| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting general conic |
| Difficulty | Standard +0.3 This is a straightforward line-conic intersection problem requiring substitution to form a quadratic, solving it, and applying the distance formula. While it involves multiple steps (5 marks typical), each step uses standard C1 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y = x + 2 \Rightarrow x^2 + 4(x+2)^2 - 2x = 35\) | M1 | Substitute \(y = \pm x \pm 2\) into \(x^2 + 4y^2 - 2x = 35\) to obtain an equation in \(x\) only |
| \(5x^2 + 14x - 19 = 0\) | M1 | Multiply out and collect terms producing 3 term quadratic in any form |
| \((5x+19)(x-1) = 0 \Rightarrow x = \ldots\) | dM1 | Solves their quadratic, usual rules, as far as \(x = \ldots\) Dependent on the first M1 i.e. a correct method for eliminating \(y\) (or \(x\) - see below) |
| \(x = -\dfrac{19}{5},\ x = 1\) | A1 for both | Both correct |
| \(y = -\dfrac{9}{5},\ y = 3\) | M1 | Substitutes back into either given equation to find a value for \(y\) |
| Coordinates are \(\left(-\dfrac{19}{5}, -\dfrac{9}{5}\right)\) and \((1, 3)\) | A1 | Correct matching pairs. Coordinates need not be given explicitly but it must be clear which \(x\) goes with which \(y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = y - 2 \Rightarrow (y-2)^2 + 4y^2 - 2(y-2) =\) | M1 | Substitutes \(x = \pm y \pm 2\) into \(x^2 + 4y^2 - 2x = 35\) |
| \(5y^2 - 6y - 27 = 0\) | M1 | Multiply out, collect terms producing 3 term quadratic in any form |
| \((5y+9)(y-3) = 0 \Rightarrow y = \ldots\) | dM1 | Solves their quadratic, usual rules, as far as \(y = \ldots\) Dependent on the first M1, correct method for eliminating \(x\) |
| \(y = -\dfrac{9}{5},\ y = 3\) | A1 for both | Both correct |
| \(x = -\dfrac{19}{5},\ x = 1\) | M1 | Substitutes back into either given equation to find a value for \(x\) |
| Coordinates are \(\left(-\dfrac{19}{5}, -\dfrac{9}{5}\right)\) and \((1,3)\) | A1 | Correct matching pairs as above |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(d^2 = \left(1 - \left(-\dfrac{19}{5}\right)\right)^2 + \left(3 - \left(-\dfrac{9}{5}\right)\right)^2\) or \(d = \sqrt{\left(1+\dfrac{19}{5}\right)^2 + \left(3+\dfrac{9}{5}\right)^2}\) | M1A1ft | M1: Use of \(d^2 = (x_1-x_2)^2 + (y_1-y_2)^2\) where neither \((x_1-x_2)\) nor \((y_1-y_2)\) are zero. A1ft: Correct ft expression for \(d\) or \(d^2\) (may be unsimplified) |
| \(d = \dfrac{24}{5}\sqrt{2}\) | A1cao | Allow \(4.8\sqrt{2}\) |
## Question 11:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y = x + 2 \Rightarrow x^2 + 4(x+2)^2 - 2x = 35$ | M1 | Substitute $y = \pm x \pm 2$ into $x^2 + 4y^2 - 2x = 35$ to obtain an equation in $x$ only |
| $5x^2 + 14x - 19 = 0$ | M1 | Multiply out and collect terms producing 3 term quadratic in any form |
| $(5x+19)(x-1) = 0 \Rightarrow x = \ldots$ | dM1 | Solves their quadratic, usual rules, as far as $x = \ldots$ **Dependent on the first M1** i.e. a correct method for eliminating $y$ (or $x$ - see below) |
| $x = -\dfrac{19}{5},\ x = 1$ | A1 for both | Both correct |
| $y = -\dfrac{9}{5},\ y = 3$ | M1 | Substitutes back into either given equation to find a value for $y$ |
| Coordinates are $\left(-\dfrac{19}{5}, -\dfrac{9}{5}\right)$ and $(1, 3)$ | A1 | Correct matching pairs. Coordinates need not be given explicitly but it must be clear which $x$ goes with which $y$ |
**Total: (6)**
---
### Alternative to part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = y - 2 \Rightarrow (y-2)^2 + 4y^2 - 2(y-2) =$ | M1 | Substitutes $x = \pm y \pm 2$ into $x^2 + 4y^2 - 2x = 35$ |
| $5y^2 - 6y - 27 = 0$ | M1 | Multiply out, collect terms producing 3 term quadratic in any form |
| $(5y+9)(y-3) = 0 \Rightarrow y = \ldots$ | dM1 | Solves their quadratic, usual rules, as far as $y = \ldots$ **Dependent on the first M1**, correct method for eliminating $x$ |
| $y = -\dfrac{9}{5},\ y = 3$ | A1 for both | Both correct |
| $x = -\dfrac{19}{5},\ x = 1$ | M1 | Substitutes back into either given equation to find a value for $x$ |
| Coordinates are $\left(-\dfrac{19}{5}, -\dfrac{9}{5}\right)$ and $(1,3)$ | A1 | Correct matching pairs as above |
---
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $d^2 = \left(1 - \left(-\dfrac{19}{5}\right)\right)^2 + \left(3 - \left(-\dfrac{9}{5}\right)\right)^2$ or $d = \sqrt{\left(1+\dfrac{19}{5}\right)^2 + \left(3+\dfrac{9}{5}\right)^2}$ | M1A1ft | M1: Use of $d^2 = (x_1-x_2)^2 + (y_1-y_2)^2$ where neither $(x_1-x_2)$ nor $(y_1-y_2)$ are zero. A1ft: Correct ft expression for $d$ or $d^2$ (may be unsimplified) |
| $d = \dfrac{24}{5}\sqrt{2}$ | A1cao | Allow $4.8\sqrt{2}$ |
**Total: (3)**
**Question Total: [9]**11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{cfc23548-bf4f-4efa-9ceb-b8d03bb1f019-16_556_1214_219_370}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The line $y = x + 2$ meets the curve $x ^ { 2 } + 4 y ^ { 2 } - 2 x = 35$ at the points $A$ and $B$ as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$ and the coordinates of $B$.
\item Find the distance $A B$ in the form $r \sqrt { 2 }$ where $r$ is a rational number.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2013 Q11 [9]}}