AS June 2022 Q4
4. The parabola \(C\) has equation \(y^2 = 10x\)
The point \(F\) is the focus of \(C\).
The point \(P\) on \(C\) has \(y\) coordinate \(q\), where \(q \gt 0\)
The tangent to \(C\) at \(P\) intersects the directrix of \(C\) at the point \(A\).
The point \(B\) lies on the directrix such that \(PB\) is parallel to the \(x\)-axis.
| Scheme | Marks | AO |
|---|---|---|
| \(\left(\dfrac{5}{2}, 0\right)\) o.e. | B1 | 2.2a |
| (1) |
Notes
B1: Deduces correct coordinates.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{5}{q}\) | B1 | 1.1b |
| At \(P\), \(x = \dfrac{q^2}{10}\) so tangent has equation \(y - q = \text{their } \dfrac{5}{q}\left(x - \dfrac{q^2}{10}\right)\) or \(q = \left(\text{their } \dfrac{5}{q}\right)\left(\dfrac{q^2}{10}\right) + c \Rightarrow c = \ldots\) to reach an equation for \(y\) | M1 | 1.1b |
| \(\Rightarrow qy - q^2 = 5x - \dfrac{q^2}{2} \Rightarrow 10x - 2qy + q^2 = 0\ *\) cso or \(\Rightarrow y = \dfrac{5}{q}x + \dfrac{q}{2} \Rightarrow 10x - 2qy + q^2 = 0\ *\) cso | A1* | 2.1 |
| (3) |
Notes
B1: Using or deriving \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{5}{q}\)
M1: Finds the equation of the tangent using the equation of a line formula with \(y_1 = q\), \(x = \dfrac{q^2}{10}\) (or clear attempt at it) and \(m = \dfrac{2 \times \text{their } \text{'}a\text{'}}{q}\).
If uses \(y = mx + c\) must find a value for \(c\) and substitute back to find an equation for the tangent
A1*: Completes correctly to the given equation, no errors seen.
| Scheme | Marks | AO |
|---|---|---|
| \(B\) is \(\left(-\dfrac{5}{2}, q\right)\) o.e. | B1 | 2.2a |
| So diagonal \(BF\) has equation \(\dfrac{y - 0}{q - 0} = \dfrac{x - \frac{5}{2}}{-\frac{5}{2} - \frac{5}{2}}\) or \(y = -\dfrac{q}{5}\left(x - \dfrac{5}{2}\right)\) | M1 | 1.1b |
| (\(AP\) is a tangent so) diagonals meet when \(10x - 2q\left(-\dfrac{q}{5}\left(x - \dfrac{5}{2}\right)\right) + q^2 = 0\) or \(x = \dfrac{2qy - q^2}{10}\) therefore \(y = -\dfrac{q}{5}\left(\dfrac{2qy - q^2}{10} - \dfrac{5}{2}\right)\) leading to \(y = \ldots\) \(\left\{y = \dfrac{25q + q^3}{50 + 2q^2}\right\}\) | dM1 | 3.1a |
| \(\Rightarrow 10x + \dfrac{2q^2}{5}x - q^2 + q^2 = 0 \Rightarrow x\left(10 + \dfrac{2q^2}{5}\right) = 0\) or \(x = \dfrac{1}{10}\left(2q\left(\dfrac{25q + q^3}{50 + 2q^2}\right) - q^2\right)\) | M1 | 1.1b |
| But \(10 + \dfrac{2q^2}{5} \gt 0\) so not zero, hence \(x = 0\), so the intersection lies on the \(y\)-axis. Or achieves \(x = 0\) (with no errors), so the intersection lies on the \(y\) axis. | A1 | 2.4 |
| (5) | ||
| (9 marks) |
Notes
B1: \(B\) is \(\left(-\dfrac{5}{2}, q\right)\) seen or used.
M1: A correct method to find the equation of the diagonal \(BF\) using their coordinates of \(F\) and \(B\)
dM1: Uses the printed answer in (b) and their equation of the diagonal \(BF\) to form an equation just involving \(x\) or solves the two diagonals simultaneously to find an expression for \(y\)
M1: Correctly factors out the \(x\) to achieve \(x(\ldots) = 0\) or uses their expression for \(y\) to find an expression for \(x\)
A1: Conclusion given including reference to \(10 + \dfrac{2q^2}{5} \neq 0\)
Alternative for the last three marks
| Scheme | Marks | AO |
|---|---|---|
| When \(x = 0\) for \(BF\) \(y = -\dfrac{q}{5}\left(-\dfrac{5}{2}\right) = \ldots\) or for \(AP\) \(2qy = q^2 \Rightarrow y = \ldots\) | M1 | 1.1b |
| For \(BF\) \(y\) intercept is \(\dfrac{q}{2}\) and for \(AP\) \(y\) intercept is \(\dfrac{q}{2}\) | M1 | 3.1a |
| Since both diagonals always cross the \(y\)-axis at the same place, their intersection must always be on the \(y\) axis. | A1 | 2.4 |
M1: Attempts to find the \(y\) intercept for at least one of the two diagonals.
M1: Finds \(y\) intercept for both diagonals in order to compare
A1: Both intercepts correct and suitable conclusion giving reference to both diagonals always crossing \(y\)-axis at same point.