A2 June 2022 Q1
1. An ellipse has equation \(\dfrac{x^2}{16} + \dfrac{y^2}{4} = 1\) and eccentricity \(e_1\)
A hyperbola has equation \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) and eccentricity \(e_2\)
Given that \(e_1 \times e_2 = 1\)
Given also that the coordinates of the foci of the ellipse are the same as the coordinates of the foci of the hyperbola,
| Scheme | Marks | AO |
|---|---|---|
| \(b^2 = a^2\left(1 - e_1^2\right) \Rightarrow 4 = 16\left(1 - e_1^2\right) \Rightarrow e_1^2 = \ldots\) | M1 | 1.1b |
| \(e_1^2 = \dfrac{3}{4}\) or \(e_1 = \dfrac{\sqrt{3}}{2}\) | A1 | 1.1b |
| E.g. \(b^2 = a^2\left(e_2^2 - 1\right) = a^2\left(\dfrac{1}{e_1^2} - 1\right) = a^2\left(\dfrac{4}{3} - 1\right)\) | dM1 | 2.1 |
| \(\Rightarrow b^2 = \dfrac{1}{3}a^2 \Rightarrow a^2 = 3b^2\ *\) cso | A1* | 1.1b |
| (4) |
Notes
M1: Uses “\(b^2 = a^2\left(1 - e_1^2\right)\)” with values for \(a\) and \(b\) to find a value for \(e_1\) or \(e_1^2\). They may just call it \(e\) and will likely use \(a\) and \(b\) before substituted, which is fine. The formula must be correct but allow slips with \(a\) and \(b\).
A1: Correct exact value for \(e_1\) or \(e_1^2\). Note: allow M1A1 here if the relevant work is seen in (b).
dM1: Dependent on previous method mark. Uses \(e_1 \times e_2 = 1\) with their \(e_1\) or \(e_1^2\) to find an expression between \(a\) and \(b\). May find an expression for \({e_2}^{(2)}\) and apply \(e_1 \times e_2 = 1\) directly or may first substitute as per scheme. Any full method.
SC: Allow M0A0dM1A0 if \(b^2 = a^2\left(1 - e_1\right)\) and \(b^2 = a^2\left(e_2 - 1\right)\) are used in an otherwise correct process.
A1*: Achieves \(a^2 = 3b^2\) with at least one intermediate unsimplified equation in \(a\) and \(b\) cso
| Scheme | Marks | AO |
|---|---|---|
| For the focus of the ellipse \((x =)\,4 \times \text{‘}\dfrac{\sqrt{3}}{2}\text{’}\) | M1 | 1.1b |
| For focus of the hyperbola \((x =)\ a \times \text{‘}\dfrac{2}{\sqrt{3}}\text{’} \Rightarrow 2\sqrt{3} = \dfrac{2a}{\sqrt{3}} \Rightarrow a = \ldots\ (= 3)\) \(\Rightarrow b^2 = \dfrac{1}{3}a^2 = \ldots\) | M1 | 3.1a |
| \(\dfrac{x^2}{9} - \dfrac{y^2}{3} = 1\) cso | A1 | 2.2a |
| (3) | ||
| (7 marks) |
Notes
M1: Uses/implies \(x\) coordinate of focus for the ellipse is \(4 \times\) their \(e_1\)
M1: For a full process to find values for \(a\) and \(b\) or their squares. E.g. for focus of hyperbola \(x = a \times\) their \(e_2 = \dfrac{a}{e_1}\) sets equal to \(4e_1\) and solves for \(a\) then attempting to use \(a^2 = 3b^2\) to obtain \(b^2\) (or \(b\)). Other methods are possible.
A1: Deduces the correct equation for the hyperbola.