FP3 June 2013 (R) Q1
1. The hyperbola \(H\) has foci at \((5, 0)\) and \((-5, 0)\) and directrices with equations \[x = \frac{9}{5} \text{ and } x = -\frac{9}{5}.\]
Find a cartesian equation for \(H\). (7)
| Scheme | Marks |
|---|---|
| Foci \((\pm 5, 0)\), Directrices \(x = \pm\dfrac{9}{5}\) | |
| \((\pm)ae = (\pm)5\) and \((\pm)\dfrac{a}{e} = (\pm)\dfrac{9}{5}\) Correct equations (ignore \(\pm\)’s) | B1 |
| so \(e = \dfrac{5}{a} \Rightarrow \dfrac{a^2}{5} = \dfrac{9}{5} \Rightarrow a^2 = 9\) or \(a = \dfrac{5}{e} \Rightarrow \dfrac{5}{e^2} = \dfrac{9}{5} \Rightarrow e = \dfrac{5}{3} \Rightarrow a = 3\) M1: Solves using an appropriate method to find \(a^2\) or \(a\) A1: \(a^2 = 9\) or \(a = (\pm)3\) | M1A1 |
| \(b^2 = a^2e^2 - a^2 \Rightarrow b^2 = 25 - 9\) so \(b^2 = 16 \quad (\Rightarrow b = 4)\) or \(b^2 = a^2(e^2 - 1) \Rightarrow b^2 = 9\left(\dfrac{25}{9} - 1\right)\) \(b^2 = 16 \quad (\Rightarrow b = 4)\) M1: Use of \(b^2 = a^2(e^2 - 1)\) to obtain a numerical value for \(b^2\) or \(b\) A1: \(b^2 = 16\) or \(b = (\pm)4\) | M1 A1 |
| So \(\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1\) M1: Use of \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) with their \(a^2\) and \(b^2\) A1: Correct hyperbola in any form. | M1 A1 |
| (7) | |
| (7 marks) |