C4 June 2008 Q4
4. A curve has equation \(3x^2 - y^2 + xy = 4\). The points \(P\) and \(Q\) lie on the curve. The gradient of the tangent to the curve is \(\tfrac{8}{3}\) at \(P\) and at \(Q\).
| Scheme | Marks |
|---|---|
| \(3x^2 - y^2 + xy = 4\) ( eqn \(*\) ) | |
| \(\left\{\xcancel{\tfrac{\mathrm{d}y}{\mathrm{d}x}\times}\right\}\quad \underline{6x - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}} + \underline{\underline{\left(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}} = \underline{0}\) | M1 B1 A1 |
| \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-6x - y}{x - 2y}\right\}\) or \(\left\{\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{6x + y}{2y - x}\right\}\) not necessarily required. | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{3} \Rightarrow \dfrac{-6x - y}{x - 2y} = \dfrac{8}{3}\) | M1* |
| giving \(-18x - 3y = 8x - 16y\) | |
| giving \(13y = 26x\) | dM1* |
| Hence, \(y = 2x \Rightarrow \underline{y - 2x = 0}\) | A1 cso |
| (6) |
Notes
M1: Differentiates implicitly to include either \(\pm ky\tfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(x\tfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\tfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\))
B1: Correct application \(\underline{\underline{(\ \ )}}\) of product rule
A1: \((3x^2 - y^2) \to \underline{\left(6x - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)}\) and \((4 \to \underline{0})\)
M1*: Substituting \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{8}{3}\) into their equation.
dM1*: Attempt to combine either terms in \(x\) or terms in \(y\) together to give either \(ax\) or \(by\).
A1 cso: simplifying to give \(\underline{y - 2x = 0}\) AG
| Scheme | Marks |
|---|---|
| At \(P\) & \(Q\), \(y = 2x\). Substituting into eqn \(*\) | |
| gives \(3x^2 - (2x)^2 + x(2x) = 4\) | M1 |
| Simplifying gives, \(x^2 = 4 \Rightarrow \underline{x = \pm 2}\) | A1 |
| \(y = 2x \Rightarrow y = \pm 4\) | |
| Hence coordinates are \(\underline{(2, 4)}\) and \(\underline{(-2, -4)}\) | A1 |
| (3) | |
| (9 marks) |
Notes
M1: Attempt replacing \(y\) by \(2x\) in at least one of the \(y\) terms in eqn \(*\)
A1: Either \(x = 2\) or \(x = -2\)
A1: Both \(\underline{(2, 4)}\) and \(\underline{(-2, -4)}\)