| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola normal equation derivation |
| Difficulty | Standard +0.3 This is a standard FP1 parabola question requiring verification of parametric form, recall of focus position, differentiation to find normal equation, and coordinate geometry for triangle area. All steps are routine applications of learned techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(y^2 = (8t)^2 = 64t^2\) and \(16x = 16 \times 4t^2 = 64t^2\); or identifies \(a=4\) and uses general coordinates \((at^2, 2at)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \((4, 0)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(y = 4x^{\frac{1}{2}},\quad \frac{dy}{dx} = 2x^{-\frac{1}{2}}\) | B1 | |
| Replaces \(x\) by \(4t^2\) to give gradient \(\left[2(4t^2)^{-\frac{1}{2}} = \frac{2}{2t} = \frac{1}{t}\right]\) | M1 | Second M1 need not be function of \(t\) |
| Uses gradient of normal \(= -\frac{1}{\text{gradient of curve}}\) \(\quad [-t]\) | M1 | |
| \(y - 8t = -t(x - 4t^2) \Rightarrow y + tx = 8t + 4t^3\) | M1 A1cso | Third M1 requires linear equation (not fraction) and should include parameter \(t\) |
| Alternatives: \(\frac{dx}{dt} = 8t\) and \(\frac{dy}{dt} = 8\), \(\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{1}{t}\); or \(2y\frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{8}{y} = \frac{8}{8t} = \frac{1}{t}\) | B1, M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| At \(N\), \(y=0\), so \(x = 8 + 4t^2\) or \(\frac{8t+4t^3}{t}\) | B1 | |
| Base \(SN = (8+4t^2) - 4 \quad(= 4 + 4t^2)\) | B1ft | Second B1 does not require simplification |
| Area of \(\triangle PSN = \frac{1}{2}(4+4t^2)(8t) = 16t(1+t^2)\) or \(16t + 16t^3\) for \(t > 0\) | M1 A1 | M1 needs correct area of triangle formula using \(\frac{1}{2}\) 'their \(SN\)' \(\times 8t\) |
## Question 6:
### Part (a)
| Working/Answer | Marks | Notes |
|---|---|---|
| $y^2 = (8t)^2 = 64t^2$ and $16x = 16 \times 4t^2 = 64t^2$; or identifies $a=4$ and uses general coordinates $(at^2, 2at)$ | B1 | |
### Part (b)
| Working/Answer | Marks | Notes |
|---|---|---|
| $(4, 0)$ | B1 | |
### Part (c)
| Working/Answer | Marks | Notes |
|---|---|---|
| $y = 4x^{\frac{1}{2}},\quad \frac{dy}{dx} = 2x^{-\frac{1}{2}}$ | B1 | |
| Replaces $x$ by $4t^2$ to give gradient $\left[2(4t^2)^{-\frac{1}{2}} = \frac{2}{2t} = \frac{1}{t}\right]$ | M1 | Second M1 need not be function of $t$ |
| Uses gradient of normal $= -\frac{1}{\text{gradient of curve}}$ $\quad [-t]$ | M1 | |
| $y - 8t = -t(x - 4t^2) \Rightarrow y + tx = 8t + 4t^3$ | M1 A1cso | Third M1 requires linear equation (not fraction) and should include parameter $t$ |
| Alternatives: $\frac{dx}{dt} = 8t$ and $\frac{dy}{dt} = 8$, $\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} = \frac{1}{t}$; or $2y\frac{dy}{dx} = 16 \Rightarrow \frac{dy}{dx} = \frac{8}{y} = \frac{8}{8t} = \frac{1}{t}$ | B1, M1 | |
### Part (d)
| Working/Answer | Marks | Notes |
|---|---|---|
| At $N$, $y=0$, so $x = 8 + 4t^2$ or $\frac{8t+4t^3}{t}$ | B1 | |
| Base $SN = (8+4t^2) - 4 \quad(= 4 + 4t^2)$ | B1ft | Second B1 does not require simplification |
| Area of $\triangle PSN = \frac{1}{2}(4+4t^2)(8t) = 16t(1+t^2)$ or $16t + 16t^3$ for $t > 0$ | M1 A1 | M1 needs correct area of triangle formula using $\frac{1}{2}$ 'their $SN$' $\times 8t$ |6. The parabola $C$ has equation $y ^ { 2 } = 16 x$.
\begin{enumerate}[label=(\alph*)]
\item Verify that the point $P \left( 4 t ^ { 2 } , 8 t \right)$ is a general point on $C$.
\item Write down the coordinates of the focus $S$ of $C$.
\item Show that the normal to $C$ at $P$ has equation
$$y + t x = 8 t + 4 t ^ { 3 }$$
The normal to $C$ at $P$ meets the $x$-axis at the point $N$.
\item Find the area of triangle $P S N$ in terms of $t$, giving your answer in its simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2009 Q6 [11]}}