| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola normal intersection problems |
| Difficulty | Standard +0.8 This is a multi-part Further Maths parabola question requiring: (a) standard normal derivation using parametric differentiation, (b) finding p from a given condition then locating directrix intersection, and (c) coordinate geometry area calculation. While systematic, it demands fluency with parametric forms, multiple algebraic steps, and careful coordinate workâmoderately challenging for FP1 but follows established patterns. |
| 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 |
|---|---|---|
| \(y = 2\sqrt{ax^2} \Rightarrow \frac{dy}{dx} = \sqrt{a}x^{-1}\) or (implicitly) \(2y\frac{dy}{dx} = 4a\) or (chain rule) \(\frac{dy}{dx} = 2a \times \frac{1}{2ap}\) | M1 | \(\frac{dy}{dx} = \pm kx^{-1}\) or \(k\frac{dy}{dx} = c\) or their \(\frac{dy}{dx}\) / their \(\frac{ds}{dx}\) |
| When \(x = ap^2\): \(\frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{ap^2}} = \frac{\sqrt{a}}{\sqrt{a}p} = \frac{1}{p}\) or \(\frac{dy}{dx} = \frac{4a}{2(2ap)} = \frac{1}{p}\) | A1 | \(\frac{dy}{dx} = \frac{1}{p}\) |
| So \(m_y = -p\) | M1 | Applies \(m_y = \frac{-1}{\text{their } m_T}\) |
| N: \(y - 2ap = -p\left(x - ap^2\right)\) | M1 | Applies \(y - 2ap = (\text{their } m_y)(x - ap^2)\) |
| N: \(y - 2ap = -px + ap^3\) | ||
| N: \(y + px = ap^3 + 2ap\) | A1 cso * | Correct solution |
| Answer | Marks | Guidance |
|---|---|---|
| \((6a, 0) \Rightarrow 0 + p(6a) = ap^3 + 2ap\) | M1 | Substitutes \(x = 6a\), \(y = 0\) into N |
| \(\Rightarrow 4a = ap^3 \Rightarrow p = 2\) | A1 | |
| \(x = -a\), \(p = 2 \Rightarrow y + 2(-a) = a(2)^3 + 2a(2)\) | dM1 | Substitutes \(x = -a\) and their \(p\) into N |
| \(\Rightarrow y = 8a + 4a + 2a = 14a \Rightarrow D(-a, 14a)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| When \(p = 2\), \(x = a(2)^2 = 4a\) | M1 | Substitutes their \(p\) into \(x = ap^2\) |
| Area\((XPD) = \frac{1}{2}(14a)(5a) = 35a^2\) | M1, A1 | Applies \(\frac{1}{2}(\text{their } 14a)(\text{their } "4a" + a)\); \(35a^2\) |
**(a)**
| $y = 2\sqrt{ax^2} \Rightarrow \frac{dy}{dx} = \sqrt{a}x^{-1}$ or (implicitly) $2y\frac{dy}{dx} = 4a$ or (chain rule) $\frac{dy}{dx} = 2a \times \frac{1}{2ap}$ | M1 | $\frac{dy}{dx} = \pm kx^{-1}$ or $k\frac{dy}{dx} = c$ or their $\frac{dy}{dx}$ / their $\frac{ds}{dx}$ |
| When $x = ap^2$: $\frac{dy}{dx} = \frac{\sqrt{a}}{\sqrt{ap^2}} = \frac{\sqrt{a}}{\sqrt{a}p} = \frac{1}{p}$ or $\frac{dy}{dx} = \frac{4a}{2(2ap)} = \frac{1}{p}$ | A1 | $\frac{dy}{dx} = \frac{1}{p}$ |
| So $m_y = -p$ | M1 | Applies $m_y = \frac{-1}{\text{their } m_T}$ |
| **N:** $y - 2ap = -p\left(x - ap^2\right)$ | M1 | Applies $y - 2ap = (\text{their } m_y)(x - ap^2)$ |
| **N:** $y - 2ap = -px + ap^3$ | | |
| **N:** $y + px = ap^3 + 2ap$ | A1 cso * | Correct solution |
**(b)**
| $(6a, 0) \Rightarrow 0 + p(6a) = ap^3 + 2ap$ | M1 | Substitutes $x = 6a$, $y = 0$ into N |
| $\Rightarrow 4a = ap^3 \Rightarrow p = 2$ | A1 | |
| $x = -a$, $p = 2 \Rightarrow y + 2(-a) = a(2)^3 + 2a(2)$ | dM1 | Substitutes $x = -a$ and their $p$ into N |
| $\Rightarrow y = 8a + 4a + 2a = 14a \Rightarrow D(-a, 14a)$ | A1 | |
**(c)**
| When $p = 2$, $x = a(2)^2 = 4a$ | M1 | Substitutes their $p$ into $x = ap^2$ |
| Area$(XPD) = \frac{1}{2}(14a)(5a) = 35a^2$ | M1, A1 | Applies $\frac{1}{2}(\text{their } 14a)(\text{their } "4a" + a)$; $35a^2$ |
---8. The parabola $C$ has equation $y ^ { 2 } = 4 a x$, where $a$ is a positive constant.
The point $P \left( a p ^ { 2 } , 2 a p \right)$ lies on the parabola $C$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the normal to $C$ at $P$ is
$$y + p x = a p ^ { 3 } + 2 a p$$
The normal to $C$ at the point $P$ meets the $x$-axis at the point $( 6 a , 0 )$ and meets the directrix of $C$ at the point $D$. Given that $p > 0$,
\item find, in terms of $a$, the coordinates of $D$.
Given also that the directrix of $C$ cuts the $x$-axis at the point $X$,
\item find, in terms of $a$, the area of the triangle XPD, giving your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q8 [12]}}