| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic translation and transformation |
| Difficulty | Standard +0.8 This is a Further Maths question on conic sections requiring completion of the square to identify a hyperbola's standard form, then describing transformations between two hyperbolas. While the algebraic manipulation is routine for FM students, identifying the correct sequence of transformations (likely translation and stretch) requires solid understanding of conic transformations. The multi-step nature and FM context place it moderately above average difficulty. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.02x Combinations of transformations: multiple transformations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Forms expression for distance of general point from line \(x = 2\) | B1 | Condone \((x-2)\) or \((2-x)\) |
| \(\sqrt{(x-5)^2 + y^2} = 2\ | x - 2\ | \) |
| \((x-5)^2 + y^2 = 4(x-2)^2\) | M1 | Forms equation in \(x\) and \(y\) for the locus; distance from \((5,0) = 2 \times\) distance from \(x=2\) |
| \(x^2 - 10x + 25 + y^2 = 4x^2 - 16x + 16\) | ||
| \(3x^2 - 6x - y^2 = 9\) | ||
| \(x^2 - 2x - \dfrac{y^2}{3} = 3\) | M1 | Simplifies equation to quadratic in \(x\) and \(y\) with coefficient of \(x\) equal to 1 |
| \((x-1)^2 - 1 - \dfrac{y^2}{3} = 3\) | ||
| \((x-1)^2 - \dfrac{y^2}{3} = 4\) | R1 | Completes a rigorous argument to show \((x-1)^2 - \dfrac{y^2}{3} = 4\) AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Translation by \(\begin{bmatrix}1\\0\end{bmatrix}\) | B1 | States horizontal translation (PI by vector with no \(y\)-component); describes translation as \(\begin{bmatrix}1\\0\end{bmatrix}\) OE |
| Stretch parallel to \(y\)-axis, scale factor \(\sqrt{3}\) | B1 | States stretch parallel to \(y\)-axis |
| Scale factor \(\sqrt{3}\) | B1F | FT their \(q\) (or \(\sqrt{q}\) if no value of \(q\) given) |
# Question 5:
## Part 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Forms expression for distance of general point from line $x = 2$ | B1 | Condone $(x-2)$ or $(2-x)$ |
| $\sqrt{(x-5)^2 + y^2} = 2\|x - 2\|$ | B1 | Forms expression for (squared) distance of general point from $(5, 0)$ |
| $(x-5)^2 + y^2 = 4(x-2)^2$ | M1 | Forms equation in $x$ and $y$ for the locus; distance from $(5,0) = 2 \times$ distance from $x=2$ |
| $x^2 - 10x + 25 + y^2 = 4x^2 - 16x + 16$ | | |
| $3x^2 - 6x - y^2 = 9$ | | |
| $x^2 - 2x - \dfrac{y^2}{3} = 3$ | M1 | Simplifies equation to quadratic in $x$ and $y$ with coefficient of $x$ equal to 1 |
| $(x-1)^2 - 1 - \dfrac{y^2}{3} = 3$ | | |
| $(x-1)^2 - \dfrac{y^2}{3} = 4$ | R1 | Completes a rigorous argument to show $(x-1)^2 - \dfrac{y^2}{3} = 4$ AG |
## Part 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Translation by $\begin{bmatrix}1\\0\end{bmatrix}$ | B1 | States horizontal translation (PI by vector with no $y$-component); describes translation as $\begin{bmatrix}1\\0\end{bmatrix}$ OE |
| Stretch parallel to $y$-axis, scale factor $\sqrt{3}$ | B1 | States stretch parallel to $y$-axis |
| Scale factor $\sqrt{3}$ | B1F | FT their $q$ (or $\sqrt{q}$ if no value of $q$ given) |
---5
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of $H _ { 1 }$ can be written in the form
$$( x - 1 ) ^ { 2 } - \frac { y ^ { 2 } } { q } = r$$
where $q$ and $r$ are integers.\\
5
\item $\quad \mathrm { H } _ { 2 }$ is the hyperbola
$$x ^ { 2 } - y ^ { 2 } = 4$$
Describe fully a sequence of two transformations which maps the graph of $H _ { 2 }$ onto the graph of $H _ { 1 }$\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2020 Q5 [9]}}