A2 June 2024 Q6
6. The ellipse \(E\) has equation
\[\frac{x^2}{25} + \frac{y^2}{9} = 1\]The hyperbola \(H\) has equation
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]where \(a\) and \(b\) are positive constants.
Given that
- the eccentricity of \(H\) is the reciprocal of the eccentricity of \(E\)
- the coordinates of the foci of \(H\) are the same as the coordinates of the foci of \(E\)
determine
(6)
| Scheme | Marks | AO |
|---|---|---|
| \(9 = 25\left(1 - e^2\right) \Rightarrow e = \ldots\left\{\dfrac{4}{5}\right\}\) | M1 | 1.1b |
| \(ae = 5 \times \dfrac{4}{5} = \ldots\{4\}\) | dM1 | 1.1b |
| A complete method to find the value of \(a\) or \(b\) e.g. \(e = \dfrac{1}{\text{their ‘}\frac{4}{5}\text{’}} = \ldots\left\{\dfrac{5}{4}\right\}\) and uses \(a = \dfrac{\text{their ‘}4\text{’}}{\text{their ‘}\frac{5}{4}\text{’}} = \ldots\left\{\dfrac{16}{5}\right\}\) | M1 | 3.1a |
| \(a = \dfrac{16}{5}\) or \(b = \dfrac{12}{5}\) | A1 | 1.1b |
| E.g. Uses \(b^2 = \left(\text{their}\dfrac{16}{5}\right)^2\left(\left(\text{their}\dfrac{5}{4}\right)^2 - 1\right)\) leading to \(b = \ldots\) | M1 | 3.1a |
| \(a = \dfrac{16}{5}\) and \(b = \dfrac{12}{5}\) | A1 | 3.2a |
| (6) | ||
| (6 marks) |
Notes
M1: A complete method to find the eccentricity of \(E\). Uses \(b^2 = a^2\left(1 - e^2\right)\) to find a value for \(e\) or \(e^2\)
Ignore references to \(\pm\) for the Ms. Note that identifying \(a\) and \(b\) correctly is part of the method.
dM1: Dependent on the previous method mark. Finds the \(x\) coordinate of the focus of \(E\), 5 multiplied by their value of \(e\) (or implied if working with squared forms).
M1: A complete method to find the value of \(a\). Finds the reciprocal of their eccentricity of \(E\) and uses their \(x\) coordinate of the focus of \(E\) divided by their reciprocal of \(e\). Note they may use \(e^2 = 1 + \dfrac{b^2}{a^2}\) and \(x_F = \sqrt{a^2 + b^2}\) for the hyperbola to form and solve suitable equations.
So for example \(a^2 + b^2 = \text{“}4\text{”}^2\) and \(\text{“}\dfrac{5}{4}\text{”}^2 = 1 + \dfrac{b^2}{a^2}\) leading to a value for \(a\) or \(b\).
A1: Correct value of \(a\) or \(b\). Accept as a decimal.
M1: A complete method to find a value of \(b\) or \(a\) if \(b\) found first. Uses \(b^2 = a^2\left(e^2 - 1\right)\) with their values for \(e\) and \(a\) to find a value for \(b\) or vice versa, allowing slips in substitution as long as the correct formula is clear.
A1: Correct values for both \(a\) and \(b\). Accept as a decimals.