| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola focus and directrix properties |
| Difficulty | Standard +0.3 This is a straightforward FP1 parabola question requiring basic knowledge of the standard form y²=4ax. Part (a) uses symmetry (trivial), part (b) substitutes into the equation (routine), and part (c) finds a line through two points using the focus at (2,0). All steps are standard textbook exercises with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03g Parametric equations: of curves and conversion to cartesian1.03h Parametric equations: in modelling contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(PQ = 12 \Rightarrow\) by symmetry \(y_P = \frac{12}{2} = 6\) | B1 | \(y = 6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^2 = 8x \Rightarrow 6^2 = 8x\) | M1 | Substitutes their \(y\)-coordinate into \(y^2 = 8x\) |
| \(x = \frac{36}{8} = \frac{9}{2}\), so \(P\) has coordinates \(\left(\frac{9}{2}, 6\right)\) | A1 oe | \(x = \frac{36}{8}\) or \(\frac{9}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Focus \(S(2, 0)\) | B1 | Focus has coordinates \((2, 0)\). Seen or implied. Can score anywhere |
| Gradient \(PS = \frac{6-0}{\frac{9}{2}-2} = \frac{6}{\frac{5}{2}} = \frac{12}{5}\) | M1 | Correct method for finding gradient of line segment \(PS\). If no gradient formula quoted and gradient incorrect, score M0. Allow if clear use of \(\frac{y_2-y_1}{x_2-x_1}\) even if coordinates 'confused' |
| \(y - 0 = \frac{12}{5}(x-2)\) or \(y - 6 = \frac{12}{5}(x - \frac{9}{2})\) | M1 | \(y - y_1 = m(x-x_1)\) with their PS gradient and their \((x_1, y_1)\). PS gradient must come from using P and S (not calculus) and must use their P or S as \((x_1, y_1)\). Or uses \(y = mx + c\) with their gradient to find \(c\), using P and S (not calculus) |
| \(l: 12x - 5y - 24 = 0\) | A1 | \(12x - 5y - 24 = 0\). Allow equivalent form e.g. \(k(12x - 5y - 24) = 0\) where \(k\) is an integer |
## Question 5:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $PQ = 12 \Rightarrow$ by symmetry $y_P = \frac{12}{2} = 6$ | B1 | $y = 6$ |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2 = 8x \Rightarrow 6^2 = 8x$ | M1 | Substitutes their $y$-coordinate into $y^2 = 8x$ |
| $x = \frac{36}{8} = \frac{9}{2}$, so $P$ has coordinates $\left(\frac{9}{2}, 6\right)$ | A1 oe | $x = \frac{36}{8}$ or $\frac{9}{2}$ |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Focus $S(2, 0)$ | B1 | Focus has coordinates $(2, 0)$. Seen or implied. Can score anywhere |
| Gradient $PS = \frac{6-0}{\frac{9}{2}-2} = \frac{6}{\frac{5}{2}} = \frac{12}{5}$ | M1 | Correct method for finding gradient of line segment $PS$. If no gradient formula quoted and gradient incorrect, score M0. Allow if clear use of $\frac{y_2-y_1}{x_2-x_1}$ even if coordinates 'confused' |
| $y - 0 = \frac{12}{5}(x-2)$ or $y - 6 = \frac{12}{5}(x - \frac{9}{2})$ | M1 | $y - y_1 = m(x-x_1)$ with their PS gradient and their $(x_1, y_1)$. **PS gradient must come from using P and S (not calculus) and must use their P or S as $(x_1, y_1)$.** Or uses $y = mx + c$ with their gradient to find $c$, using P and S (not calculus) |
| $l: 12x - 5y - 24 = 0$ | A1 | $12x - 5y - 24 = 0$. Allow equivalent form e.g. $k(12x - 5y - 24) = 0$ where $k$ is an integer |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5e512646-962b-424b-af5f-a6c6b332e0c9-06_732_654_258_646}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the parabola $C$ with equation $y ^ { 2 } = 8 x$. The point $P$ lies on $C$, where $y > 0$, and the point $Q$ lies on $C$, where $y < 0$ The line segment $P Q$ is parallel to the $y$-axis.
Given that the distance $P Q$ is 12 ,
\begin{enumerate}[label=(\alph*)]
\item write down the $y$-coordinate of $P$,
\item find the $x$-coordinate of $P$.
Figure 1 shows the point $S$ which is the focus of $C$.\\
The line $l$ passes through the point $P$ and the point $S$.
\item Find an equation for $l$ in the form $a x + b y + c = 0$, where $a$, $b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q5 [7]}}