FP1 June 2017 Q7
7. The parabola \(C\) has equation \(y^2 = 4ax\), where \(a\) is a constant and \(a > 0\)
The point \(Q(aq^2, 2aq)\), \(q > 0\), lies on the parabola \(C\).
The tangent to \(C\) at the point \(Q\) meets the \(x\)-axis at the point \(X\left(-\dfrac{1}{4}a, 0\right)\) and meets the directrix of \(C\) at the point \(D\).
Given that the point \(F\) is the focus of the parabola \(C\),
| Scheme | Marks |
|---|---|
| \(y^2 = 4ax,\) at \(Q(aq^2, 2aq)\) | |
| \(y = 2\sqrt{a}\,x^{\frac{1}{2}} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{a}\,x^{-\frac{1}{2}}\) or \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4a\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2a \times \dfrac{1}{2aq}\) | M1 |
| When \(x = aq^2\), \(m_T = \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\sqrt{a}}{\sqrt{aq^2}} = \dfrac{\sqrt{a}}{\sqrt{a}\,q} = \dfrac{1}{q}\) or when \(y = 2aq\), \(m_T = \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4a}{2(2aq)} = \dfrac{1}{q}\) | A1 |
| T: \(y - 2aq = \dfrac{1}{q}\left(x - aq^2\right)\) | dM1 |
| T: \(qy - 2aq^2 = x - aq^2\) | |
| T: \(qy = x + aq^2\ *\) | A1 * |
| (4) |
Notes
M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm k\,x^{-\frac{1}{2}}\) or \(ky\dfrac{\mathrm{d}y}{\mathrm{d}x} = c\) or \(\dfrac{\text{their } \frac{\mathrm{d}y}{\mathrm{d}q}}{\text{their } \frac{\mathrm{d}x}{\mathrm{d}q}}\)
A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{q}\)
dM1: Applies \(y - 2aq = \left(\text{their } m_T\right)\left(x - aq^2\right)\) or \(y = \left(\text{their } m_T\right)x + c\) and an attempt to find \(c\) with gradient from calculus.
A1 *: cso
| Scheme | Marks |
|---|---|
| \(X\left(-\tfrac{1}{4}a, 0\right) \Rightarrow 0 = -\tfrac{1}{4}a + aq^2\) | M1 |
| \(\Rightarrow \left\{q^2 = \dfrac{1}{4} \Rightarrow q = -\dfrac{1}{2}\ (\text{reject})\right\}\ q = \dfrac{1}{2}\) | A1 |
| So, \(\dfrac{1}{2}y = -a + a\left(\dfrac{1}{2}\right)^2\) | M1 |
| giving, \(y = -\dfrac{3a}{2}\). So \(D\left(-a, -\tfrac{3}{2}a\right)\) o.e. | A1 |
| (4) |
Notes
M1: Substitutes \(x = -\tfrac{1}{4}a\) and \(y = 0\) into T
A1: \(q = \dfrac{1}{2}\) oe
M1: Substitutes their "\(q = \tfrac{1}{2}\)" and \(x = -a\) in T or finds \(y_D = \dfrac{1}{q}\left(-a + aq^2\right)\)
A1: \(D\left(-a, -\tfrac{3}{2}a\right)\) o.e.
| Scheme | Marks |
|---|---|
| Way 1 \(\{\text{focus } F(a, 0)\}\) | |
| \(\text{Area}(FXD) = \dfrac{1}{2}\left(\dfrac{5a}{4}\right)\left(\dfrac{3a}{2}\right) = \dfrac{15a^2}{16}\) | M1 A1 cso |
| (2) | |
| (10 marks) |
Notes
M1: Applies \(\dfrac{1}{2}\left(\text{their } |FX|\right)\left(\text{their } |y_D|\right)\). If their \(\left|y_D = \dfrac{1}{q}\left(-a + aq^2\right)\right|\) then require an attempt to sub for \(q\) to award M.
A1 cso: \(\dfrac{15a^2}{16}\) or \(0.9375a^2\)
Do not award M1 if area of wrong triangle found e.g. \(\dfrac{1}{2}.2a.\dfrac{3a}{2} = \dfrac{3a^2}{2}\)
(c) Way 2
| Scheme | Marks |
|---|---|
| \(\text{Area}(FXD) = \dfrac{1}{2}\begin{vmatrix} a & -\frac{1}{4}a & -a & a \\ 0 & 0 & -\frac{3}{2}a & 0 \end{vmatrix}\) | |
| \(= \dfrac{1}{2}\left|\left(0 + \dfrac{3}{8}a^2 + 0\right) - \left(0 + 0 - \dfrac{3}{2}a^2\right)\right| = \dfrac{15}{16}a^2\) | M1 A1cao |
| (2) |
M1: A correct attempt to apply the shoelace method.
A1cao: \(\dfrac{15a^2}{16}\) or \(0.9375a^2\)
(c) Way 3
| Scheme | Marks |
|---|---|
| Rectangle – triangle 1 – triangle 2 \(= 2a.\dfrac{3a}{2} - \dfrac{1}{2}.\dfrac{3a}{4}.\dfrac{3a}{2} - \dfrac{1}{2}.2a.\dfrac{3a}{2} = 3a^2 - \dfrac{9a^2}{16} - \dfrac{3a^2}{2}\) | M1 |
| \(\dfrac{15a^2}{16}\) or \(0.9375a^2\) | A1cao |
(c) Way 4
| Scheme | Marks |
|---|---|
| Attempts sine rule using appropriate choice from \(FX = \dfrac{5a}{4},\ FD = \dfrac{5a}{2},\ DX = \dfrac{3\sqrt{5}a}{4},\ \sin F = \dfrac{3}{5},\ \sin X = \dfrac{2}{\sqrt{5}}\) | M1 |
| \(\dfrac{15a^2}{16}\) or \(0.9375a^2\) | A1cao |
M1: Uses Area \(= \dfrac{1}{2}ab\sin C\)