| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola normal equation derivation |
| Difficulty | Moderate -0.3 This is a standard Further Maths parabola question requiring recall of focus/directrix formulas (comparing y²=16x to y²=4ax) and routine differentiation to find the normal equation. Part (a) is direct recall, part (b) involves implicit differentiation and point-slope form—straightforward techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Focus \((4, 0)\) | B1 | |
| Directrix \(x + 4 = 0\) | M1 | \(x +\) "4" \(= 0\) or \(x = -\) "4" |
| A1 | \(x + 4 = 0\) or \(x = -4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 4x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = 2x^{-\frac{1}{2}}\) | M1 | \(\frac{dy}{dx} = kx^{-\frac{1}{2}}\), or \(ky\frac{dy}{dx} = c\), or their \(\frac{dy}{dt} \times \left(\frac{1}{\text{their } \frac{dx}{dt}}\right)\) |
| \(\frac{dy}{dx} = 2x^{-\frac{1}{2}}\) or \(2y\frac{dy}{dx} = 16\) or \(\frac{dy}{dx} = 8 \cdot \frac{1}{8t}\) | A1 | Correct differentiation |
| At \(P\), gradient of normal \(= -t\) | A1 | Correct normal gradient with no errors seen |
| \(y - 8t = -t(x - 4t^2)\) | M1 | Applies \(y - 8t =\) their \(m_N(x - 4t^2)\); their \(m_N\) must be different from their \(m_T\) and must be a function of \(t\) |
| \(y + tx = 8t + 4t^3\) | A1 | cso given answer |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Focus $(4, 0)$ | B1 | |
| Directrix $x + 4 = 0$ | M1 | $x +$ "4" $= 0$ or $x = -$ "4" |
| | A1 | $x + 4 = 0$ or $x = -4$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 4x^{\frac{1}{2}} \Rightarrow \frac{dy}{dx} = 2x^{-\frac{1}{2}}$ | M1 | $\frac{dy}{dx} = kx^{-\frac{1}{2}}$, or $ky\frac{dy}{dx} = c$, or their $\frac{dy}{dt} \times \left(\frac{1}{\text{their } \frac{dx}{dt}}\right)$ |
| $\frac{dy}{dx} = 2x^{-\frac{1}{2}}$ or $2y\frac{dy}{dx} = 16$ or $\frac{dy}{dx} = 8 \cdot \frac{1}{8t}$ | A1 | Correct differentiation |
| At $P$, gradient of normal $= -t$ | A1 | Correct normal gradient with no errors seen |
| $y - 8t = -t(x - 4t^2)$ | M1 | Applies $y - 8t =$ their $m_N(x - 4t^2)$; **their $m_N$ must be different from their $m_T$ and must be a function of $t$** |
| $y + tx = 8t + 4t^3$ | A1 | cso **given answer** |
---3. A parabola $C$ has cartesian equation $y ^ { 2 } = 16 x$. The point $P \left( 4 t ^ { 2 } , 8 t \right)$ is a general point on $C$.
\begin{enumerate}[label=(\alph*)]
\item Write down the coordinates of the focus $F$ and the equation of the directrix of $C$.
\item Show that the equation of the normal to $C$ at $P$ is $y + t x = 8 t + 4 t ^ { 3 }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 2012 Q3 [8]}}