| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Parabola area calculations |
| Difficulty | Challenging +1.2 This is a standard Further Pure 1 parabola question requiring implicit differentiation to find the tangent equation (routine technique), then finding intersection of two tangents and using area formula. While it involves multiple steps and algebraic manipulation, the techniques are all standard FP1 material with no novel insight required. The parametric form and tangent equation are textbook results, making this slightly above average difficulty for A-level but routine for Further Maths students. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2a}{2at} = \frac{1}{t}\) or \(2y\frac{\mathrm{d}y}{\mathrm{d}x}=4a \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4a}{2(2at)}=\frac{1}{t}\) | B1 | Correct gradient. |
| \(y - 2at = \frac{1}{t}(x - at^2)\) | M1 | Uses gradient to form tangent equation at \((at^2, 2at)\). |
| \(yt = x + at^2\)* | A1\* | Correct equation shown. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equations of tangents: \(yk = x+2k^2\) and/or \(-yk = x+2k^2\) | B1 | Both tangent equations stated (using \(a=2\), \(t=k\) and \(t=-k\)). |
| \(yk = x+2k^2\) and \(-yk = x+2k^2 \Rightarrow x = \ldots\{-2k^2\},\ y=\ldots\{0\}\) | M1 | Solves simultaneously to find intersection point. |
| \(\frac{8k\times(2k^2 - \text{their }(-2k^2))}{2} = 432\) leading to \(k=\ldots\) | M1 | Finds area of triangle, sets equal to 432 and solves for \(k\). |
| \(k = 3\) | A1 | Correct value. |
| \((18,12)\) and \((18,-12)\) | A1 | Both points correct. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct derivative \(\frac{dy}{dx} = \frac{1}{t}\) obtained from correct calculus method | B1 | Just stating \(\frac{dy}{dx} = \frac{1}{t}\) with no supporting working is B0 |
| Uses \(y - 2at =\) their \(\frac{1}{t}(x - at^2)\), or uses \(y = \frac{1}{t}x + c\) with \((at^2, 2at)\) to find \(c\) | M1 | May have just stated \(\frac{dy}{dx} = \frac{1}{t}\) for this mark |
| Achieves printed equation with no errors | A1* | Allow if \(\frac{dy}{dx} = \frac{1}{t}\) stated without working; B0M1A1 possible |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Deduces correct equation for one of the tangents | B1 | Must identify \(a=2\) or \(at^2 = -2k^2\) at some stage; accept one correct tangent |
| Finds intersection of tangent with \(x\)-axis, or solves both tangent equations simultaneously for at least \(x\)-coordinate | M1 | May be in terms of \(a\) and \(t\) |
| Complete method to find \(k\): \(\frac{8k \times (2k^2 - \text{their} - 2k^2)}{2} = 432\) and solves | M1 | Alt: \(432 = \frac{1}{2} |
| \(k = 3\) | A1 | |
| Correct coordinates \((18, 12)\) and \((18, -12)\) | A1 | Do not be concerned with labelling |
# Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2a}{2at} = \frac{1}{t}$ or $2y\frac{\mathrm{d}y}{\mathrm{d}x}=4a \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{4a}{2(2at)}=\frac{1}{t}$ | **B1** | Correct gradient. |
| $y - 2at = \frac{1}{t}(x - at^2)$ | **M1** | Uses gradient to form tangent equation at $(at^2, 2at)$. |
| $yt = x + at^2$* | **A1\*** | Correct equation shown. |
---
# Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equations of tangents: $yk = x+2k^2$ and/or $-yk = x+2k^2$ | **B1** | Both tangent equations stated (using $a=2$, $t=k$ and $t=-k$). |
| $yk = x+2k^2$ and $-yk = x+2k^2 \Rightarrow x = \ldots\{-2k^2\},\ y=\ldots\{0\}$ | **M1** | Solves simultaneously to find intersection point. |
| $\frac{8k\times(2k^2 - \text{their }(-2k^2))}{2} = 432$ leading to $k=\ldots$ | **M1** | Finds area of triangle, sets equal to 432 and solves for $k$. |
| $k = 3$ | **A1** | Correct value. |
| $(18,12)$ and $(18,-12)$ | **A1** | Both points correct. |
# Question (a) - Parametric Tangent
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct derivative $\frac{dy}{dx} = \frac{1}{t}$ obtained from correct calculus method | B1 | Just stating $\frac{dy}{dx} = \frac{1}{t}$ with no supporting working is B0 |
| Uses $y - 2at =$ their $\frac{1}{t}(x - at^2)$, or uses $y = \frac{1}{t}x + c$ with $(at^2, 2at)$ to find $c$ | M1 | May have just stated $\frac{dy}{dx} = \frac{1}{t}$ for this mark |
| Achieves printed equation with no errors | A1* | Allow if $\frac{dy}{dx} = \frac{1}{t}$ stated without working; B0M1A1 possible |
---
# Question (b) - Tangent Intersection and Area
| Working/Answer | Mark | Guidance |
|---|---|---|
| Deduces correct equation for one of the tangents | B1 | Must identify $a=2$ or $at^2 = -2k^2$ at some stage; accept one correct tangent |
| Finds intersection of tangent with $x$-axis, or solves both tangent equations simultaneously for at least $x$-coordinate | M1 | May be in terms of $a$ and $t$ |
| Complete method to find $k$: $\frac{8k \times (2k^2 - \text{their} - 2k^2)}{2} = 432$ and solves | M1 | Alt: $432 = \frac{1}{2}|\overrightarrow{DC} \times \overrightarrow{DB}| = \frac{1}{2}\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 4k^2 & 4k & 0\\ 4k^2 & -4k & 0\end{vmatrix} = \frac{1}{2}|-32k^3\mathbf{k}| = 16k^3$ |
| $k = 3$ | A1 | |
| Correct coordinates $(18, 12)$ and $(18, -12)$ | A1 | Do not be concerned with labelling |
---\begin{enumerate}
\item The parabola $P$ has equation $y ^ { 2 } = 4 a x$, where $a$ is a positive constant.
\end{enumerate}
The point $A \left( a t ^ { 2 } , 2 a t \right)$, where $t \neq 0$, lies on $P$.\\
(a) Use calculus to show that an equation of the tangent to $P$ at $A$ is
$$y t = x + a t ^ { 2 }$$
The point $B \left( 2 k ^ { 2 } , 4 k \right)$ and the point $C \left( 2 k ^ { 2 } , - 4 k \right)$, where $k$ is a constant, lie on $P$.\\
The tangent to $P$ at $B$ and the tangent to $P$ at $C$ intersect at the point $D$.\\
Given that the area of the triangle $B C D$ is 432\\
(b) determine the coordinates of $B$ and the coordinates of $C$.
\hfill \mbox{\textit{Edexcel FP1 2024 Q8 [8]}}