| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 1 (Further AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Standard +0.3 This is a structured multi-part polar coordinates question with standard techniques: finding points on curves (parts a,b), calculating intersection points and using the triangle area formula (part c), and sketching a related cardioid (part d). While polar coordinates is a Further Maths topic, these are routine applications requiring recall of polar curve properties rather than novel problem-solving, making it slightly easier than average. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = 2a(1 + \sin 0) = 2a\), so \(M = (2a, 0)\) | B1 (1.1b) | Condone \((0, 2a)\) after seeing \(r = 2a\) and \(\theta = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(\sin\theta = 1\) | M1 (3.1a) | PI by \(r = 4a\) or \(\theta = \frac{\pi}{2}\) |
| \(r = 2a(1+1) = 4a\), so \(N = \left(4a, \frac{\pi}{2}\right)\) | A1 (1.1b) | Condone \(\left(\frac{\pi}{2}, 4a\right)\) after \(r=4a\) and \(\theta=\frac{\pi}{2}\); condone \(\left(\frac{\pi}{2}, 4a\right)\) if \((0,2a)\) penalised in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3a = 2a(1+\sin\theta) \Rightarrow \frac{3}{2} = 1 + \sin\theta\) | M1 (3.1a) | Forms equation by equating the two polar equations |
| \(\sin\theta = \frac{1}{2}\), so \(\theta = \frac{\pi}{6}\), \(\theta = \frac{5\pi}{6}\) | A1 (1.1b) | At least one correct \(\theta\) value for \(P\) or \(Q\); implied by correct area |
| \(x = 3a\cos\frac{\pi}{6} = \frac{3\sqrt{3}a}{2}\) | M1 (1.1a) | Calculates \(x\)-coordinate of \(P\) or \(Q\), ie \(x = 3a\cos\theta\); implied by \(6a\cos\theta\) for length \(PQ\); or calculates \(y\)-coordinate ie \(y = 3a\sin\theta\) |
| \(y = 3a\sin\frac{\pi}{6} = \frac{3a}{2}\) | M1 (2.2a) | Calculates required area (or half), ie \(\frac{1}{2} \times (y_N - y_P) \times 6a\cos\theta\) |
| \(\text{area} = \frac{1}{2} \times \frac{6\sqrt{3}a}{2} \times \left(4a - \frac{3a}{2}\right) = \frac{3\sqrt{3}a}{2} \times \frac{5a}{2} = \frac{15}{4}\sqrt{3}a^2\) | A1 (3.2a) | Correct area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape in all four quadrants with a cusp at the pole | B1 (1.1b) | |
| Point \((4a, 0)\) marked on diagram (intercept on initial line) | B1 (1.1b) | Condone omission of \((0,0)\) |
## Question 17(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 2a(1 + \sin 0) = 2a$, so $M = (2a, 0)$ | B1 (1.1b) | Condone $(0, 2a)$ after seeing $r = 2a$ and $\theta = 0$ |
---
## Question 17(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $\sin\theta = 1$ | M1 (3.1a) | PI by $r = 4a$ or $\theta = \frac{\pi}{2}$ |
| $r = 2a(1+1) = 4a$, so $N = \left(4a, \frac{\pi}{2}\right)$ | A1 (1.1b) | Condone $\left(\frac{\pi}{2}, 4a\right)$ after $r=4a$ and $\theta=\frac{\pi}{2}$; condone $\left(\frac{\pi}{2}, 4a\right)$ if $(0,2a)$ penalised in (a) |
---
## Question 17(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3a = 2a(1+\sin\theta) \Rightarrow \frac{3}{2} = 1 + \sin\theta$ | M1 (3.1a) | Forms equation by equating the two polar equations |
| $\sin\theta = \frac{1}{2}$, so $\theta = \frac{\pi}{6}$, $\theta = \frac{5\pi}{6}$ | A1 (1.1b) | At least one correct $\theta$ value for $P$ or $Q$; implied by correct area |
| $x = 3a\cos\frac{\pi}{6} = \frac{3\sqrt{3}a}{2}$ | M1 (1.1a) | Calculates $x$-coordinate of $P$ or $Q$, ie $x = 3a\cos\theta$; implied by $6a\cos\theta$ for length $PQ$; or calculates $y$-coordinate ie $y = 3a\sin\theta$ |
| $y = 3a\sin\frac{\pi}{6} = \frac{3a}{2}$ | M1 (2.2a) | Calculates required area (or half), ie $\frac{1}{2} \times (y_N - y_P) \times 6a\cos\theta$ |
| $\text{area} = \frac{1}{2} \times \frac{6\sqrt{3}a}{2} \times \left(4a - \frac{3a}{2}\right) = \frac{3\sqrt{3}a}{2} \times \frac{5a}{2} = \frac{15}{4}\sqrt{3}a^2$ | A1 (3.2a) | Correct area |
---
## Question 17(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape in all four quadrants with a cusp at the pole | B1 (1.1b) | |
| Point $(4a, 0)$ marked on diagram (intercept on initial line) | B1 (1.1b) | Condone omission of $(0,0)$ |17 The curve $C _ { 1 }$ has polar equation $r = 2 a ( 1 + \sin \theta )$ for $- \pi < \theta \leq \pi$ where $a$ is a positive constant.\\
\includegraphics[max width=\textwidth, alt={}, center]{f7e7c21b-6e72-4c20-92fc-ba0336a11136-22_469_830_402_605}
The point $M$ lies on $C _ { 1 }$ and the initial line.\\
17
\begin{enumerate}[label=(\alph*)]
\item Write down, in terms of $a$, the polar coordinates of $M$
17
\item $\quad N$ is the point on $C _ { 1 }$ that is furthest from the pole $O$\\
Find, in terms of $a$, the polar coordinates of $N$\\
17
\item The curve $C _ { 2 }$ has polar equation $r = 3 a$ for $- \pi < \theta \leq \pi$ $C _ { 2 }$ intersects $C _ { 1 }$ at points $P$ and $Q$
Show that the area of triangle $N P Q$ can be written in the form
$$m \sqrt { 3 } a ^ { 2 }$$
where $m$ is a rational number to be determined.\\
17
\item On the initial line below, sketch the graph of $r = 2 a ( 1 + \cos \theta )$ for $- \pi < \theta \leq \pi$ Include the polar coordinates, in terms of $a$, of any intersection points with the initial line.\\[0pt]
[2 marks]\\
\includegraphics[max width=\textwidth, alt={}, center]{f7e7c21b-6e72-4c20-92fc-ba0336a11136-24_65_657_1425_991}\\
\includegraphics[max width=\textwidth, alt={}, center]{f7e7c21b-6e72-4c20-92fc-ba0336a11136-25_2492_1721_217_150}
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 1 2021 Q17 [12]}}