| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic translation and transformation |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on standard ellipse transformations. Part (a) requires basic sketching and finding intercepts by substitution. Part (b) involves applying a simple stretch transformation rule. Part (c) requires completing the square (a routine FP1 technique) to identify the translation vector. All parts follow standard procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x)1.03d Circles: equation (x-a)^2+(y-b)^2=r^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct ellipse shape drawn | B1 | Must be closed curve |
| \((0, 2)\) and \((0, -2)\) marked on y-axis | B1 | Accept \(y\)-intercepts \(\pm 2\) |
| \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\) marked on x-axis | B1 | Accept \(x\)-intercepts \(\pm\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Replacement \(y \to \frac{y}{2}\) used | M1 | Scale factor 2 parallel to \(y\)-axis means replace \(y\) with \(\frac{y}{2}\) |
| \(\frac{x^2}{3} + \frac{y^2}{16} = 1\) | A1 | |
| Simplified to \(16x^2 + 3y^2 = 48\) or equivalent | A1 | Must be simplified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Complete the square or expand translation: replace \(x\) with \((x-a)\), \(y\) with \((y-b)\) | M1 | |
| \(\frac{(x-a)^2}{3} + \frac{(y-b)^2}{4} = 1\) expanded | M1 | |
| Comparing \(4x^2 + 3y^2 - 8x + 6y = 5\): factor out to get \(4(x^2 - 2x) + 3(y^2 + 2y) = 5\) | M1 | |
| \(4(x-1)^2 - 4 + 3(y+1)^2 - 3 = 5 \Rightarrow \frac{(x-1)^2}{3} + \frac{(y+1)^2}{4} = 1\) | A1 | |
| \(a = 1, \quad b = -1\) | A1 | Both correct |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct ellipse shape drawn | B1 | Must be closed curve |
| $(0, 2)$ and $(0, -2)$ marked on y-axis | B1 | Accept $y$-intercepts $\pm 2$ |
| $(\sqrt{3}, 0)$ and $(-\sqrt{3}, 0)$ marked on x-axis | B1 | Accept $x$-intercepts $\pm\sqrt{3}$ |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Replacement $y \to \frac{y}{2}$ used | M1 | Scale factor 2 parallel to $y$-axis means replace $y$ with $\frac{y}{2}$ |
| $\frac{x^2}{3} + \frac{y^2}{16} = 1$ | A1 | |
| Simplified to $16x^2 + 3y^2 = 48$ or equivalent | A1 | Must be simplified |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Complete the square or expand translation: replace $x$ with $(x-a)$, $y$ with $(y-b)$ | M1 | |
| $\frac{(x-a)^2}{3} + \frac{(y-b)^2}{4} = 1$ expanded | M1 | |
| Comparing $4x^2 + 3y^2 - 8x + 6y = 5$: factor out to get $4(x^2 - 2x) + 3(y^2 + 2y) = 5$ | M1 | |
| $4(x-1)^2 - 4 + 3(y+1)^2 - 3 = 5 \Rightarrow \frac{(x-1)^2}{3} + \frac{(y+1)^2}{4} = 1$ | A1 | |
| $a = 1, \quad b = -1$ | A1 | Both correct |
---6 An ellipse $E$ has equation
$$\frac { x ^ { 2 } } { 3 } + \frac { y ^ { 2 } } { 4 } = 1$$
\begin{enumerate}[label=(\alph*)]
\item Sketch the ellipse $E$, showing the coordinates of the points of intersection of the ellipse with the coordinate axes.
\item The ellipse $E$ is stretched with scale factor 2 parallel to the $y$-axis.
Find and simplify the equation of the curve after the stretch.
\item The original ellipse, $E$, is translated by the vector $\left[ \begin{array} { l } a \\ b \end{array} \right]$. The equation of the translated ellipse is
$$4 x ^ { 2 } + 3 y ^ { 2 } - 8 x + 6 y = 5$$
Find the values of $a$ and $b$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2009 Q6 [11]}}