| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola normal intersection problems |
| Difficulty | Standard +0.8 This is a multi-step Further Maths parabola problem requiring: (1) implicit differentiation to find the normal equation, (2) knowledge of parabola focus properties (y²=4ax form), (3) coordinate geometry to find intersection points, and (4) triangle area calculation. While the techniques are standard for FP1, the combination of conic section theory with coordinate geometry and the multi-part nature elevates this above average A-level difficulty. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y = 6x^2\), so \(\frac{dy}{dx} = 3x^{-\frac{1}{2}}\) | M1 | |
| Gradient when \(x = 4\) is \(\frac{3}{4}\) and gradient of normal is \(-\frac{2}{3}\) | M1 A1 | |
| So equation of normal is \((y - 12) = -\frac{2}{3}(x-4)\) (or \(3y + 2x = 44\)) | M1 A1 | (5) |
| (b) \(S\) is at point \((9,0)\); \(N\) is at \((22,0)\), found by substituting y=0 into their part (a) | B1; B1ft |
| Answer | Marks | Guidance |
|---|---|---|
| So area is \(\frac{1}{2} \times 12 \times (22-9) = 78\) | M1 A1 cao | (4) [9] |
| Answer | Marks |
|---|---|
| \(x = 9t^2, y = 18t \Rightarrow \frac{dx}{dt} = 18t, \frac{dy}{dt} = 18 \Rightarrow \frac{dy}{dx} = \frac{1}{t}\) |
(a) $y = 6x^2$, so $\frac{dy}{dx} = 3x^{-\frac{1}{2}}$ | M1 |
Gradient when $x = 4$ is $\frac{3}{4}$ and gradient of normal is $-\frac{2}{3}$ | M1 A1 |
So equation of normal is $(y - 12) = -\frac{2}{3}(x-4)$ (or $3y + 2x = 44$) | M1 A1 | (5)
(b) $S$ is at point $(9,0)$; $N$ is at $(22,0)$, found by substituting y=0 into their part (a) | B1; B1ft |
Both B marks can be implied or on diagram.
So area is $\frac{1}{2} \times 12 \times (22-9) = 78$ | M1 A1 cao | (4) [9]
**Alternatives:**
First M1 for $ky\frac{dy}{dx} = 36$ or for
$x = 9t^2, y = 18t \Rightarrow \frac{dx}{dt} = 18t, \frac{dy}{dt} = 18 \Rightarrow \frac{dy}{dx} = \frac{1}{t}$ |
**Notes:**
(a) First M for $\frac{dy}{dx} = ax^{-\frac{1}{2}}$
Second M for substituting x=4 (or y=12 or t=2/3 if alternative used) into their gradient and applying negative reciprocal.
First A for $-\frac{2}{3}$
Third M for $y - y_1 = m(x - x_1)$ or $y = mx + c$ and attempt to substitute a changed gradient AND $(4,12)$
Second A for $3y + 2x = 44$ or any equivalent equation
(b) M for Area=$\frac{1}{2}$ base x height and attempt to substitute including their numerical '(22-9)' or equivalent complete method to find area of triangle $PSN$.
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7833e9c0-4a73-4ac6-8a77-51a5489e0614-10_624_716_210_614}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the parabola with equation $y ^ { 2 } = 36 x$.\\
The point $P ( 4,12 )$ lies on the parabola.
\begin{enumerate}[label=(\alph*)]
\item Find an equation for the normal to the parabola at $P$.
This normal meets the $x$-axis at the point $N$ and $S$ is the focus of the parabola, as shown in Figure 1.
\item Find the area of triangle $P S N$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2013 Q9 [9]}}