| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Asymptotes after transformation |
| Difficulty | Standard +0.3 This is a straightforward transformation question requiring students to apply a horizontal translation, use the given condition to find k, then state the transformed asymptotes. While it involves multiple steps, each is routine: finding k from a point condition, then applying the standard transformation x → x-k to asymptote equations. This is slightly easier than average as it's a standard textbook exercise with no novel insight required. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Hyperbola shape with two branches opening left/right | B1 | Must show correct shape |
| x-intercepts at \((\pm 3, 0)\) stated or marked on graph | B1 | Both values required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Translation gives \(\frac{(x-k)^2}{9} - \frac{y^2}{16} = 1\) | M1 | Replacing \(x\) with \(x-k\) |
| Substituting \((0,0)\): \(\frac{k^2}{9} - 0 = 1\), so \(k = \pm 3\), thus \(k = -3\) | M1 | Using the condition that curve passes through origin |
| Centre of \(C_2\) is \((-3, 0)\) | A1 | |
| Asymptotes of \(C_1\) are \(y = \pm \frac{4}{3}x\), so asymptotes of \(C_2\) are \(y = \pm \frac{4}{3}(x+3)\) | A1 | Both equations required: \(y = \frac{4}{3}(x+3)\) and \(y = -\frac{4}{3}(x+3)\) |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Hyperbola shape with two branches opening left/right | B1 | Must show correct shape |
| x-intercepts at $(\pm 3, 0)$ stated or marked on graph | B1 | Both values required |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Translation gives $\frac{(x-k)^2}{9} - \frac{y^2}{16} = 1$ | M1 | Replacing $x$ with $x-k$ |
| Substituting $(0,0)$: $\frac{k^2}{9} - 0 = 1$, so $k = \pm 3$, thus $k = -3$ | M1 | Using the condition that curve passes through origin |
| Centre of $C_2$ is $(-3, 0)$ | A1 | |
| Asymptotes of $C_1$ are $y = \pm \frac{4}{3}x$, so asymptotes of $C_2$ are $y = \pm \frac{4}{3}(x+3)$ | A1 | Both equations required: $y = \frac{4}{3}(x+3)$ and $y = -\frac{4}{3}(x+3)$ |
---\begin{enumerate}[label=(\alph*)]
\item Sketch the curve $C _ { 1 }$, stating the values of its intercepts with the coordinate axes.
\item The curve $C _ { 1 }$ is translated by the vector $\left[ \begin{array} { l } k \\ 0 \end{array} \right]$, where $k < 0$, to give a curve $C _ { 2 }$.
Given that $C _ { 2 }$ passes through the origin $( 0,0 )$, find the equations of the asymptotes of $C _ { 2 }$.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2015 Q6 [5]}}