AS June 2024 Paper 1 Q16
16 The curve \(C\) has the polar equation
\[r = \frac{2}{\sqrt{\cos^2\theta + 4\sin^2\theta}} \qquad -\pi \lt \theta \leqslant \pi\](a) Show that the Cartesian equation of \(C\) can be written as\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
where \(a\) and \(b\) are positive integers to be determined. [4 marks]
(b) Hence sketch the graph of \(C\) on the axes below.
Indicate the value of any intercepts of the curve with the axes. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Correctly removes the square root sign. Or Correctly isolates \(r^2\) (or \(r\)) if moving from the Cartesian form to the polar form. | M1 | 3.1a |
| Correctly uses \(x = r\cos\theta\) or \(y = r\sin\theta\) | M1 | 1.1a |
| Correctly uses \(x = r\cos\theta\) and \(y = r\sin\theta\) | M1 | 1.1a |
| Completes a reasoned argument to obtain \(\dfrac{x^2}{2^2} + \dfrac{y^2}{1^2} = 1\) Accept \(\dfrac{x^2}{4} + y^2 = 1\) if \(a = 2\) and \(b = 1\) seen. | A1 | 3.2a |
| (4) |
Typical solution
\[r = \frac{2}{\sqrt{\cos^2\theta + 4\sin^2\theta}}\]\[r^2 = \frac{4}{\cos^2\theta + 4\sin^2\theta}\]\[r^2\cos^2\theta + 4r^2\sin^2\theta = 4\]\[(r\cos\theta)^2 + 4(r\sin\theta)^2 = 4\]\[x^2 + 4y^2 = 4\]\[\frac{x^2}{4} + y^2 = 1\]\[\frac{x^2}{2^2} + \frac{y^2}{1^2} = 1\]| Scheme | Marks | AO |
|---|---|---|
| Draws an ellipse centred on the origin. Accept a circle. | M1 | 1.1a |
| Identifies the correct intercepts at 2 and \(-2\) and 1 and \(-1\) FT their \(a\) and \(b\) Accept \(\pm a\) and \(\pm b\) if no values found in part (a) | A1F | 1.1b |
| (2) | ||
| (6 marks) |
Typical solution
