| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic translation and transformation |
| Difficulty | Standard +0.3 This is a structured multi-part question on hyperbolas covering standard techniques: finding asymptotes (routine formula), sketching (basic visualization), applying translation (direct substitution), solving simultaneous equations (algebraic manipulation), and making a connection between translated and original curves. While it requires multiple steps and some algebraic care in part (c)(ii), each component uses well-practiced FP1 methods without requiring novel insight or particularly challenging problem-solving. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Mark |
| (a) | \(y = \pm\frac{1}{4}x\) | B1 |
| (b) | 2-branch curve with branches in correct regions above and below x-axis; Curve approaching asymptotes | B1 B1 |
| B1 | ||
| (c)(i) | \(\frac{(x+3)^2}{9} - y^2 = 1\) | M1 A1 |
| (c)(ii) | \(\frac{(x+3)^2}{9} - x^2 = 1\) | M1 |
| \(x^2 + 6x + 9 = 9(x^2 + 1)\) | A1F | Correct expansion of \((x + 3)^2\) equated to \(9(x^2 + 1)\) OE ft; [OE in y] |
| \(8x^2 - 6x = 0\) (\(8x^2 = 6x\)) | A1F | Ft on error (\(x - 3\)) for \((x + 3)\) in (c)(i) which gives \(8x^2 + 6x = 0\) (\(8x^2 = -6x\)) [OE in y] |
| Points are \((0, 0)\), \(\left(\frac{3}{4}, \frac{3}{4}\right)\) | A1 | 4 marks; Both. ACF |
| (d) | M1 | |
| Points are \((3, 0)\), \(\left(3\frac{3}{4}, -\frac{3}{4}\right)\) | A1F | 2 marks; Ft on c's (c)(ii) coordinates for the two points; If not deduced then M0A0 |
| Part | Answer/Working | Mark | Guidance |
|------|---|---|---|
| (a) | $y = \pm\frac{1}{4}x$ | B1 | 1 mark; ACF Need both |
| (b) | 2-branch curve with branches in correct regions above and below x-axis; Curve approaching asymptotes | B1 B1 | B1 B1 3 marks total; Coords $(\pm3, 0)$, as only points of intersection with coordinate axes, indicated. Condone −3 and +3 marked on x-axis at points of intersection as $(\pm3, 0)$ indicated |
| | | B1 | |
| (c)(i) | $\frac{(x+3)^2}{9} - y^2 = 1$ | M1 A1 | 2 marks; Replacing x by either $x + 3$ or $x - 3$ ACF |
| (c)(ii) | $\frac{(x+3)^2}{9} - x^2 = 1$ | M1 | Substitution into c's (c)(i) eqn of y = x to eliminate y or of x = y to eliminate x |
| | $x^2 + 6x + 9 = 9(x^2 + 1)$ | A1F | Correct expansion of $(x + 3)^2$ equated to $9(x^2 + 1)$ OE ft; [OE in y] |
| | $8x^2 - 6x = 0$ ($8x^2 = 6x$) | A1F | Ft on error ($x - 3$) for $(x + 3)$ in (c)(i) which gives $8x^2 + 6x = 0$ ($8x^2 = -6x$) [OE in y] |
| | Points are $(0, 0)$, $\left(\frac{3}{4}, \frac{3}{4}\right)$ | A1 | 4 marks; Both. ACF |
| (d) | | M1 | Adding 3 to c's (c)(ii) two x-coords keeping y-coordinates unchanged |
| | Points are $(3, 0)$, $\left(3\frac{3}{4}, -\frac{3}{4}\right)$ | A1F | 2 marks; Ft on c's (c)(ii) coordinates for the two points; If not deduced then M0A0 |7 A hyperbola $H$ has equation
$$\frac { x ^ { 2 } } { 9 } - y ^ { 2 } = 1$$
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of $H$.
\item The asymptotes of $H$ are shown in the diagram opposite. On the same diagram, sketch the hyperbola $H$. Indicate on your sketch the coordinates of the points of intersection of $H$ with the coordinate axes.
\item The hyperbola $H$ is now translated by the vector $\left[ \begin{array} { r } - 3 \\ 0 \end{array} \right]$.
\begin{enumerate}[label=(\roman*)]
\item Write down the equation of the translated curve.
\item Calculate the coordinates of the two points of intersection of the translated curve with the line $y = x$.
\end{enumerate}\item From your answers to part (c)(ii), deduce the coordinates of the points of intersection of the original hyperbola $H$ with the line $y = x - 3$.\\
\includegraphics[max width=\textwidth, alt={}, center]{f9345653-d426-4350-bf1d-901506211078-4_675_1157_1932_495}
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q7 [12]}}