| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find normal equation |
| Difficulty | Standard +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 algebraic manipulation to eliminate the parameter. All steps are routine C4 material with no novel insight required, 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 |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{dx}{dt} = -\frac{1}{t^2}\), \(\frac{dy}{dt} = 1 - \frac{1}{2t^2}\) | B1B1 | |
| \(\frac{dy}{dx} = \frac{1-\frac{1}{2t^2}}{-\frac{1}{t^2}} \left(= \frac{2t^2-1}{-2}\right)\) | M1 | Their \(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\); condone 1 slip |
| (correct answer) | A1 | CSO; ISW |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = \frac{1}{x} + \frac{x}{2}\) | (B1) | |
| \(\frac{dy}{dx} = -\frac{1}{x^2} + \frac{1}{2}\) | (B1) | |
| Substitute \(x = \frac{1}{t}\) | (M1) | |
| \(\frac{dy}{dx} = -t^2 + \frac{1}{2}\) | (A1) | CSO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(t=1\): \(\frac{dy}{dx} = -\frac{1}{2}\) | M1 | Substitute \(t=1\) in \(\frac{f(t)}{g(t)} \neq k\) |
| \(m_T = -\frac{1}{2} \Rightarrow m_n = 2\) | B1F | F on \(m_T \neq 0\); if \(t \to\) numerical later |
| \((x,y) = \left(1, \frac{3}{2}\right)\) | B1 | PI \(\frac{3}{2} = m(\times 1) + c\) |
| \(\left(y - \frac{3}{2}\right) = 2(x-1)\) or \(y = 2x+c\), \(c=-\frac{1}{2}\) | A1 | ISW, CSO (a) and (b) all correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = \frac{1}{t} + \frac{1}{2} \times \frac{1}{t}\); use \(t = \frac{1}{x}\) to eliminate \(t\) | M1 | Attempt to use \(t=\frac{1}{x}\), or equivalent |
| \(= \frac{1}{x} + \frac{x}{2}\) | A1 | |
| \(2xy = 2 + x^2 \Rightarrow x^2 - 2xy + 2 = 0\) | A1 | Correct algebra to AG with \(k=2\); allow \(k=2\) stated |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\left(\frac{1}{t}\right)^2 - 2\left(\frac{1}{t}\right)\left(t+\frac{1}{2t}\right)\) and \(xy = \frac{1}{t}\left(t+\frac{1}{2t}\right)\) | (M1) | Substitute and multiply out |
| \(= -2\) and \(= 1 + \frac{x^2}{2}\) | (A1) | Eliminate \(t\) |
| \(\Rightarrow x^2 - 2xy + 2 = 0\) | (A1) | Conclusion, \(k=2\) |
# Question 2:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{dx}{dt} = -\frac{1}{t^2}$, $\frac{dy}{dt} = 1 - \frac{1}{2t^2}$ | B1B1 | |
| $\frac{dy}{dx} = \frac{1-\frac{1}{2t^2}}{-\frac{1}{t^2}} \left(= \frac{2t^2-1}{-2}\right)$ | M1 | Their $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$; condone 1 slip |
| (correct answer) | A1 | CSO; ISW |
**Alternative:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $y = \frac{1}{x} + \frac{x}{2}$ | (B1) | |
| $\frac{dy}{dx} = -\frac{1}{x^2} + \frac{1}{2}$ | (B1) | |
| Substitute $x = \frac{1}{t}$ | (M1) | |
| $\frac{dy}{dx} = -t^2 + \frac{1}{2}$ | (A1) | CSO |
**Total: 4 marks**
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $t=1$: $\frac{dy}{dx} = -\frac{1}{2}$ | M1 | Substitute $t=1$ in $\frac{f(t)}{g(t)} \neq k$ |
| $m_T = -\frac{1}{2} \Rightarrow m_n = 2$ | B1F | F on $m_T \neq 0$; if $t \to$ numerical later |
| $(x,y) = \left(1, \frac{3}{2}\right)$ | B1 | PI $\frac{3}{2} = m(\times 1) + c$ |
| $\left(y - \frac{3}{2}\right) = 2(x-1)$ or $y = 2x+c$, $c=-\frac{1}{2}$ | A1 | ISW, CSO (a) and (b) all correct |
**Total: 4 marks**
## Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $y = \frac{1}{t} + \frac{1}{2} \times \frac{1}{t}$; use $t = \frac{1}{x}$ to eliminate $t$ | M1 | Attempt to use $t=\frac{1}{x}$, or equivalent |
| $= \frac{1}{x} + \frac{x}{2}$ | A1 | |
| $2xy = 2 + x^2 \Rightarrow x^2 - 2xy + 2 = 0$ | A1 | Correct algebra to AG with $k=2$; allow $k=2$ stated |
**Alternative:**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\left(\frac{1}{t}\right)^2 - 2\left(\frac{1}{t}\right)\left(t+\frac{1}{2t}\right)$ and $xy = \frac{1}{t}\left(t+\frac{1}{2t}\right)$ | (M1) | Substitute and multiply out |
| $= -2$ and $= 1 + \frac{x^2}{2}$ | (A1) | Eliminate $t$ |
| $\Rightarrow x^2 - 2xy + 2 = 0$ | (A1) | Conclusion, $k=2$ |
**Total: 3 marks**
---2 A curve is defined by the parametric equations
$$x = \frac { 1 } { t } , \quad y = t + \frac { 1 } { 2 t }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$.
\item Find an equation of the normal to the curve at the point where $t = 1$.
\item Show that the cartesian equation of the curve can be written in the form
$$x ^ { 2 } - 2 x y + k = 0$$
where $k$ is an integer.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2009 Q2 [11]}}