A2 June 2024 Paper 1 Q8
8 The ellipse \(E\) has equation
\[x^2 + \frac{y^2}{9} = 1\]The line with equation \(y = mx + 4\) is a tangent to \(E\)
Without using differentiation show that \(m = \pm\sqrt{7}\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Forms an equation by eliminating \(y\) or \(x\) | M1 | 1.1a |
| Obtains a correct three-term quadratic equation. PI correct discriminant | A1 | 1.1b |
| Forms a quadratic equation in \(m\) or \(m^2\) using \(b^2 - 4ac = 0\) | M1 | 2.2a |
| Completes fully correct working to obtain \(m = \pm\sqrt{7}\) AG | R1 | 2.1 |
| (4 marks) |
Typical solution
\[x^2 + \frac{(mx + 4)^2}{9} = 1\]\[9x^2 + m^2x^2 + 8mx + 16 = 9\]\[(9 + m^2)x^2 + 8mx + 7 = 0\]For a tangent, \(b^2 - 4ac = 0\)
\[64m^2 - 28(9 + m^2) = 0\]\[36m^2 = 252\]\[m^2 = 7\]\[m = \pm\sqrt{7}\]