| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola focus and directrix properties |
| Difficulty | Challenging +1.2 This is a Further Maths FP1 question requiring parametric coordinate manipulation and the perpendicularity condition. Students must find the normal equation, substitute Q's coordinates to get a relationship between p and q, then use the perpendicularity of OP and OQ to establish p² = 2. While it involves multiple algebraic steps and coordinate geometry insight, it follows standard parabola techniques taught in FP1 with clear signposting of what to prove. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^2 = 4ax \Rightarrow 2y\frac{dy}{dx} = 4a\) | M1 | |
| \(\frac{dy}{dx} = \frac{2a}{y}\); gradient of normal is \(\frac{-y}{2a} = -p\) | A1 | |
| Equation of normal: \(y - 2ap = -p(x - ap^2)\) | M1 | |
| Normal passes through \(Q(aq^2, 2aq)\): \(2aq + apq^2 = 2ap + ap^3\) | M1 | |
| Grad \(OP \times\) Grad \(OQ = -1 \Rightarrow \frac{2ap}{ap^2}\cdot\frac{2aq}{aq^2} = -1\) | M1 | |
| \(q = \frac{-4}{p}\) | A1 | |
| \(2a\left(\frac{-4}{p}\right) + ap\left(\frac{16}{p^2}\right) = 2ap + ap^3 \Rightarrow p^4 + 2p^2 - 8 = 0\) | M1 | |
| \((p^2-2)(p^2+4) = 0 \Rightarrow p^2 = \ldots\) | M1 | |
| Hence (as \(p^2+4 \neq 0\)), \(p^2 = 2\) | A1* | Correct solution, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| First three marks as above | M1, A1, M1 | 2.1, 1.1b, 1.1b |
| Solves \(y^2 = 4ax\) and normal simultaneously to find \(x_Q = ap^2 + 4a + \frac{4a}{p^2}\) or \(y_Q = -2ap - \frac{4a}{p}\) | M1 | 3.1a |
| Forms relationship between \(p\) and \(q\): either \(y_Q = 2a\left(-p - \frac{2}{p}\right) \Rightarrow q = -p - \frac{2}{p}\) or \(x_Q = a\left(p + \frac{2}{p}\right)^2 \Rightarrow q = \pm\left(p + \frac{2}{p}\right)\) | M1 | 2.1 |
| \(q = -p - \frac{2}{p}\) (if \(x\) coordinate used, correct root must be clearly identified) | A1 | 1.1b |
| Grad \(OP \times\) Grad \(OQ = -1 \Rightarrow \frac{2ap}{ap^2} \times \frac{2aq}{aq^2} = -1 \Rightarrow q = -\frac{4}{p}\) | M1 | 2.1 |
| Sets \(q = -p - \frac{2}{p} = -\frac{4}{p}\) and solves to give \(p^2 = \ldots\) | M1 | 1.1b |
| Hence \(p^2 = 2\) (only), as \(q = p + \frac{2}{p} = -\frac{4}{p}\) gives no solution | A1* | 1.1b |
| Total: (9) | (9 marks) |
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^2 = 4ax \Rightarrow 2y\frac{dy}{dx} = 4a$ | M1 | |
| $\frac{dy}{dx} = \frac{2a}{y}$; gradient of normal is $\frac{-y}{2a} = -p$ | A1 | |
| Equation of normal: $y - 2ap = -p(x - ap^2)$ | M1 | |
| Normal passes through $Q(aq^2, 2aq)$: $2aq + apq^2 = 2ap + ap^3$ | M1 | |
| Grad $OP \times$ Grad $OQ = -1 \Rightarrow \frac{2ap}{ap^2}\cdot\frac{2aq}{aq^2} = -1$ | M1 | |
| $q = \frac{-4}{p}$ | A1 | |
| $2a\left(\frac{-4}{p}\right) + ap\left(\frac{16}{p^2}\right) = 2ap + ap^3 \Rightarrow p^4 + 2p^2 - 8 = 0$ | M1 | |
| $(p^2-2)(p^2+4) = 0 \Rightarrow p^2 = \ldots$ | M1 | |
| Hence (as $p^2+4 \neq 0$), $p^2 = 2$ | A1* | Correct solution, no errors |
# Question 5 (Alternative 2):
| Answer/Working | Mark | Guidance |
|---|---|---|
| First three marks as above | M1, A1, M1 | 2.1, 1.1b, 1.1b |
| Solves $y^2 = 4ax$ and normal simultaneously to find $x_Q = ap^2 + 4a + \frac{4a}{p^2}$ or $y_Q = -2ap - \frac{4a}{p}$ | M1 | 3.1a |
| Forms relationship between $p$ and $q$: **either** $y_Q = 2a\left(-p - \frac{2}{p}\right) \Rightarrow q = -p - \frac{2}{p}$ **or** $x_Q = a\left(p + \frac{2}{p}\right)^2 \Rightarrow q = \pm\left(p + \frac{2}{p}\right)$ | M1 | 2.1 |
| $q = -p - \frac{2}{p}$ (if $x$ coordinate used, correct root must be clearly identified) | A1 | 1.1b |
| Grad $OP \times$ Grad $OQ = -1 \Rightarrow \frac{2ap}{ap^2} \times \frac{2aq}{aq^2} = -1 \Rightarrow q = -\frac{4}{p}$ | M1 | 2.1 |
| Sets $q = -p - \frac{2}{p} = -\frac{4}{p}$ and solves to give $p^2 = \ldots$ | M1 | 1.1b |
| Hence $p^2 = 2$ (only), as $q = p + \frac{2}{p} = -\frac{4}{p}$ gives no solution | A1* | 1.1b |
| **Total: (9)** | | **(9 marks)** |
---\begin{enumerate}
\item The normal to the parabola $y ^ { 2 } = 4 a x$ at the point $P \left( a p ^ { 2 } , 2 a p \right)$ passes through the parabola again at the point $Q \left( a q ^ { 2 } , 2 a q \right)$.
\end{enumerate}
The line $O P$ is perpendicular to the line $O Q$, where $O$ is the origin.\\
Prove that $p ^ { 2 } = 2$
\hfill \mbox{\textit{Edexcel FP1 Q5 [9]}}