| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola normal equation derivation |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring parametric differentiation, normal equation derivation, simultaneous equations, and area calculation using the focus. While the techniques are standard FP1 material, the extended nature (4 parts building on each other) and the geometric insight needed for part (d) elevate it above average difficulty. It's more challenging than typical C1-C4 questions but not exceptionally hard for FP1. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 2a^{\frac{1}{2}}x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}\); or \(y^2 = 4ax \Rightarrow 2y\frac{dy}{dx} = 4a\); or \(\frac{dy}{dx} = \frac{dy}{dp}\cdot\frac{dp}{dx} = 2a\cdot\frac{1}{2ap}\) | M1 | \(\frac{dy}{dx} = kx^{-\frac{1}{2}}\); or \(ky\frac{dy}{dx} = c\); or their \(\frac{dy}{dp} \times \left(\frac{1}{\text{their } \frac{dx}{dp}}\right)\) |
| \(\frac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}\) or \(2y\frac{dy}{dx} = 4a\) or \(\frac{dy}{dx} = 2a\cdot\frac{1}{2ap}\) | A1 | Correct differentiation |
| At \(P\), gradient of normal \(= -p\) | A1 | Correct normal gradient with no errors seen |
| \(y - 2ap = -p(x - ap^2)\) | M1 | Applies \(y - 2ap = m_N(x - ap^2)\) or \(y = (m_N)x + c\) using \(x = ap^2\) and \(y = 2ap\) to find \(c\); \(m_N\) must be different from \(m_T\) and must be a function of \(p\) |
| \(y + px = 2ap + ap^3\) * | A1* | cso — given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y - px = -2ap - ap^3\) | B1 | oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 0 \Rightarrow x = 2a + ap^2\) | M1A1 | M1: \(y = 0\) in either normal or solves simultaneously to find \(x\); A1: \(y = 0\) and correct \(x\) coordinate |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(S\) is \((a, 0)\) | B1 | Can be implied below |
| Area \(SPQP' = \frac{1}{2} \times (``2a + ap^2\text{''} - a) \times 2ap \times 2\) | M1 | Correct method for the area of the quadrilateral |
| \(= 2a^2p(1 + p^2)\) | A1 | Any equivalent form |
## Question 7:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 2a^{\frac{1}{2}}x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}$; or $y^2 = 4ax \Rightarrow 2y\frac{dy}{dx} = 4a$; or $\frac{dy}{dx} = \frac{dy}{dp}\cdot\frac{dp}{dx} = 2a\cdot\frac{1}{2ap}$ | M1 | $\frac{dy}{dx} = kx^{-\frac{1}{2}}$; or $ky\frac{dy}{dx} = c$; or their $\frac{dy}{dp} \times \left(\frac{1}{\text{their } \frac{dx}{dp}}\right)$ |
| $\frac{dy}{dx} = a^{\frac{1}{2}}x^{-\frac{1}{2}}$ or $2y\frac{dy}{dx} = 4a$ or $\frac{dy}{dx} = 2a\cdot\frac{1}{2ap}$ | A1 | Correct differentiation |
| At $P$, gradient of normal $= -p$ | A1 | Correct normal gradient with no errors seen |
| $y - 2ap = -p(x - ap^2)$ | M1 | Applies $y - 2ap = m_N(x - ap^2)$ or $y = (m_N)x + c$ using $x = ap^2$ and $y = 2ap$ to find $c$; $m_N$ must be different from $m_T$ and must be a function of $p$ |
| $y + px = 2ap + ap^3$ * | A1* | cso — **given answer** |
**(5 marks)**
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y - px = -2ap - ap^3$ | B1 | oe |
**(1 mark)**
### Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 0 \Rightarrow x = 2a + ap^2$ | M1A1 | M1: $y = 0$ in either normal or solves simultaneously to find $x$; A1: $y = 0$ and correct $x$ coordinate |
**(2 marks)**
### Part (d)
| Answer | Mark | Guidance |
|--------|------|----------|
| $S$ is $(a, 0)$ | B1 | Can be implied below |
| Area $SPQP' = \frac{1}{2} \times (``2a + ap^2\text{''} - a) \times 2ap \times 2$ | M1 | Correct method for the area of the quadrilateral |
| $= 2a^2p(1 + p^2)$ | A1 | Any equivalent form |
**(3 marks; Total: 11)**
---7. The parabola $C$ has cartesian equation $y ^ { 2 } = 4 a x , a > 0$
The points $P \left( a p ^ { 2 } , 2 a p \right)$ and $P ^ { \prime } \left( a p ^ { 2 } , - 2 a p \right)$ lie on $C$.
\begin{enumerate}[label=(\alph*)]
\item Show that an equation of the normal to $C$ at the point $P$ is
$$y + p x = 2 a p + a p ^ { 3 }$$
\item Write down an equation of the normal to $C$ at the point $P ^ { \prime }$.
The normal to $C$ at $P$ meets the normal to $C$ at $P ^ { \prime }$ at the point $Q$.
\item Find, in terms of $a$ and $p$, the coordinates of $Q$.
Given that $S$ is the focus of the parabola,
\item find the area of the quadrilateral $S P Q P ^ { \prime }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2014 Q7 [11]}}