| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find tangent equation |
| Difficulty | Moderate -0.3 This is a standard C4 parametric equations question requiring routine application of the chain rule for dy/dx, finding a tangent equation, and algebraic manipulation to verify a Cartesian form. All techniques are textbook exercises with no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dt} = \frac{-2}{t^2}\), \(\frac{dx}{dt} = -4\) | M1A1 | |
| \(\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{1}{\frac{dx}{dt}} = \frac{1}{2t^2}\) | m1, A1F | Use chain rule; Follow on use of chain rule (if \(f(t)\)); Or eliminate \(t\): M1 \(y=f(x)\) attempt to differentiate M1A1 chain rule; A1F reintroduce \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t=2\), \(m_T = \frac{1}{8}\) | B1F | follow on gradient (possibly used later) |
| \(x=-5\), \(y=2\) | B1 | |
| \(y - 2 = \frac{1}{8}(x+5)\) | M1 | Their \((x,y), m\) |
| \(x - 8y + 21 = 0\) | A1F | Ft on \((x,y)\) and \(m\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x - 3 = -4t\), \(y - 1 = \frac{2}{t}\) | M1 | PI |
| \((x-3)(y-1) = -4t \times \frac{2}{t} = (-8)\) | M1, A1 | Attempt to eliminate \(t\); AG convincingly obtained |
## Question 2:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dt} = \frac{-2}{t^2}$, $\frac{dx}{dt} = -4$ | M1A1 | |
| $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{1}{\frac{dx}{dt}} = \frac{1}{2t^2}$ | m1, A1F | Use chain rule; Follow on use of chain rule (if $f(t)$); Or eliminate $t$: M1 $y=f(x)$ attempt to differentiate M1A1 chain rule; A1F reintroduce $t$ |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t=2$, $m_T = \frac{1}{8}$ | B1F | follow on gradient (possibly used later) |
| $x=-5$, $y=2$ | B1 | |
| $y - 2 = \frac{1}{8}(x+5)$ | M1 | Their $(x,y), m$ |
| $x - 8y + 21 = 0$ | A1F | Ft on $(x,y)$ and $m$ |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x - 3 = -4t$, $y - 1 = \frac{2}{t}$ | M1 | PI |
| $(x-3)(y-1) = -4t \times \frac{2}{t} = (-8)$ | M1, A1 | Attempt to eliminate $t$; AG convincingly obtained |
---2 A curve is defined by the parametric equations
$$x = 3 - 4 t \quad y = 1 + \frac { 2 } { t }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
\item Find the equation of the tangent to the curve at the point where $t = 2$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\item Verify that the cartesian equation of the curve can be written as
$$( x - 3 ) ( y - 1 ) + 8 = 0$$
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2006 Q2 [11]}}