| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic translation and transformation |
| Difficulty | Challenging +1.2 This is a standard Further Maths conic sections question requiring completion of the square to identify transformations and find asymptotes. Part (a) involves algebraic manipulation to rewrite C₂ in standard form, then identifying the translation and stretch. Part (b) uses the standard asymptote formula for hyperbolas. While it requires multiple steps and careful algebra, the techniques are routine for Further Maths students with no novel problem-solving required. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{x^2}{25} - \frac{y^2}{4} = 1\) — Deduces stretch only along \(y\)-axis required | M1 | |
| Stretch s.f. \(\frac{1}{2}\) along \(y\)-axis gives \(\frac{x^2}{25} - \frac{(2y)^2}{4} = 1 \Rightarrow \frac{x^2}{25} - y^2 = 1\) — Deduces translation with two non-zero components needed | M1 | Or a translation parallel to \(x\)-axis and a translation parallel to \(y\)-axis |
| Translation by vector \(\begin{bmatrix}3\\-4\end{bmatrix}\) gives \(\frac{(x-3)^2}{25} - (y+4)^2 = 1\) | A1 | Correct stretch and/or correct translation (accept \(\begin{bmatrix}3\\-8\end{bmatrix}\)) |
| Stretch, scale factor \(\frac{1}{2}\), along \(y\)-axis followed by translation by vector \(\begin{bmatrix}3\\-4\end{bmatrix}\) | E1 | States correct order of completely correct transformations (accept any correct sequence) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Asymptotes of \(C_1\): \(y = \pm\frac{2x}{5}\); Stretch \(\Rightarrow 2y = \frac{2x}{5} \Rightarrow 5y = x\); Translation \(\Rightarrow 5(y+4) = x-3\), giving \(5y - x + 23 = 0\) | M1 | Obtains correct asymptotes of \(C_1\), or uses correct method to obtain both asymptotes of \(C_2\) directly |
| \(y = -\frac{2x}{5}\); Stretch \(\Rightarrow 2y = -\frac{2x}{5} \Rightarrow 5y = -x\); Translation \(\Rightarrow 5(y+4) = -x-3\), giving \(5y + x + 17 = 0\) | M1 | Correctly applies their sequence of at least two different types of transformations to one asymptote |
| \(5y - x + 23 = 0\) and \(5y + x + 17 = 0\) | A1 | Both correct results |
## Question 10(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{x^2}{25} - \frac{y^2}{4} = 1$ — Deduces stretch only along $y$-axis required | M1 | |
| Stretch s.f. $\frac{1}{2}$ along $y$-axis gives $\frac{x^2}{25} - \frac{(2y)^2}{4} = 1 \Rightarrow \frac{x^2}{25} - y^2 = 1$ — Deduces translation with two non-zero components needed | M1 | Or a translation parallel to $x$-axis and a translation parallel to $y$-axis |
| Translation by vector $\begin{bmatrix}3\\-4\end{bmatrix}$ gives $\frac{(x-3)^2}{25} - (y+4)^2 = 1$ | A1 | Correct stretch and/or correct translation (accept $\begin{bmatrix}3\\-8\end{bmatrix}$) |
| Stretch, scale factor $\frac{1}{2}$, along $y$-axis followed by translation by vector $\begin{bmatrix}3\\-4\end{bmatrix}$ | E1 | States correct order of completely correct transformations (accept any correct sequence) |
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## Question 10(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Asymptotes of $C_1$: $y = \pm\frac{2x}{5}$; Stretch $\Rightarrow 2y = \frac{2x}{5} \Rightarrow 5y = x$; Translation $\Rightarrow 5(y+4) = x-3$, giving $5y - x + 23 = 0$ | M1 | Obtains correct asymptotes of $C_1$, or uses correct method to obtain both asymptotes of $C_2$ directly |
| $y = -\frac{2x}{5}$; Stretch $\Rightarrow 2y = -\frac{2x}{5} \Rightarrow 5y = -x$; Translation $\Rightarrow 5(y+4) = -x-3$, giving $5y + x + 17 = 0$ | M1 | Correctly applies their sequence of at least two different types of transformations to one asymptote |
| $5y - x + 23 = 0$ and $5y + x + 17 = 0$ | A1 | Both correct results |
---10 The curve $C _ { 1 }$ has equation
$$\frac { x ^ { 2 } } { 25 } - \frac { y ^ { 2 } } { 4 } = 1$$
The curve $C _ { 2 }$ has equation
$$x ^ { 2 } - 25 y ^ { 2 } - 6 x - 200 y - 416 = 0$$
10
\begin{enumerate}[label=(\alph*)]
\item Find a sequence of transformations that maps the graph of $C _ { 1 }$ onto the graph of $C _ { 2 }$ [4 marks]\\
10
\item Find the equations of the asymptotes to $C _ { 2 }$\\
Give your answers in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2022 Q10 [7]}}