 5)$ | B1 F | ft on $x, y$ and gradient; if $y = mx + c$ used, $c$ must be found correctly and the equation must be re-written |
**5(b)**
| either $x - y = \frac{2}{t^2}$ or both of $x - y = 4t$ and $x + y = \frac{2}{t^2}$ | M1 | |
| $\frac{2}{(x-y)} = \left(\frac{x+y}{4}\right)^2$ | M1 | eliminate $t$ |
| $32 = (x-y)(x+y)^2$ | A1 | or $(x-y)(x+y)^2 = \frac{2}{t^2} \times(4t)^2 = 32$; $k = 32$ alone, no marks |
**Total: 10 marks**
---5 A curve is defined by the parametric equations $x = 2 t + \frac { 1 } { t ^ { 2 } } , \quad y = 2 t - \frac { 1 } { t ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item At the point $P$ on the curve, $t = \frac { 1 } { 2 }$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of $P$.
\item Find an equation of the tangent to the curve at $P$.
\end{enumerate}\item Show that the cartesian equation of the curve can be written as
$$( x - y ) ( x + y ) ^ { 2 } = k$$
where $k$ is an integer.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q5 [10]}}