| Date | May 2017 | Marks available | 5 | Reference code | 17M.1.hl.TZ0.8 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Determine | Question number | 8 | Adapted from | N/A |
Question
The normal at the point \({\text{T}}(a{t^2},{\text{ }}2at),{\text{ }}t \ne 0\), on the parabola \({y^2} = 4ax\) meets the parabola again at the point \({\text{S}}(a{s^2},{\text{ }}2as)\).
Show that \({t^2} + st + 2 = 0\).
Given that \({\rm{S\hat OT}}\) is a right-angle, where O is the origin, determine the possible values of \(t\).
Markscheme
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\frac{{{\text{d}}y}}{{{\text{d}}t}}}}{{\frac{{{\text{d}}x}}{{{\text{d}}t}}}}\) (M1)
\( = \frac{{2a}}{{2at}} = \frac{1}{t}\) A1
the gradient of the normal \( = - t\) (A1)
the equation of the normal at T is \(y - 2at = - t(x - a{t^2})\) A1
substituting the coordinates of S, M1
\(2as - 2at = - t(a{s^2} - a{t^2})\)
\(2a(s - t) = - at(s - t)(s + t)\) A1
\(2 = - t(s + t) = - st - {t^2}\) A1
\({t^2} + st + 2 = 0\) AG
[7 marks]
gradient of \({\text{OT}} = \frac{{2at}}{{a{t^2}}} = \frac{2}{t}\) A1
gradient of \({\text{OS}} = \frac{{2as}}{{a{s^2}}} = \frac{2}{s}\) (A1)
the condition for perpendicularity is \(\frac{2}{t} \times \frac{2}{s} = - 1\) M1
\({t^2} - 4 + 2 = 0\) A1
\(t = \pm \sqrt 2 \) A1
[5 marks]
