June 2024 Paper 3 Q6
6 The curve \(C\) is defined, for \(0 \leqslant t \lt 2\pi\), by the parametric equations
\(x = 4k + k\sin t,\quad y = 2 + 4\cos t,\)
where \(k\) is a constant.
You are given that \(C\) is a circle.
| Scheme | Marks | AO |
|---|---|---|
| \(x = 4k + k\sin t,\ y = 2 + 4\cos t\) \(\sin t = \dfrac{x - 4k}{k},\ \cos t = \dfrac{y - 2}{4}\) and use of \(\sin^2 t + \cos^2 t = 1\) | M1 | 3.1a |
| \(\dfrac{(x - 4k)^2}{k^2} + \dfrac{(y - 2)^2}{16} = 1\) | A1 | 1.1 |
| [2] |
Notes
M1: Re-arranges to obtain both \(\sin t = \dfrac{x \pm 4k}{k}\) and \(\cos t = \dfrac{y \pm 2}{4}\) and use \(\sin^2 t + \cos^2 t = 1\) to eliminate \(t\)
A1: Allow any correct un-simplified cartesian form not involving trigonometric terms
ISW once a correct answer seen
SC for part (a): If M0 awarded then SC B1 for \(y = 2 + 4\cos\left(\arcsin\left(\dfrac{x - 4k}{k}\right)\right)\) or \(x = 4k + k\sin\left(\arccos\left(\dfrac{y - 2}{4}\right)\right)\) or \(\arcsin\left(\dfrac{x - 4k}{k}\right) = \arccos\left(\dfrac{y - 2}{4}\right)\) (or any correct form in terms of trig. functions – allow \(\cos^{-1}\) for \(\arccos\) etc.)
If considering one branch only e.g. \(y\left(= 2 + 4\sqrt{1 - \sin^2 t}\right) = 2 + 4\sqrt{1 - \left(\dfrac{x - 4k}{k}\right)^2}\) this scores M1 only (or equivalent expression for \(x\))
If considering both branches e.g. \(y\left(= 2 \pm 4\sqrt{1 - \sin^2 t}\right) = 2 \pm 4\sqrt{1 - \left(\dfrac{x - 4k}{k}\right)^2}\) then this scores M1 A1 (or equivalent expression for \(x\))
Note a (common) correct answer for M1 A1 is: \(x^2 + y^2 = 16k^2 + 8k(x - 4k) + (x - 4k)^2 + 4 + 4(y - 2) + (y - 2)^2\)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(C\) is a circle \(\Rightarrow k^2 = 16\) | M1 | 3.1a |
| \(r = 4\) | A1 | 1.1 |
| [2] | ||
| (ii) \((16, 2), (-16, 2)\) | B2 | 2.2a 2.2a |
| [2] |
Notes
(b)(i)
M1: Setting \(c\) and \(d\) equal in their \(\dfrac{(x \pm a)^2}{c} + \dfrac{(y \pm b)^2}{d} = 1\) or stating that \(y_{\min} = -2\) and \(y_{\max} = 6\)
Possibly implied by correct value for \(r\) www
A1: \(r = \pm 4\) is A0 unless replaced with positive \(r\) only
(b)(ii) B2: B1 for either one correct centre or both \(x\)-coordinates correct
These marks are dependent on \(r = 4\) from correct working or www