AS June 2023 Paper 1 Q11
11 A point has Cartesian coordinates \((x, y)\) and polar coordinates \((r, \theta)\) where \(r \geqslant 0\) and \(-\pi \lt \theta \leqslant \pi\)
(a) Express \(r\) in terms of \(x\) and \(y\) [1 mark]
(b) Express \(x\) in terms of \(r\) and \(\theta\) [1 mark]
(c) The curve \(C_1\) has the polar equation\[r(2 + \cos\theta) = 1 \qquad -\pi \lt \theta \leqslant \pi\]
(i) Show that the Cartesian equation of \(C_1\) can be written as\[ay^2 = (1 + bx)(1 + x)\]
where \(a\) and \(b\) are integers to be determined. [4 marks]
(ii) The curve \(C_2\) has the Cartesian equation\[ax^2 = (1 + by)(1 + y)\]
where \(a\) and \(b\) take the same values as in part (c)(i).
Describe fully a single transformation that maps the curve \(C_1\) onto the curve \(C_2\) [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes a correct expression for \(r\) | B1 | 1.2 |
| (1) |
Typical solution
\[r = \sqrt{x^2 + y^2}\]| Scheme | Marks | AO |
|---|---|---|
| Writes a correct expression for \(x\) | B1 | 1.2 |
| (1) |
Typical solution
\[x = r\cos\theta\]| Scheme | Marks | AO |
|---|---|---|
| (i) Substitutes their (a) and (b) to form an equation in \(x\) and \(y\) only. | M1 | 3.1a |
| Correctly removes the square root from their equation. Must be an equation in terms of \(x\) and \(y\) only. | M1 | 1.1a |
| Obtains a correct equation without roots. | A1 | 1.1b |
| Obtains the correct equation in the required form. | A1 | 3.2a |
| (4) | ||
| (ii) Identifies a reflection. Must also specify a line of reflection (which could be wrong for this mark). or Identifies a rotation about \((0, 0)\) | M1 | 1.1a |
| Fully describes a correct transformation. Accept \(90^\circ\) rotation about \((0, 0)\) If a rotation direction is included, it must be anticlockwise. | A1 | 1.1b |
| (2) | ||
| (8 marks) |
Typical solution
(i)
\[\begin{aligned} & r(2 + \cos\theta) = 1 \\ \Rightarrow \ & 2r + r\cos\theta = 1 \\ \Rightarrow \ & 2\sqrt{x^2 + y^2} + x = 1 \\ \Rightarrow \ & 2\sqrt{x^2 + y^2} = 1 - x \\ \Rightarrow \ & 4(x^2 + y^2) = (1 - x)^2 \\ \Rightarrow \ & 4x^2 + 4y^2 = 1 - 2x + x^2 \\ \Rightarrow \ & 4y^2 = 1 - 2x - 3x^2 \\ \Rightarrow \ & 4y^2 = (1 - 3x)(1 + x)\end{aligned}\](ii)
\(y\) is replaced with \(x\), and vice versa
Reflection in \(y = x\)