A2 June 2023 Q4
4. The ellipse \(E\) has equation
\[\frac{x^2}{16} + \frac{y^2}{9} = 1\]The points \(P(4\cos\theta, 3\sin\theta)\) and \(Q(4\cos\theta, -3\sin\theta)\) lie on \(E\) where \(0 \lt \theta \lt \dfrac{\pi}{2}\)
The line \(l_1\) is the normal to \(E\) at the point \(P\)
The line \(l_2\) passes through the origin and the point \(Q\)
The lines \(l_1\) and \(l_2\) intersect at the point \(R\)
| Scheme | Marks | AO |
|---|---|---|
| \(b^2 = a^2\left(1 - e^2\right) \Rightarrow 9 = 16\left(1 - e^2\right) \Rightarrow e^2 = \ldots\) | M1 | 1.1b |
| \(e = \dfrac{\sqrt{7}}{4}\) | A1 | 1.1b |
| (2) |
Notes
M1: Uses the correct eccentricity formula and the given equation to find a value for \(e\) or \(e^2\).
A1: Correct exact value \(\dfrac{\sqrt{7}}{4}\), must be simplified. Must reject the negative value, so A0 if \(-\dfrac{\sqrt{7}}{4}\) also included.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1 \Rightarrow \dfrac{x}{8} + \dfrac{2yy^{\prime}}{9} = 0 \Rightarrow y^{\prime} = -\dfrac{9x}{16y} = -\dfrac{36\cos\theta}{48\sin\theta}\) or \(x = 4\cos\theta,\ y = 3\sin\theta \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3\cos\theta}{4\sin\theta}\) | M1 A1 | 2.1 1.1b |
| \(m_N = \dfrac{4\sin\theta}{3\cos\theta} \Rightarrow y - 3\sin\theta = \dfrac{4\sin\theta}{3\cos\theta}(x - 4\cos\theta)\) | M1 | 1.1b |
| \(3y\cos\theta - 9\sin\theta\cos\theta = 4x\sin\theta - 16\sin\theta\cos\theta\) \(\Rightarrow 4x\sin\theta - 3y\cos\theta = 7\sin\theta\cos\theta\ *\) | A1* | 2.1 |
| (4) |
Notes
M1: Starts the process of establishing the gradient of the normal by adopting a suitable method to differentiate such as using implicit differentiation or parametric differentiation to obtain an expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\). The substitution for \(x\) and \(y\) may happen later, but must be in terms of \(\theta\) when used in the equation for normal.
A1: Correct unsimplified \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(\theta\). May be implied (if substitution happens later).
M1: Applies the negative reciprocal to their gradient and uses this and the coordinates of \(P\) to establish an equation of the normal. If using \(y = mx + c\) they must proceed as far as finding \(c\).
A1*: Fully correct proof with sufficient working shown. Must be an intermediate step between the initial equation and the final answer. Do not penalise minor notational slips such as missing some \(\theta\)’s if the working is clear, but award A0 if they are consistently missed.
| Scheme | Marks | AO |
|---|---|---|
| \(y = -\dfrac{3\sin\theta}{4\cos\theta}x\) or \(y + 3\sin\theta = -\dfrac{3\sin\theta}{4\cos\theta}(x - 4\cos\theta)\) | B1 | 2.2a |
| \(4x\sin\theta + \dfrac{9x\sin\theta\cos\theta}{4\cos\theta} = 7\sin\theta\cos\theta \Rightarrow x = \ldots\) or \(-\dfrac{16y\sin\theta\cos\theta}{3\sin\theta} - 3y\cos\theta = 7\sin\theta\cos\theta \Rightarrow y = \ldots\) | M1 | 3.1a |
| \(x = \dfrac{28\cos\theta}{25}\) or \(y = -\dfrac{21\sin\theta}{25}\) | A1 | 1.1b |
| \(x = \dfrac{28\cos\theta}{25}\) and \(y = -\dfrac{21\sin\theta}{25}\) | A1 | 1.1b |
| (4) |
Notes
B1: Deduces the correct equation of \(OQ\). Need not be simplified - may use point \(Q\).
M1: Proceeds to solve their \(OQ\) line simultaneously with the given \(l_1\) to find either the \(x\) or the \(y\) coordinate of the intersection.
A1: One correct coordinate simplified or unsimplified.
A1: Both correct simplified coordinates. (Accept decimal equivalents.)
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{28\cos\theta}{25},\ y = -\dfrac{21\sin\theta}{25} \Rightarrow a = \dfrac{28}{25}, b = -\dfrac{21}{25}\) \(\left(\dfrac{21}{25}\right)^2 = \left(\dfrac{28}{25}\right)^2\left(1 - e^2\right) \Rightarrow \dfrac{441}{625} = \dfrac{784}{625}\left(1 - e^2\right) \Rightarrow e = \ldots\) | M1 | 3.1a |
| Of form \((a\cos\phi, b\sin\phi)\) (where \(\phi = -\theta\)) so an ellipse, and \(e = \dfrac{\sqrt{7}}{4}\) as required. | A1cso | 1.1b |
| (2) | ||
| (12 marks) |
Notes
M1: Uses their coordinates of \(R\) (of an appropriate form) correctly to determine the values of “\(a\)” and “\(b\)” for the second ellipse and uses their values with the correct eccentricity formula in an attempt to show that the eccentricities are the same.
A1cso: Must have had correct coordinates in (c), a statement or deduction (or work shown) to verify the locus of \(R\) is an ellipse (conclusion not needed, e.g. accept if the equation is given to show it is an ellipse), and proceeds to show that the value of \(e\) (or \(e^2\) - do not penalise inclusion of the negative value a second time) is the same as that obtained in part (a) or equivalent work. Note that the correct eccentricity can arise from coordinates of \(R\) where the denominator was incorrect, but these will score A0.