June 2022 Paper 3 Q5
5 A curve is defined implicitly by the equation \(2x^2 + 3xy + y^2 + 2 = 0\).
(a) Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{4x+3y}{3x+2y}\). [3]
(b) In this question you must show detailed reasoning.
Find the coordinates of the stationary points of the curve. [4]
Find the coordinates of the stationary points of the curve. [4]
| Scheme | Marks | AO |
|---|---|---|
| \(4x + 3x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 | 1.1a |
| A1 | 1.1 | |
| \(3x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4x - 3y\) | ||
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{4x+3y}{3x+2y}\) | E1 | 2.1 |
| [3] |
Notes
M1: Attempt at implicit differentiation.
Either \(3x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 3y\) or \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) correct. Condone \(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) for M1
A1: All correct
E1: AG
At least one step of working needed to achieve convincing completion.
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow y = -\dfrac{4x}{3}\) or \(x = -\dfrac{3y}{4}\) | M1 | 3.1a |
| \(2x^2 - 4x^2 + \dfrac{16x^2}{9} + 2 = 0\) Or \(\dfrac{9}{8}y^2 - \dfrac{9}{4}y^2 + y^2 + 2 = 0\) | M1 | 1.1 |
| \(\dfrac{2x^2}{9} = 2 \Rightarrow x = \pm 3\) Or \(-\dfrac{1}{8}y^2 = -2 \Rightarrow y = \pm 4\) | A1 | 1.1 |
| \((3, -4)\) and \((-3, 4)\) | A1 | 1.1 |
| [4] |
Notes
M1: Finding \(y\) in terms of \(x\) (or vice versa)
Condone sign errors for M1
M1: Substitution into equation of curve to get equation in \(x\) (or \(y\))
A1: Both values – some working needed
nfww (eg sign error in first line)
A1: Both points as coordinates. Dep on M2A1.
A0 if extra values