A2 June 2019 Paper 1 Q7
7 A curve has cartesian equation \((x^2 + y^2)^2 = 2c^2xy\), where \(c\) is a positive constant.
(a) Show that the polar equation of the curve is \(r^2 = c^2\sin 2\theta\). [2]
(b) Sketch the curves \(r = c\sqrt{\sin 2\theta}\) and \(r = -c\sqrt{\sin 2\theta}\) for \(0 \leqslant \theta \leqslant \frac{1}{2}\pi\). [3]
(c) Find the area of the region enclosed by one of the loops in part (b). [3]
| Scheme | Marks | AO |
|---|---|---|
| \((x^2 + y^2)^2 = 2c^2xy \Rightarrow (r^2)^2 = 2c^2r\cos\theta r\sin\theta\) | M1 | 1.1b |
| \(\Rightarrow r^2 = 2c^2\cos\theta\sin\theta = c^2\sin 2\theta\,*\) | A1 | 2.2a |
| [2] |
Notes
M1: substituting for \(r^2\), \(x\) and \(y\)
A1: NB AG
| Scheme | Marks | AO |
|---|---|---|
 | B1 B1* B1dep | 1.1b 1.1b 2.5 |
| [3] |
Notes
B1: one loop shown; allow sep diags
B1*: both shown (no extras), in correct quadrant
B1dep: \(-\)ve \(r\) with broken line; dep B1*
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle A = \int_0^{\frac{\pi}{2}} \frac{1}{2}c^2\sin 2\theta\,\mathrm{d}\theta\) | B1 | 1.1a |
| \(= \left[-\dfrac{1}{4}c^2\cos 2\theta\right]_0^{\frac{\pi}{2}}\) | B1 | 1.1b |
| \(= \dfrac{1}{2}c^2\) | B1cao | 1.1b |
| [3] |
Notes
B1: condone missing \(\mathrm{d}\theta\); limits soi
B1: \(\displaystyle\int \sin 2\theta\,\mathrm{d}\theta = -\frac{1}{2}\cos 2\theta\)