FP3 June 2015 Q8
8. The ellipse \(E\) has equation \(x^2 + 4y^2 = 4\)
A chord of an ellipse is a line segment joining two points on the ellipse.
The set of midpoints of the parallel chords of \(E\) with gradient \(m\), where \(m\) is a constant, lie on a straight line \(l\).
| Scheme | Marks |
|---|---|
| \(b^2 = a^2(1 - e^2) \Rightarrow e^2 = \dfrac{3}{4}\) or \(e = \dfrac{\sqrt{3}}{2}\) NB \(a = 2\), \(b = 1\) | M1A1 |
| Foci: \((\pm ae, 0) \Rightarrow (\pm\sqrt{3}, 0)\) | B1 |
| Directrices: \(x = \pm\frac{a}{e} \Rightarrow x = \pm\frac{4}{\sqrt{3}}\) | B1 |
| (4) |
Notes
M1: Uses a correct eccentricity formula to find a value for e or e2
A1: \(e^2 = \frac{3}{4}\) or \(e = \frac{\sqrt{3}}{2}\) (allow \(e = \pm\frac{\sqrt{3}}{2}\))
B1: Both correct as coordinates
B1: Both directrices correct seen as equations. Accept un-simplified e.g. \(x = \pm\dfrac{2}{\sqrt{3}/2}\)
| Scheme | Marks |
|---|---|
| \(PF_1 = ePN_1\) and \(PF_2 = ePN_2\) | M1 |
| \(PF_1 + PF_2 = e(PN_1 + PN_2) = eN_1N_2\) | dM1A1 |
| \(= 4\) ** | A1** |
| (4) |
Notes
M1: Use of definition of ellipse for either \(PF_1\) or \(PF_2\)
dM1: \((\text{their } e) \times 2\left(\text{their } \frac{4}{\sqrt{3}}\right)\) Dependent on the previous method mark
A1: \(\frac{\sqrt{3}}{2} \times \left(2 \times \frac{4}{\sqrt{3}}\right)\)
A1**: cso
(b) Alternative 1: Using \(P(2\cos\theta, \sin\theta)\) (Must be of this form)
| Scheme | Marks |
|---|---|
| \(PF_1 = \sqrt{(2\cos\theta - \sqrt{3})^2 + \sin^2\theta}\) \(PF_2 = \sqrt{(2\cos\theta + \sqrt{3})^2 + \sin^2\theta}\) | M1 |
| \(PF_1 = \sqrt{(2 - \sqrt{3}\cos\theta)^2}\) and \(PF_2 = \sqrt{(2 + \sqrt{3}\cos\theta)^2}\) | dM1 |
| \(\left|PF_1\right| + \left|PF_2\right| = 2 - \sqrt{3}\cos\theta + 2 + \sqrt{3}\cos\theta\) | A1 |
| \(= 4\) ** | A1** |
M1: Correct use of Pythagoras for either \(PF_1\) or \(PF_2\)
dM1: Obtains both \({PF_1}^2 = \left(\sqrt{3}\cos\theta - \sqrt{p^2 + 1}\right)^2\) and \({PF_2}^2 = \left(\sqrt{3}\cos\theta + \sqrt{p^2 + 1}\right)^2\) where \(p\) is the \(x\)-coordinate of a focus. Dependent on the previous method mark
A1: \(2 - \sqrt{3}\cos\theta + 2 + \sqrt{3}\cos\theta\). Note that if \(\sqrt{3}\cos\theta - 2\) is obtained correctly, it must become \(2 - \sqrt{3}\cos\theta\) to score any A marks
A1**: cso
(b) Alternative 2: Using \(P\left(x, \sqrt{\dfrac{4 - x^2}{4}}\right)\) (Must be of this form) or \(P\left(\sqrt{4 - 4y^2}, y\right)\)
| Scheme | Marks |
|---|---|
| \(PF_1 = \sqrt{(x - \sqrt{3})^2 + \dfrac{4 - x^2}{4}}\quad PF_2 = \sqrt{(x + \sqrt{3})^2 + \dfrac{4 - x^2}{4}}\) | M1 |
| \(PF_1 = \sqrt{\left(2 + \dfrac{\sqrt{3}}{2}x\right)^2}\) and \(PF_2 = \sqrt{\left(2 - \dfrac{\sqrt{3}}{2}x\right)^2}\) | dM1 |
| \(\left|PF_1\right| + \left|PF_2\right| = 2 - \dfrac{\sqrt{3}}{2}x + 2 + \dfrac{\sqrt{3}}{2}x\) | A1 |
| \(= 4\) ** | A1** |
M1: Correct use of Pythagoras for either \(PF_1\) or \(PF_2\)
dM1: Obtains both \({PF_1}^2 = \left(\dfrac{\sqrt{3}}{2}x - \sqrt{p^2 + 1}\right)^2\) and \({PF_2}^2 = \left(\dfrac{\sqrt{3}}{2}x + \sqrt{p^2 + 1}\right)^2\) where \(p\) is the \(x\)-coordinate of the foci. Dependent on the previous method mark
A1: \(2 - \dfrac{\sqrt{3}}{2}x + 2 + \dfrac{\sqrt{3}}{2}x\)
A1**: cso
| Scheme | Marks |
|---|---|
| Using chord as \(y = mx + c\) | |
| \(\dfrac{x^2}{4} + (mx + c)^2 = 1\) | M1 |
| \((1 + 4m^2)x^2 + 8mcx + 4(c^2 - 1) = 0\) | A1 |
| \(x = \dfrac{1}{2}(\text{sum of roots}) = \dfrac{-4mc}{1 + 4m^2}\) | M1 |
| \(\Rightarrow c = -\dfrac{(1 + 4m^2)x}{4m}\) | A1 |
| So \(y = mx - \dfrac{(1 + 4m^2)x}{4m}\left(= -\dfrac{1}{4m}x\right)\) | ddM1A1 |
| (6) | |
| (14 marks) |
Notes
M1: Substitutes the equation of a straight line with gradient \(m\) into the equation of the ellipse
A1: Correct quadratic in \(x\) with terms collected
M1: Attempts \(\dfrac{1}{2}\)(sum of roots)
A1: Correct expression for \(c\) in terms of \(m\) and \(x\)
ddM1: Substitutes back into \(y = mx + c\) Depends on both previous method marks
A1: Correct equation
Or for last 3 marks
| Scheme | Marks |
|---|---|
| \(x = \dfrac{-4mc}{1 + 4m^2} \Rightarrow y = \dfrac{-4m^2c}{1 + 4m^2} + c\left(= \dfrac{c}{1 + 4m^2}\right)\) | A1 |
| \(y = -\dfrac{1}{4m}x\) | ddM1A1 |
A1: Correct \(y\)-coordinate in terms of \(m\) and \(c\).
ddM1: Obtains \(y\) in terms of \(x\) and \(m\) Depends on both previous method marks
A1: Correct equation
Alternative: Using factor formulae
Let ends of the chord be \((2\cos\alpha, \sin\alpha)\) and \((2\cos\beta, \sin\beta)\) (Must be of this form)
| Scheme | Marks |
|---|---|
| \(\left(\cos\alpha + \cos\beta, \dfrac{\sin\alpha + \sin\beta}{2}\right) = \left(2\cos\left(\dfrac{\alpha + \beta}{2}\right)\cos\left(\dfrac{\alpha - \beta}{2}\right),\ \sin\left(\dfrac{\alpha + \beta}{2}\right)\cos\left(\dfrac{\alpha - \beta}{2}\right)\right)\) M1: Attempt mid-point and uses factor formulae A1: Correct mid-point | M1A1 |
| \(m = \dfrac{\sin\beta - \sin\alpha}{2\cos\beta - 2\cos\alpha} = \dfrac{2\cos\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right)}{-4\sin\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right)}\left(= -\dfrac{1}{2}\cot\left(\dfrac{\alpha + \beta}{2}\right)\right)\) M1: Attempt gradient and uses factor formulae A1: Correct gradient | M1A1 |
| \(y = \dfrac{\sin\left(\frac{\alpha + \beta}{2}\right)}{2\cos\left(\frac{\alpha + \beta}{2}\right)}x\) and \(m = \dfrac{\cos\left(\frac{\alpha + \beta}{2}\right)}{-2\sin\left(\frac{\alpha + \beta}{2}\right)} \Rightarrow y = -\dfrac{1}{4m}x\) ddM1: Uses the mid-point and gradient to establish an equation connecting \(y\), \(m\) and \(x\) Dependent on both previous method marks A1: Correct equation | ddM1A1 |
Special Case
| Scheme | Marks |
|---|---|
| \(x^2 + 4y^2 = 4 \Rightarrow 2x + 8y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{x}{4y}\), so \(m = -\dfrac{x}{4y}\left(y = -\dfrac{1}{4m}x\right)\) Attempts like these that include further explanation should be sent to review. | M1A1 First 2 marks on ePEN |