C4 June 2005 Q2
2. A curve has equation
\[x^2 + 2xy - 3y^2 + 16 = 0.\]
Find the coordinates of the points on the curve where \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\). (7)
| Scheme | Marks |
|---|---|
| \(2x + \left(2x\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y\right) - 6y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | M1 (A1) A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow x + y = 0\) or equivalent | M1 |
| Eliminating either variable and solving for at least one value of \(x\) or \(y\). \(y^2 - 2y^2 - 3y^2 + 16 = 0\) or the same equation in \(x\) | M1 |
| \(y = \pm 2\) or \(x = \pm 2\) | A1 |
| \((2, -2), (-2, 2)\) | A1 |
| (7) | |
| (7 marks) |
Notes
Note: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x + y}{3y - x}\)
Alternative
| \(3y^2 - 2xy - (x^2 + 16) = 0\) \(y = \dfrac{2x \pm \surd(16x^2 + 192)}{6}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{3} \pm \dfrac{1}{3}\cdot\dfrac{8x}{\surd(16x^2 + 192)}\) | M1 A1± A1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{8x}{\surd(16x^2 + 192)} = \pm 1\) | M1 |
| \(64x^2 = 16x^2 + 192\) \(x = \pm 2\) | M1 A1 |
| \((2, -2), (-2, 2)\) | A1 |
| [7] |