| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parametric point verification |
| Difficulty | Standard +0.3 This is a straightforward parametric parabola question requiring standard techniques: (a) finding the normal gradient using implicit differentiation and solving a simple equation, (b) finding the normal equation and substituting into the parabola equation. All steps are routine FP1 material with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{2a}{2at}\) or \(\frac{dy}{dx} = 2\sqrt{a}\times\frac{1}{2}x^{-\frac{1}{2}}\) or \(\frac{dy}{dx} = \frac{2a}{y}\) | B1 | 1.1b |
| Finds perpendicular gradient, sets equal to 2, uses \((at^2, 2at)\): \(-t=2 \Rightarrow t=-2\) | M1 | 2.1 |
| \(t=-2\) (correct, no errors seen) | A1* | 1.1b |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=-2 \Rightarrow\) point \((4a, -4a)\); correct equation of normal: \(y+4a = 2(x-4a)\) | B1 | 1.1b |
| Substituting \(x=9\) and forming 3TQ in \(\sqrt{a}\) or \(a\): e.g. \(2x-12a=\sqrt{4ax} \Rightarrow 18-12a=6\sqrt{a} \Rightarrow 12a+6\sqrt{a}-18=0\), or \(144a^2-468a+324=0\) | M1 | 3.1a |
| Solves 3TQ to find value of \(a\) (\(\sqrt{a}=1\) and \(-1.5\), or \(a=1\) and \(\frac{9}{4}\)) | dM1 | 1.1b |
| Correct possible values for \(a\) | A1 | 1.1b |
| \(a=\frac{9}{4}\) leads to \(x=9\) but \(P\) is other intersection point; \(a=1\) gives \(P(4,-4)\); \(\sqrt{a}\) cannot be negative, therefore \(a=1\) | A1 | 2.4 |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=-2 \Rightarrow (4a,-4a)\); normal: \(y+4a=2(x-4a)\) | B1 | 1.1b |
| \(2x-12a^2=4ax \Rightarrow 4x^2-48ax+144a^2=4ax \Rightarrow x^2-13ax+36a^2=0\) | M1 | 3.1a |
| \((x-4a)(x-9a)=0 \Rightarrow x=4a, 9a\) | dM1, A1 | 1.1b |
| \(x=4a\) is \(P\) so other point is \(x=9a\), so \(9=9a \Rightarrow a=1\) | A1 | 2.4 |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t=-2 \Rightarrow (4a,-4a)\); normal: \(y+4a=2(x-4a)\) | B1 | 1.1b |
| \(2at+4a=2(at^2-4a) \Rightarrow 2t+4=2t^2-8 \Rightarrow 2t^2-2t-12=0\) | M1 | 3.1a |
| \(t^2-t-6=0 \Rightarrow (t+2)(t-3)=0 \Rightarrow t=\text{"3"}\) | dM1, A1 | 1.1b |
| \(t=-2\) is \(P\), other point at \(t=3\): \(9=a\times3^2 \Rightarrow a=1\) | A1 | 2.4 |
| (5 marks) |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{2a}{2at}$ or $\frac{dy}{dx} = 2\sqrt{a}\times\frac{1}{2}x^{-\frac{1}{2}}$ or $\frac{dy}{dx} = \frac{2a}{y}$ | B1 | 1.1b |
| Finds perpendicular gradient, sets equal to 2, uses $(at^2, 2at)$: $-t=2 \Rightarrow t=-2$ | M1 | 2.1 |
| $t=-2$ (correct, no errors seen) | A1* | 1.1b |
| **(3 marks)** | | |
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=-2 \Rightarrow$ point $(4a, -4a)$; correct equation of normal: $y+4a = 2(x-4a)$ | B1 | 1.1b |
| Substituting $x=9$ and forming 3TQ in $\sqrt{a}$ or $a$: e.g. $2x-12a=\sqrt{4ax} \Rightarrow 18-12a=6\sqrt{a} \Rightarrow 12a+6\sqrt{a}-18=0$, or $144a^2-468a+324=0$ | M1 | 3.1a |
| Solves 3TQ to find value of $a$ ($\sqrt{a}=1$ and $-1.5$, or $a=1$ and $\frac{9}{4}$) | dM1 | 1.1b |
| Correct possible values for $a$ | A1 | 1.1b |
| $a=\frac{9}{4}$ leads to $x=9$ but $P$ is other intersection point; $a=1$ gives $P(4,-4)$; $\sqrt{a}$ cannot be negative, therefore $a=1$ | A1 | 2.4 |
| **(5 marks)** | | |
**Alt 1 to (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=-2 \Rightarrow (4a,-4a)$; normal: $y+4a=2(x-4a)$ | B1 | 1.1b |
| $2x-12a^2=4ax \Rightarrow 4x^2-48ax+144a^2=4ax \Rightarrow x^2-13ax+36a^2=0$ | M1 | 3.1a |
| $(x-4a)(x-9a)=0 \Rightarrow x=4a, 9a$ | dM1, A1 | 1.1b |
| $x=4a$ is $P$ so other point is $x=9a$, so $9=9a \Rightarrow a=1$ | A1 | 2.4 |
| **(5 marks)** | | |
**Alt 2 to (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t=-2 \Rightarrow (4a,-4a)$; normal: $y+4a=2(x-4a)$ | B1 | 1.1b |
| $2at+4a=2(at^2-4a) \Rightarrow 2t+4=2t^2-8 \Rightarrow 2t^2-2t-12=0$ | M1 | 3.1a |
| $t^2-t-6=0 \Rightarrow (t+2)(t-3)=0 \Rightarrow t=\text{"3"}$ | dM1, A1 | 1.1b |
| $t=-2$ is $P$, other point at $t=3$: $9=a\times3^2 \Rightarrow a=1$ | A1 | 2.4 |
| **(5 marks)** | | |\begin{enumerate}
\item The parabola $C$ has equation $y ^ { 2 } = 4 a x$ where $a$ is a positive constant.
\end{enumerate}
The point $P \left( a t ^ { 2 } , 2 a t \right) , t \neq 0$, lies on $C$\\
The normal to $C$ at $P$ is parallel to the line with equation $y = 2 x$\\
(a) For the point $P$, show that $t = - 2$
The normal to $C$ at $P$ intersects $C$ again when $x = 9$\\
(b) Determine the value of $a$, giving a reason for your answer.
\hfill \mbox{\textit{Edexcel FP1 AS 2023 Q6 [8]}}