| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parametric point verification |
| Difficulty | Standard +0.3 This is a straightforward FP1 conic sections question requiring substitution of parametric equations into a linear equation, solving a quadratic, and finding coordinates. The algebraic manipulation is routine and the question structure is standard for this topic, making it slightly easier than average overall. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(H \equiv L \Rightarrow 6\left(\frac{3}{t}\right) = 4(3t) - 15\) | M1 A1 | An attempt to substitute \(x = 3t\) and \(y = \frac{3}{t}\) into \(L\). Correct equation in \(t\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow 4t^2 - 5t - 6 = 0\) | A1 cso | Correct solution only, involving at least one intermediate step to given answer. |
| (b) \((t-2)(4t+3) = \{0\}\) | M1 | A valid attempt at solving the quadratic. |
| \(\Rightarrow t = 2, -\frac{3}{4}\) | A1 | Both \(t = 2\) and \(t = -\frac{3}{4}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 3(2) = 6, \quad y = \frac{3}{2} \Rightarrow \left(6, \frac{3}{2}\right)\) | M1 | An attempt to use one of their \(t\)-values to find one of either \(x\) or \(y\). |
| A1 | One set of coordinates correct or both \(x\)-values are correct. |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 3\left(-\frac{3}{4}\right) = -\frac{9}{4}, \quad y = \frac{3}{-\frac{3}{}} = -4 \Rightarrow \left(-\frac{9}{4}, -4\right)\) | A1 | Both sets of values correct. |
| [5] |
**Ignore part labels and mark part (a) and part (b) together**
**(a)** $H \equiv L \Rightarrow 6\left(\frac{3}{t}\right) = 4(3t) - 15$ | M1 A1 | An attempt to substitute $x = 3t$ and $y = \frac{3}{t}$ into $L$. Correct equation in $t$.
$\Rightarrow 18 = 12t^2 - 15t \Rightarrow 12t^2 - 15t - 18 = 0$
$\Rightarrow 4t^2 - 5t - 6 = 0$ | A1 cso | Correct solution only, involving at least one intermediate step to given answer. | [3]
**(b)** $(t-2)(4t+3) = \{0\}$ | M1 | A valid attempt at solving the quadratic.
$\Rightarrow t = 2, -\frac{3}{4}$ | A1 | Both $t = 2$ and $t = -\frac{3}{4}$
When $t = 2$:
$x = 3(2) = 6, \quad y = \frac{3}{2} \Rightarrow \left(6, \frac{3}{2}\right)$ | M1 | An attempt to use one of their $t$-values to find one of either $x$ or $y$.
| A1 | One set of coordinates correct or both $x$-values are correct.
When $t = -\frac{3}{4}$:
$x = 3\left(-\frac{3}{4}\right) = -\frac{9}{4}, \quad y = \frac{3}{-\frac{3}{}} = -4 \Rightarrow \left(-\frac{9}{4}, -4\right)$ | A1 | Both sets of values correct.
| [5]
**Alt Method:** An attempt to eliminate either $x$ or $y$ from $xy = 9$ and $6y = 4x - 15$
1st M1: A full method to obtain a quadratic equation in either $x$ or $y$.
1st A1: For either $4x^2 - 15x - 54 = 0$ or $6y^2 + 15y - 36 = 0$ or equivalent.
2nd M1: A valid attempt at solving the quadratic.
2nd A1: For either $x = 6, -\frac{9}{4}$ or $y = \frac{3}{2}, -4$
3rd A1: Both $\left(6, \frac{3}{2}\right)$ and $\left(-\frac{9}{4}, -4\right)$.
**Total: [8]**
---\includegraphics{figure_1}
Figure 1 shows a rectangular hyperbola $H$ with parametric equations
$$x = 3t, \quad y = \frac{3}{t}, \quad t \neq 0$$
The line $L$ with equation $6y = 4x - 15$ intersects $H$ at the point $P$ and at the point $Q$ as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that $L$ intersects $H$ where $4t^2 - 5t - 6 = 0$ [3]
\item Hence, or otherwise, find the coordinates of points $P$ and $Q$. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q5 [8]}}