A2 October 2021 Paper 1 Q14
14 A curve has polar equation \(r = a(\cos\theta + 2\sin\theta)\), where \(a\) is a positive constant and \(0 \leqslant \theta \leqslant \pi\).
(a) Determine the polar coordinates of the point on the curve which is furthest from the pole. [7]
(b)
(i) Show that the curve is a circle whose radius should be specified. [6]
(ii) Write down the polar coordinates of the centre of the circle. [1]
| Scheme | Marks | AO |
|---|---|---|
| Let \(\cos\theta + 2\sin\theta = R\cos(\theta - \alpha)\) \(= R\cos\theta\cos\alpha + R\sin\theta\sin\alpha\) \(R\cos\alpha = 1,\ R\sin\alpha = 2\) | M1A1 | 3.1a 1.1 |
| \(R = \sqrt{5}\) | M1 | 1.1 |
| \(\tan\alpha = 2\) \(\alpha = 1.107\) | A1 | 1.1 |
| so \(r = \sqrt{5}a\cos(\theta - 1.107)\) \(r\) is maximum when \(\cos(\theta - 1.107) = 1\) | M1 | 3.1a |
| \(\Rightarrow \theta = 1.107\) rad | A1 | 1.1 |
| polar coordinates are \(\left[\sqrt{5}a,\ 1.107\right]\) | A1 | 1.1 |
| [7] |
Notes
\(\alpha = 1.107\): or \(\arctan(2)\)
\(\theta = 1.107\) rad: or \(\arctan(2)\)
Alternative solution
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = a(-\sin\theta + 2\cos\theta)\) | M1A1 |
| \(r\) is maximum when \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta} = 0\) | M1 |
| \(\Rightarrow -2\cos\theta + \sin\theta = 0\) \(\Rightarrow \tan\theta = 2 \Rightarrow \theta = 1.107\) | M1 |
| when \(\theta = 1.107\), \(r = \sqrt{5}a\) or \(2.236a\) | A1B1 |
| polar coordinates are \([2.236a,\ 1.107]\) | B1 |
| [7] |
B1: 2.24 or better
| Scheme | Marks | AO |
|---|---|---|
| (i) \(r^2 = ar\cos\theta + 2ar\sin\theta\) | M1 M1 | 3.1a 3.1a |
| \(\Rightarrow x^2 + y^2 = ax + 2ay\) | A1 | 1.1 |
| \(\Rightarrow \left(x - \frac{1}{2}a\right)^2 + (y - a)^2 = \frac{5}{4}a^2\) | M1 | 2.1 |
| This is the cartesian equation of a circle radius \(\frac{1}{2}\sqrt{5}a\) | A1 A1 | 2.2a 3.2a |
| [6] | ||
| (ii) centre \(\left[\frac{1}{2}\sqrt{5}a,\ 1.107\right]\) | B1 | 1.1 |
| [1] |
Notes
(b)(i)
M1: attempt to find cartesian eqn
M1: multiplying by \(r\)
M1: completing the square