Higher November 2018 Paper 3 Q27
27 The line \(\quad y = 3x + p \quad\) and the circle \(\quad x^2 + y^2 = 53 \quad\) intersect at points \(A\) and \(B\).
\(p\) is a positive integer.
(a) Show that the \(x\)-coordinates of points \(A\) and \(B\) satisfy the equation
\(10x^2 + 6px + p^2 - 53 = 0\) [3 marks]
(b) The coordinates of \(A\) are \((2, 7)\)
Work out the coordinates of \(B\).
You must show your working. [5 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(x^2 + (3x + p)^2 = 53\) | M1 | oe |
| \(9x^2 + 3xp + 3xp + p^2\) or \(9x^2 + 6xp + p^2\) | M1 | Expands \((3x + p)^2\) correctly |
| \(x^2 + (3x + p)^2 = 53\) and \(x^2 + 9x^2 + 3xp + 3xp + p^2 = 53\) and \(10x^2 + 6px + p^2 - 53 = 0\) or \(x^2 + (3x + p)^2 = 53\) and \(x^2 + 9x^2 + 6xp + p^2 = 53\) and \(10x^2 + 6px + p^2 - 53 = 0\) | A1 |
| Answer | Mark | Comments |
|---|---|---|
| \(7 = 3 \times 2 + p\) or \(7 = 6 + p\) or \(p = 1\) | M1 | oe Substitutes \(x = 2\) into given equation \(10(2)^2 + 6p(2) + p^2 - 53 = 0\) or \(p^2 + 12p - 13 = 0\) or \((p - 1)(p + 13)\) or \(p = 1\) (and \(p = {-13}\)) |
| \(10x^2 + 6x + 1 - 53\ (= 0)\) or \(10x^2 + 6x - 52\ (= 0)\) or \(5x^2 + 3x - 26\ (= 0)\) | M1dep | oe equation Substitutes their \(p\) into given equation |
| \((5x + 13)(x - 2)\) or \(\dfrac{-3 \pm \sqrt{3^2 - 4 \times 5 \times {-26}}}{2 \times 5}\) or \(-\dfrac{3}{10} \pm \sqrt{\dfrac{529}{100}}\) | M1 | oe Correct factorisation of their 3-term quadratic or correct substitution in formula for their 3-term quadratic or correct completion of square to expression for \(x\) |
| (\(x =\)) \(-2.6\) | A1 | oe |
| \(({-2.6}, {-6.8})\) | A1 | oe |
Additional guidance
| After scoring first M1, they substitute \(p = {-13}\) \((p - 1)(p + 13)\) or \(p = 1\) (and \(p = {-13}\)) | M1 |
| \(10x^2 - 78x + 169 - 53 = 0\) or \(10x^2 - 78x + 116 = 0\) or \(5x^2 - 39x + 58 = 0\) | M1dep |
| \((5x - 29)(x - 2)\) or \(\dfrac{-{-39} \pm \sqrt{({-39})^2 - 4 \times 5 \times 58}}{2 \times 5}\) or \(\dfrac{39}{10} \pm \sqrt{\dfrac{361}{100}}\) | M1dep A0 A0 |