Higher November 2024 Paper 1 Q20
20 C is the circle with equation \(x^2 + y^2 = 4\)
Find an equation of the tangent to C at the point \((p, 1)\) where \(p \gt 0\)
Give your answer in the form \(y + \sqrt{a}x = b\) where \(a\) and \(b\) are integers.
You must show all your working. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| \(y + \sqrt{3}x = 4\) | P1 | for process to find the value of \(p\), eg \(\sqrt{4 - 1^2}\ (= \sqrt{3})\) |
| P1 | for a start of a process to find gradient of tangent, eg gradient of normal/radius \(= \dfrac{1}{p}\) or \(\dfrac{1}{\text{``}\sqrt{3}\text{''}}\) or \(\dfrac{1}{[p]}\) or for gradient of tangent \(= -p\) or \(-\text{``}\sqrt{3}\text{''}\) or \(-[p]\) | |
| P1 | (dep P1) for substituting \((\text{``}\sqrt{3}\text{''}, 1)\) into \(y = \text{``}{-}\sqrt{3}\text{''}x + c\) or for \(y - 1 = \text{``}{-}\sqrt{3}\text{''}(x - \text{``}\sqrt{3}\text{''})\) oe or for \(1 = -p \times p + c\) or for substituting \(([p], 1)\) into \(y = -[p]x + c\) or for substituting \((\text{``}\sqrt{3}\text{''}, 1)\) into \(y = -\dfrac{1}{[m]}x + c\) | |
| A1 | for \(y + \sqrt{3}x = 4\) |
Additional guidance
May occur later in the process
Where \([p]\) is their stated value of \(p\)
Where \([m]\) is clearly their gradient of the normal/radius
A correct answer with no supportive working gets 0 marks