| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse locus problems |
| Difficulty | Challenging +1.3 This is a structured Further Maths ellipse question requiring parametric differentiation, finding tangent/normal equations, and deriving a locus through coordinate geometry. While it involves multiple steps and Further Maths content (making it harder than typical A-level), the techniques are standard for F3 and the question provides significant scaffolding through parts (a)-(c). The locus derivation requires algebraic manipulation but follows a predictable pattern for this topic. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -\frac{3\cos\theta}{4\sin\theta}\) or \(\frac{2x}{16} + \frac{2y}{9}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{18x}{32y}\) | M1 | Uses correct method and finds expression for \(\frac{dy}{dx}\) of correct form (sign and coefficient slips only) |
| \(\frac{dy}{dx} = -\frac{3\cos\theta}{4\sin\theta}\) e.g. \(-\frac{3}{4}\cot\theta\) | A1 | Any correct derivative in terms of \(\theta\) only |
| \(y - 3\sin\theta = -\frac{3\cos\theta}{4\sin\theta}(x - 4\cos\theta)\) or \(y = -\frac{3\cos\theta}{4\sin\theta}x + c \Rightarrow c = ...\left\{\frac{12\sin^2\theta + 12\cos^2\theta}{4\sin\theta}\right\}\) | M1 | Applies correct straight line method using any gradient in terms of \(\theta\). If \(y=mx+c\) must substitute coordinates correctly and reach \(c=...\). M0 if normal gradient used |
| \(\Rightarrow 4y\sin\theta - 12\sin^2\theta = -3x\cos\theta + 12\cos^2\theta\) | M1 | Multiplies through to remove fraction to obtain equation with trig expressions in sin and cos only |
| \(\Rightarrow 3x\cos\theta + 4y\sin\theta = 12(\cos^2\theta + \sin^2\theta) = 12\) | A1* | \(\sin^2\theta\) and \(\cos^2\theta\) must be seen somewhere in working. Accept \(\sin^2\theta + \cos^2\theta = 1\) in side-working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y - 3\sin\theta = \frac{4\sin\theta}{3\cos\theta}(x - 4\cos\theta)\) e.g. \(4x\sin\theta - 3y\cos\theta = 7\sin\theta\cos\theta\) | M1 | Applies correct straight line method with negative reciprocal of tangent gradient. If \(y=mx+c\) coordinates must be substituted correctly and \(c=...\) reached |
| \(\Rightarrow 3\sin\theta = \frac{4\sin\theta}{3\cos\theta}\cdot 4\cos\theta + c \Rightarrow c = ...\left\{\frac{-7\sin\theta\cos\theta}{3\cos\theta}\right\}\) | A1 | Any correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A\) is \(\left(\frac{4}{\cos\theta}, 0\right)\) | B1 | Any correct \(x\)-axis intercept of tangent. Allow e.g. \(\{x=\}\frac{12}{3\cos\theta}\), \(4\sec\theta\). Could be on diagram or implied by midpoint |
| \(x = 0 \Rightarrow y - 3\sin\theta = -\frac{16}{3}\sin\theta \Rightarrow B\) is \(\left(0, -\frac{7}{3}\sin\theta\right)\) | M1 | Sets \(x=0\) in normal equation (changed gradient) and finds \(y\). Allow just \(-\frac{7}{3}\sin\theta\) |
| Midpoint \(M\) of \(AB\) is \(\left(\frac{2}{\cos\theta}, -\frac{7}{6}\sin\theta\right)\) | A1 | Any correct midpoint, any equivalents as \(x=...,\ y=...\) |
| \(\sin^2\theta + \cos^2\theta = 1 \Rightarrow \left(-\frac{6}{7}y\right)^2 + \left(\frac{2}{x}\right)^2 = 1\) | M1 | Uses \(\sin^2\theta + \cos^2\theta = 1\) to obtain equation in \(x\) and \(y\) only. May follow incorrect or no attempt at midpoint |
| \(\Rightarrow \frac{36}{49}y^2 + \frac{4}{x^2} = 1 \Rightarrow 36x^2y^2 + 49\times 4 = 49x^2\) \(\Rightarrow x^2(49 - 36y^2) = 196\) | dM1 A1 | dM1: Rearranges to form \(x^2(p \pm qy^2) = r\). Requires all previous M marks. A1: Correct equation |
## Question 6:
### Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -\frac{3\cos\theta}{4\sin\theta}$ or $\frac{2x}{16} + \frac{2y}{9}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{18x}{32y}$ | M1 | Uses correct method and finds expression for $\frac{dy}{dx}$ of correct form (sign and coefficient slips only) |
| $\frac{dy}{dx} = -\frac{3\cos\theta}{4\sin\theta}$ e.g. $-\frac{3}{4}\cot\theta$ | A1 | Any correct derivative in terms of $\theta$ only |
| $y - 3\sin\theta = -\frac{3\cos\theta}{4\sin\theta}(x - 4\cos\theta)$ or $y = -\frac{3\cos\theta}{4\sin\theta}x + c \Rightarrow c = ...\left\{\frac{12\sin^2\theta + 12\cos^2\theta}{4\sin\theta}\right\}$ | M1 | Applies correct straight line method using any gradient in terms of $\theta$. If $y=mx+c$ must substitute coordinates correctly and reach $c=...$. M0 if normal gradient used |
| $\Rightarrow 4y\sin\theta - 12\sin^2\theta = -3x\cos\theta + 12\cos^2\theta$ | M1 | Multiplies through to remove fraction to obtain equation with trig expressions in sin and cos only |
| $\Rightarrow 3x\cos\theta + 4y\sin\theta = 12(\cos^2\theta + \sin^2\theta) = 12$ | A1* | $\sin^2\theta$ and $\cos^2\theta$ must be seen somewhere in working. Accept $\sin^2\theta + \cos^2\theta = 1$ in side-working |
**Total: 5 marks**
### Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y - 3\sin\theta = \frac{4\sin\theta}{3\cos\theta}(x - 4\cos\theta)$ e.g. $4x\sin\theta - 3y\cos\theta = 7\sin\theta\cos\theta$ | M1 | Applies correct straight line method with negative reciprocal of tangent gradient. If $y=mx+c$ coordinates must be substituted correctly and $c=...$ reached |
| $\Rightarrow 3\sin\theta = \frac{4\sin\theta}{3\cos\theta}\cdot 4\cos\theta + c \Rightarrow c = ...\left\{\frac{-7\sin\theta\cos\theta}{3\cos\theta}\right\}$ | A1 | Any correct equation |
**Total: 2 marks**
### Part 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A$ is $\left(\frac{4}{\cos\theta}, 0\right)$ | B1 | Any correct $x$-axis intercept of tangent. Allow e.g. $\{x=\}\frac{12}{3\cos\theta}$, $4\sec\theta$. Could be on diagram or implied by midpoint |
| $x = 0 \Rightarrow y - 3\sin\theta = -\frac{16}{3}\sin\theta \Rightarrow B$ is $\left(0, -\frac{7}{3}\sin\theta\right)$ | M1 | Sets $x=0$ in normal equation (changed gradient) and finds $y$. Allow just $-\frac{7}{3}\sin\theta$ |
| Midpoint $M$ of $AB$ is $\left(\frac{2}{\cos\theta}, -\frac{7}{6}\sin\theta\right)$ | A1 | Any correct midpoint, any equivalents as $x=...,\ y=...$ |
| $\sin^2\theta + \cos^2\theta = 1 \Rightarrow \left(-\frac{6}{7}y\right)^2 + \left(\frac{2}{x}\right)^2 = 1$ | M1 | Uses $\sin^2\theta + \cos^2\theta = 1$ to obtain equation in $x$ and $y$ only. May follow incorrect or no attempt at midpoint |
| $\Rightarrow \frac{36}{49}y^2 + \frac{4}{x^2} = 1 \Rightarrow 36x^2y^2 + 49\times 4 = 49x^2$ $\Rightarrow x^2(49 - 36y^2) = 196$ | dM1 A1 | dM1: Rearranges to form $x^2(p \pm qy^2) = r$. Requires all previous M marks. A1: Correct equation |
**Total: 6 marks**
---\begin{enumerate}
\item The ellipse $E$ has equation $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$
\end{enumerate}
The point $P ( 4 \cos \theta , 3 \sin \theta )$ lies on $E$.\\
(a) Use calculus to show that an equation of the tangent to $E$ at $P$ is
$$3 x \cos \theta + 4 y \sin \theta = 12$$
(b) Determine an equation for the normal to $E$ at $P$.
The tangent to $E$ at $P$ meets the $x$-axis at the point $A$.\\
The normal to $E$ at $P$ meets the $y$-axis at the point $B$.\\
(c) Show that the locus of the midpoint of $A$ and $B$ as $\theta$ varies has equation
$$x ^ { 2 } \left( p - q y ^ { 2 } \right) = r$$
where $p , q$ and $r$ are integers to be determined.
\hfill \mbox{\textit{Edexcel F3 2023 Q6 [13]}}