A2 June 2023 Paper 1 Q14
14 The curve \(C\) has polar equation
\[r = \frac{4}{5 + 3\cos\theta} \qquad (-\pi \lt \theta \leqslant \pi)\](a) Show that \(r\) takes values in the range \(\dfrac{1}{k} \leqslant r \leqslant k\), where \(k\) is an integer. [2 marks]
(b) Find the Cartesian equation of \(C\) in the form \(y^2 = \mathrm{f}(x)\) [4 marks]
(c) The ellipse \(E\) has equation\[y^2 + \frac{16x^2}{25} = 1\]
Find the transformation that maps the graph of \(E\) onto \(C\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses the range of \(\cos\theta\) to obtain max and min values of r or obtains \(k = 2\) | M1 | 2.4 |
| Completes a reasoned argument to obtain the correct inequality | R1 | 2.1 |
| (2) |
Typical solution
\(r\) is minimum when \(\cos\theta = 1\), and maximum when \(\cos\theta = -1\)
\[\text{So } \frac{1}{2} \leqslant r \leqslant 2\]| Scheme | Marks | AO |
|---|---|---|
| Uses \(x = r\cos\theta\) | M1 | 3.1a |
| Uses \(r^2 = x^2 + y^2\) or \(r = \sqrt{x^2 + y^2}\) | M1 | 3.1a |
| Rearranges to make \(ky^2\) the subject, having correctly removed any square root in their equation. Equation must be in terms of \(x\) and \(y\) only. | M1 | 1.1a |
| Obtains a correct result in required form ISW | A1 | 1.1b |
| (4) |
Typical solution
\[\begin{aligned} 5r + 3r\cos\theta &= 4 \\ 5r &= 4 - 3r\cos\theta \\ 5\sqrt{x^2 + y^2} &= 4 - 3x \\ 25(x^2 + y^2) &= (4 - 3x)^2 \\ 25y^2 &= 16 - 24x - 16x^2 \\ y^2 &= \tfrac{1}{25}(16 - 24x - 16x^2) \end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Completes the square to bring together all the \(x\) terms | M1 | 3.1a |
| Rearranges equation to obtain a form that enables them to deduce a transformation | M1 | 1.1a |
| Deduces that a horizontal translation is required. PI a vector \(\begin{bmatrix} p \\ 0 \end{bmatrix}\) | M1 | 2.2a |
| Deduces a translation by \(\begin{bmatrix} -\frac{3}{4} \\ 0 \end{bmatrix}\) as the only transformation | A1 | 2.2a |
| (4) | ||
| (10 marks) |
Typical solution
\[\begin{aligned} y^2 &= -\frac{16}{25}\left(x^2 + \frac{3}{2}x - 1\right) \\ &= -\frac{16}{25}\left\{\left(x + \frac{3}{4}\right)^2 - \frac{25}{16}\right\} \\ &= -\frac{16}{25}\left(x + \frac{3}{4}\right)^2 + 1 \end{aligned}\]\[y^2 + \frac{16}{25}\left(x + \frac{3}{4}\right)^2 = 1\]Starting from \(y^2 + \dfrac{16x^2}{25} = 1\):
The required transformation is a translation by \(\begin{bmatrix} -\frac{3}{4} \\ 0 \end{bmatrix}\)