A2 June 2025 Q8
8. The ellipse \(E\) has equation
\[\frac{x^2}{64} + \frac{y^2}{36} = 1\]The line \(l\) has equation \(y = mx + c\) where \(m\) and \(c\) are constants.
Hence, given that the line \(l\)
- is a tangent to \(E\)
- passes through the point \((10, 20)\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x^2}{64} + \dfrac{y^2}{36} = 1,\ y = mx + c\) \(\Rightarrow \dfrac{x^2}{64} + \dfrac{(mx + c)^2}{36} = 1 \Rightarrow \dfrac{x^2}{64} + \dfrac{m^2x^2 + 2cmx + c^2}{36} = 1\) | M1 | 1.1b |
| \(\Rightarrow 36x^2 + 64m^2x^2 + 128cmx + 64c^2 = 2304\) \(\Rightarrow \left(16m^2 + 9\right)x^2 + 32cmx + 16c^2 - 576 = 0\) | A1 | 2.1 |
| (2) |
Notes
M1: Attempts to solve simultaneously by substituting \(y = mx + c\) into \(\dfrac{x^2}{64} + \dfrac{y^2}{36} = 1\) and attempts to multiply out \((mx + c)^2\) to obtain an equation in \(x\), \(m\) and \(c\)
A1: Proceeds to obtain the equation in the required form with the correct value of \(k\)
| Scheme | Marks | AO |
|---|---|---|
| \(b^2 - 4ac = 0 \Rightarrow (32cm)^2 - 4\left(16m^2 + 9\right)\left(16c^2 - 576\right) = 0\) \(\left(\Rightarrow 36864m^2 - 576c^2 + 20736 = 0,\ 64m^2 - c^2 + 36 = 0\right)\) | M1 | 3.1a |
| \(x = 10, y = 20 \Rightarrow 20 = 10m + c\) \(\Rightarrow 64m^2 - (20 - 10m)^2 + 36 = 0\) or \(c^2 - 64\left(\dfrac{20 - c}{10}\right)^2 - 36 = 0\) | M1 | 3.1a |
| \(36m^2 - 400m + 364 = 0\) o.e. \(324m^2 - 3600m + 3276 = 0\) or \(\Rightarrow 36c^2 + 2560c - 29200 = 0\) o.e. | A1 | 1.1b |
| \(36m^2 - 400m + 364 = 0 \Rightarrow m = 1, \dfrac{91}{9} \Rightarrow c = \ldots\) or \(\Rightarrow 36c^2 + 2560c - 29200 = 0 \Rightarrow c = 10, -\dfrac{730}{9} \Rightarrow m = \ldots\) | dM1 | 1.1b |
| \(y = x + 10, \qquad y = \dfrac{91}{9}x - \dfrac{730}{9}\) o.e. | A1 | 1.1b |
| (5) | ||
| (7 marks) |
Notes
M1: Realises that for \(l\) to be a tangent, the discriminant must be zero and applies this to their equation correctly, may contain \(k\)
M1: Uses the other condition that \(l\) passes through \((10, 20)\) to eliminate \(c\) or \(m\) to obtain an equation in \(c\) or \(m\) only following an attempt at setting their discriminant \(= 0\)
A1: Correct 3TQ in \(c\) or \(m\)
dM1: Dependent on previous M. For a complete method to obtain at least one set of values for \(c\) and \(m\). I.e. Finds values for \(m\) or \(c\) from a 3TQ (no need to check the roots) and attempts to find the value of the other constant
A1: Both correct equations in any form