| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Ellipse locus problems |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring implicit differentiation to find the normal, coordinate geometry to find intercepts and midpoint, then parametric elimination to find a locus. While multi-step, each technique is standard for F3, though the algebraic manipulation in part (b) requires care and persistence across several steps. |
| Spec | 1.03d Circles: equation (x-a)^2+(y-b)^2=r^21.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{6\cos\theta}{-8\sin\theta}\) or \(\frac{2x}{64}+\frac{2y}{36}\frac{dy}{dx}=0\) or \(\frac{dy}{dx}=\frac{1}{4}\times\frac{1}{2}(576-9x^2)^{-\frac{1}{2}}\times-18x\) | B1 | Correct statement for, or involving, \(\frac{dy}{dx}\) |
| \(m_T = -\frac{3\cos\theta}{4\sin\theta} \Rightarrow m_N = -\frac{1}{m_T} = \frac{4\sin\theta}{3\cos\theta}\) | M1 | Finds \(\frac{dy}{dx}\) in terms of \(\theta\) and applies perpendicular condition |
| \(y - 6\sin\theta = \frac{4\sin\theta}{3\cos\theta}(x-8\cos\theta)\) | dM1 | Uses their normal gradient and \(P\) to find equation of normal |
| \(\Rightarrow 4x\sin\theta - 3y\cos\theta = 14\sin\theta\cos\theta\)* | A1* | Correct answer from correct work with at least one intermediate step, no errors |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A\) is \(\left(\frac{7}{2}\cos\theta, 0\right)\) and \(B\) is \(\left(0, -\frac{14}{3}\sin\theta\right)\) | B1 | Correct coordinates for \(A\) and \(B\) (or correct intercepts). Allow unsimplified |
| \(M\) is \(\left(\frac{\frac{7}{2}\cos\theta}{2}, \frac{-\frac{14}{3}\sin\theta}{2}\right) = \left(\frac{7}{4}\cos\theta, -\frac{7}{3}\sin\theta\right)\) | M1 | Uses \(A\) and \(B\) to attempt midpoint \(M\). May be implied by at least one correct coordinate |
| \(\sin^2\theta+\cos^2\theta=1 \Rightarrow \left(-\frac{3}{7}y\right)^2+\left(\frac{4}{7}x\right)^2=1\) | dM1, A1 | Uses \(\sin^2\theta+\cos^2\theta=1\) with their \(M\) to form equation in \(x\) and \(y\) only. Depends on previous mark. Correct unsimplified equation |
| \(\Rightarrow 16x^2+9y^2=49\) | A1 | Correct equation in required form. Allow any integer multiple |
| (5) |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{6\cos\theta}{-8\sin\theta}$ or $\frac{2x}{64}+\frac{2y}{36}\frac{dy}{dx}=0$ or $\frac{dy}{dx}=\frac{1}{4}\times\frac{1}{2}(576-9x^2)^{-\frac{1}{2}}\times-18x$ | B1 | Correct statement for, or involving, $\frac{dy}{dx}$ |
| $m_T = -\frac{3\cos\theta}{4\sin\theta} \Rightarrow m_N = -\frac{1}{m_T} = \frac{4\sin\theta}{3\cos\theta}$ | M1 | Finds $\frac{dy}{dx}$ in terms of $\theta$ and applies perpendicular condition |
| $y - 6\sin\theta = \frac{4\sin\theta}{3\cos\theta}(x-8\cos\theta)$ | dM1 | Uses their normal gradient and $P$ to find equation of normal |
| $\Rightarrow 4x\sin\theta - 3y\cos\theta = 14\sin\theta\cos\theta$* | A1* | Correct answer from correct work with at least one intermediate step, no errors |
| | **(4)** | |
# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A$ is $\left(\frac{7}{2}\cos\theta, 0\right)$ and $B$ is $\left(0, -\frac{14}{3}\sin\theta\right)$ | B1 | Correct coordinates for $A$ and $B$ (or correct intercepts). Allow unsimplified |
| $M$ is $\left(\frac{\frac{7}{2}\cos\theta}{2}, \frac{-\frac{14}{3}\sin\theta}{2}\right) = \left(\frac{7}{4}\cos\theta, -\frac{7}{3}\sin\theta\right)$ | M1 | Uses $A$ and $B$ to attempt midpoint $M$. May be implied by at least one correct coordinate |
| $\sin^2\theta+\cos^2\theta=1 \Rightarrow \left(-\frac{3}{7}y\right)^2+\left(\frac{4}{7}x\right)^2=1$ | dM1, A1 | Uses $\sin^2\theta+\cos^2\theta=1$ with their $M$ to form equation in $x$ and $y$ only. **Depends on previous mark.** Correct unsimplified equation |
| $\Rightarrow 16x^2+9y^2=49$ | A1 | Correct equation in required form. Allow any integer multiple |
| | **(5)** | |3. The ellipse $E$ has equation
$$\frac { x ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 36 } = 1$$
The line $l$ is the normal to $E$ at the point $P ( 8 \cos \theta , 6 \sin \theta )$.
\begin{enumerate}[label=(\alph*)]
\item Using calculus, show that an equation for $l$ is
$$4 x \sin \theta - 3 y \cos \theta = 14 \sin \theta \cos \theta$$
The line $l$ meets the $x$-axis at the point $A$ and meets the $y$-axis at the point $B$.\\
The point $M$ is the midpoint of $A B$.
\item Determine a Cartesian equation for the locus of $M$ as $\theta$ varies, giving your answer in the form $a x ^ { 2 } + b y ^ { 2 } = c$ where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2021 Q3 [9]}}