FP1 June 2018 Q7
7. The parabola \(C\) has equation \(y^2 = 4ax\), where \(a\) is a positive constant.
The point \(S\) is the focus of \(C\).
The straight line \(l\) passes through the point \(S\) and meets the directrix of \(C\) at the point \(D\).
Given that the \(y\) coordinate of \(D\) is \(\dfrac{24a}{5}\),
The point \(P(ak^2, 2ak)\), where \(k\) is a positive constant, lies on the parabola \(C\).
Given that the line segment \(SP\) is perpendicular to \(l\),
| Scheme | Marks |
|---|---|
| \(y^2 = 4ax,\ S(a, 0),\ D\left(-a, \dfrac{24a}{5}\right),\ P(ak^2, 2ak)\) | |
| \(m_l = \dfrac{\frac{24a}{5} - 0}{-a - a}\ \left\{= \dfrac{\frac{24a}{5} - 0}{-2a} = -\dfrac{12}{5}\right\}\) \(\dfrac{y - \frac{24a}{5}}{0 - \frac{24a}{5}} = \dfrac{x - -a}{a - -a}\) or \(\dfrac{y - 0}{\frac{24a}{5} - 0} = \dfrac{x - a}{-a - a}\) | M1 |
| \(l: y - 0 = -\dfrac{12}{5}(x - a) \Rightarrow 5y = -12x + 12a\) \(l: 12x + 5y = 12a\quad (*)\) | A1 * |
| (2) |
Notes
M1: Uses \(S(a, 0)\) and \(D\left(\text{their "}-a\text{"}, \dfrac{24a}{5}\right)\) to find an expression for the gradient of \(l\) or applies the formula \(\dfrac{y - y_1}{y_2 - y_1} = \dfrac{x - x_1}{x_2 - x_1}\). Can be un-simplified or simplified.
A1 *: Correct solution only leading to \(12x + 5y = 12a\). No errors seen.
ALT (a)
| Scheme | Marks |
|---|---|
| \(y = mx + c\) At \(S\), \(0 = ma + c\) At \(D\), \(\dfrac{24a}{5} = -ma + c\) \(\Rightarrow c = \dfrac{12a}{5},\ m = -\dfrac{12}{5}\) | M1 |
| \(y = -\dfrac{12}{5}x + \dfrac{12a}{5} \Rightarrow 12x + 5y = 12a\ *\) | A1* |
| (2) |
M1: Uses \(S(a, 0)\) and \(D\left(\text{their "}-a\text{"}, \dfrac{24a}{5}\right)\) to find 2 simultaneous equations and solves to achieve \(c = \ldots, m = \ldots\)
A1*: Correct solution only leading to \(12x + 5y = 12a\)
| Scheme | Marks |
|---|---|
| \(m_{SP} = \dfrac{2ak}{ak^2 - a}\ \left\{= \dfrac{2k}{k^2 - 1}\right\}\) | M1 |
| \(m_l = -\left(\dfrac{ak^2 - a}{2ak}\right)\) or \(m_{SP} = -\dfrac{1}{(-\frac{12}{5})}\ \left\{= \dfrac{5}{12}\right\}\) | M1 |
| So \(\left\{\dfrac{2k}{k^2 - 1} = \dfrac{5}{12} \Rightarrow\right\}\ 24k = 5k^2 - 5\) | A1 |
| \(\left\{5k^2 - 24k - 5 = 0 \Rightarrow\right\}\ (k - 5)(5k + 1) = 0 \Rightarrow k = \ldots\) | M1 |
| \(\{\text{As } k > 0, \text{ so } k = 5\} \Rightarrow (25a, 10a)\) | M1 A1 |
| (6) | |
| (8 marks) |
Notes
M1: Attempts to find the gradient of \(SP\)
M1: Some evidence of applying \(m_1m_2 = -1\)
A1: Correct 3TQ in terms of \(k\) in any form.
M1: Attempt to solve their 3TQ for \(k\)
M1: Uses their \(k\) to find \(P\)
A1: \((25a, 10a)\)
ALT 1 (b)
| Scheme | Marks |
|---|---|
| \(SP\): \(y - 0 = \dfrac{5}{12}(x - a)\) | M1 M1 |
| \(\left\{y^2 = 4ax \Rightarrow\right\}\ \left(\dfrac{5}{12}(x - a)\right)^2 = 4ax\) | |
| \(25(x^2 - 2ax + a^2) = 576ax\) | |
| \(25x^2 - 626ax + 25a^2 = 0\) | A1 |
| \((25x - a)(x - 25a) = 0 \Rightarrow x = \ldots\) | M1 |
| \(x = \dfrac{a}{25} \Rightarrow y = \dfrac{5}{12}\left(\dfrac{a}{25} - a\right)\ \left\{= -\dfrac{2a}{5}\right\}\) \(x = 25a \Rightarrow y = \dfrac{5}{12}(25a - a)\ \{= 10a\}\) | M1 |
| \(\{\text{As } k > 0,\} \Rightarrow (25a, 10a)\) | A1 |
| (6) |
M1: \(y - 0 = m_{SP}(x - a)\)
M1: \(m_{SP} = -\dfrac{1}{(-\frac{12}{5})}\ \left\{= \dfrac{5}{12}\right\}\)
Can sub for \(x\) and achieve \(\dfrac{12}{5}y + a\)
A1: Correct 3TQ in terms of \(a\) and \(x\) or \(5y^2 - 48ay - 20a^2 = 0\)
M1: Attempt to solve their 3TQ for \(x\)
M1: Uses their \(x\) to find \(y\)
A1: \((25a, 10a)\)
ALT 2 (b)
| Scheme | Marks |
|---|---|
| \(0 = m_{SP}a + c\) | M1 |
| \(m_{SP} = -\dfrac{1}{(-\frac{12}{5})}\ \left\{= \dfrac{5}{12}\right\}\) | M1 |
| \(y = \dfrac{5}{12}x - \dfrac{5}{12}a\) | |
| At \(P\), \(2ak = \dfrac{5}{12}ak^2 - \dfrac{5}{12}a\) | A1 |
| then as part (b) |
M1: Subs \(S\) into \(y = m_{SP}x + c\) to find \(c\)
M1: Some evidence of applying \(m_1m_2 = -1\)
A1: Correct 3TQ in terms of \(k\)