FP3 June 2014 Q6
6. [In this question you may use the appropriate trigonometric identities of the pink Mathematical Formulae and Statistical Tables.]
The points \(P(3\cos\alpha, 2\sin\alpha)\) and \(Q(3\cos\beta, 2\sin\beta)\), where \(\alpha \neq \beta\), lie on the ellipse with equation \[\frac{x^2}{9} + \frac{y^2}{4} = 1\]
(a) Show the equation of the chord \(PQ\) is \[\frac{x}{3}\cos\frac{(\alpha + \beta)}{2} + \frac{y}{2}\sin\frac{(\alpha + \beta)}{2} = \cos\frac{(\alpha - \beta)}{2}\] (4)
(b) Write down the coordinates of the mid-point of \(PQ\). (1)
Given that the gradient, \(m\), of the chord \(PQ\) is a constant,
(c) show that the centre of the chord lies on a line \[y = -kx\] expressing \(k\) in terms of \(m\). (5)
| Scheme | Marks |
|---|---|
| In this question condone the use of \(a\) and/or \(b\) for \(\alpha\) and \(\beta\) | |
| Gradient \(m = \dfrac{2\sin\beta - 2\sin\alpha}{3\cos\beta - 3\cos\alpha}\) Correct attempt at chord gradient – do not allow slips unless a correct method is clear | M1 |
| \(y - 2\sin\alpha = m(x - 3\cos\alpha)\) or \(y - 2\sin\beta = m(x - 3\cos\beta)\) or \(y = mx + c\) and attempts to find \(c\) using \(P\) or \(Q\) A correct straight line method using their chord gradient and the point \(P\) or the point \(Q\) | M1 |
| \(y - 2\sin\alpha = \dfrac{2\sin\beta - 2\sin\alpha}{3\cos\beta - 3\cos\alpha}(x - 3\cos\alpha)\) \(y - 2\sin\beta = \dfrac{2\sin\beta - 2\sin\alpha}{3\cos\beta - 3\cos\alpha}(x - 3\cos\beta)\) \(y - 2\sin\alpha = \dfrac{4\cos\frac{\alpha + \beta}{2}\sin\frac{\beta - \alpha}{2}}{-6\sin\frac{\alpha + \beta}{2}\sin\frac{\beta - \alpha}{2}}(x - 3\cos\alpha)\) \(y = \dfrac{2\sin\beta - 2\sin\alpha}{3\cos\beta - 3\cos\alpha}x + 2\sin\alpha - \dfrac{3\cos\alpha(2\sin\beta - 2\sin\alpha)}{3\cos\beta - 3\cos\alpha}\) \(y = -\dfrac{2\cos\frac{\alpha + \beta}{2}}{3\sin\frac{\alpha + \beta}{2}}x + 2\sin\alpha + \dfrac{2\cos\alpha\cos\frac{\alpha + \beta}{2}}{\sin\frac{\alpha + \beta}{2}}\) A correct equation for the chord in any form. | A1 |
| \(3y\sin\frac{1}{2}(\alpha + \beta) + 2x\cos\frac{1}{2}(\alpha + \beta) = 6(\cos\alpha\cos\frac{1}{2}(\alpha + \beta) + \sin\alpha\sin\frac{1}{2}(\alpha + \beta))\) or \(3y\sin\frac{1}{2}(\alpha + \beta) + 2x\cos\frac{1}{2}(\alpha + \beta) = 6(\cos\beta\cos\frac{1}{2}(\alpha + \beta) + \sin\beta\sin\frac{1}{2}(\alpha + \beta))\) | |
| \(\dfrac{x}{3}\cos\frac{1}{2}(\alpha + \beta) + \dfrac{y}{2}\sin\frac{1}{2}(\alpha + \beta) = \cos\frac{1}{2}(\alpha - \beta)\) **ag** | A1cso |
| (4) |
Notes
This is cso – there must no errors in applying the factor formulae and sufficient working must be shown to justify the printed answer but allow \(\cos\beta\cos\frac{1}{2}(\alpha + \beta) + \sin\beta\sin\frac{1}{2}(\alpha + \beta) = \cos\dfrac{\alpha - \beta}{2}\)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{3\cos\alpha + 3\cos\beta}{2}, \dfrac{2\sin\alpha + 2\sin\beta}{2}\right)\) or \((3\cos\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha), 2\sin\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha))\) or \((3\cos\frac{1}{2}(\alpha + \beta)\cos\frac{1}{2}(\alpha - \beta), 2\sin\frac{1}{2}(\alpha + \beta)\cos\frac{1}{2}(\alpha - \beta))\) Correct coordinates of mid-point in any form Coordinates must be in this order but condone outer brackets missing | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| Centre of chord is \((3\cos\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha), 2\sin\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha))\) Attempt factor formulae on both coordinates of mid-point at any stage in (c) May be implied by their \(\pm\dfrac{y}{x}\) below | M1 |
| \(\pm\dfrac{y}{x} = \pm\dfrac{2\sin\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha)}{3\cos\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha)}\left(= \pm\dfrac{2\sin\frac{1}{2}(\beta + \alpha)}{3\cos\frac{1}{2}(\beta + \alpha)}\right)\) Or \(\pm\dfrac{x}{y} = \pm\dfrac{3\cos\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha)}{2\sin\frac{1}{2}(\beta + \alpha)\cos\frac{1}{2}(\beta - \alpha)}\left(= \pm\dfrac{3\cos\frac{1}{2}(\beta + \alpha)}{2\sin\frac{1}{2}(\beta + \alpha)}\right)\) M1: Obtains an expression for \(k\) or \(-k\) or \(\dfrac{1}{k}\) or \(-\dfrac{1}{k}\) Dependent on the previous M1 (factor formulae must have been used) A1: Correct expression in any form | dM1A1 |
| \(m = -\dfrac{2\cos\frac{1}{2}(\beta + \alpha)}{3\sin\frac{1}{2}(\beta + \alpha)}\) Must be seen or used in (c) | B1 |
| \(\dfrac{\sin\frac{1}{2}(\beta + \alpha)}{\cos\frac{1}{2}(\beta + \alpha)} = -\dfrac{2}{3m}\) So \(\dfrac{y}{x} = \dfrac{2}{3}\left(-\dfrac{2}{3m}\right) \Rightarrow k = \dfrac{4}{9m}\) | A1cso |
| (5) | |
| (10 marks) |