C4 January 2009 Q7
7.
The curve \(C\) shown in Figure 3 has parametric equations
\[x = t^3 - 8t,\quad y = t^2\]
where \(t\) is a parameter. Given that the point \(A\) has parameter \(t = -1\),
The line \(l\) is the tangent to \(C\) at \(A\).
The line \(l\) also intersects the curve at the point \(B\).
| Scheme | Marks |
|---|---|
| At \(A\), \(x = -1 + 8 = 7\) & \(y = (-1)^2 = 1 \Rightarrow A(7, 1)\) | B1 |
| (1) |
Notes
B1: \(A(7, 1)\)
7. (a) It is acceptable for a candidate to write \(x = 7,\ y = 1,\) to gain B1.
| Scheme | Marks |
|---|---|
| \(x = t^3 - 8t,\quad y = t^2,\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 3t^2 - 8,\quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = 2t\) | |
| \(\therefore\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2t}{3t^2 - 8}\) | M1 A1 |
| At \(A\), \(\mathrm{m}(\mathbf{T}) = \underline{\dfrac{2(-1)}{3(-1)^2 - 8}} = \underline{\dfrac{-2}{3 - 8}} = \underline{\dfrac{-2}{-5}} = \underline{\dfrac{2}{5}}\) | A1 |
| \(\mathbf{T}\colon\ y - (\text{their } 1) = m_T\left(x - (\text{their } 7)\right)\) or \(1 = \tfrac{2}{5}(7) + c \Rightarrow c = 1 - \tfrac{14}{5} = -\tfrac{9}{5}\) Hence \(\mathbf{T}\colon\ y = \tfrac{2}{5}x - \tfrac{9}{5}\) | dM1 |
| gives \(\mathbf{T}\colon\ \underline{2x - 5y - 9 = 0}\) AG | A1 cso |
| (5) |
Notes
M1: Their \(\tfrac{\mathrm{d}y}{\mathrm{d}t}\) divided by their \(\tfrac{\mathrm{d}x}{\mathrm{d}t}\) A1: Correct \(\tfrac{\mathrm{d}y}{\mathrm{d}x}\)
A1: Substitutes for \(t\) to give any of the four underlined oe:
dM1: Finding an equation of a tangent with their point and their tangent gradient or finds c and uses \(y = (\text{their gradient})x + {}\)“\(c\)”.
A1 cso: \(\underline{2x - 5y - 9 = 0}\)
| Scheme | Marks |
|---|---|
| \(2(t^3 - 8t) - 5t^2 - 9 = 0\) | M1 |
| \(2t^3 - 5t^2 - 16t - 9 = 0\) | |
| \((t + 1)\left\{(2t^2 - 7t - 9) = 0\right\}\) \((t + 1)\left\{(t + 1)(2t - 9) = 0\right\}\) | dM1 |
| \(\left\{t = -1 \text{ (at } A)\right\}\ t = \tfrac{9}{2}\) at \(B\) | A1 |
| \(x = \left(\tfrac{9}{2}\right)^3 - 8\left(\tfrac{9}{2}\right) = \tfrac{729}{8} - 36 = \tfrac{441}{8} = 55.125\) or awrt 55.1 \(y = \left(\tfrac{9}{2}\right)^2 = \tfrac{81}{4} = 20.25\) or awrt 20.3 | ddM1 A1 A1 |
| Hence \(B\left(\tfrac{441}{8}, \tfrac{81}{4}\right)\) | |
| (6) | |
| (12 marks) |
Notes
M1: Substitution of both \(x = t^3 - 8t\) and \(y = t^2\) into \(\mathbf{T}\)
dM1: A realisation that \((t + 1)\) is a factor.
A1: \(t = \tfrac{9}{2}\)
ddM1: Candidate uses their value of \(t\) to find either the \(x\) or \(y\) coordinate
A1: One of either \(x\) or \(y\) correct. A1: Both \(x\) and \(y\) correct. awrt
Note: dM1 denotes a method mark which is dependent upon the award of the previous method mark. ddM1 denotes a method mark which is dependent upon the award of the previous two method marks. Oe or equivalent.
Aliter 7. (c) Way 2
| Scheme | Marks |
|---|---|
| \(x = t^3 - 8t = t(t^2 - 8) = t(y - 8)\) | |
| So, \(x^2 = t^2(y - 8)^2 = y(y - 8)^2\) | |
| \(2x - 5y - 9 = 0 \Rightarrow 2x = 5y + 9 \Rightarrow 4x^2 = (5y + 9)^2\) | |
| Hence, \(4y(y - 8)^2 = (5y + 9)^2\) | M1 |
| \(4y(y^2 - 16y + 64) = 25y^2 + 90y + 81\) \(4y^3 - 64y^2 + 256y = 25y^2 + 90y + 81\) \(4y^3 - 89y^2 + 166y - 81 = 0\) | |
| \((y - 1)(y - 1)(4y - 81) = 0\) | dM1 A1 |
| \(y = \tfrac{81}{4} = 20.25\) (or awrt 20.3) | |
| \(x^2 = \tfrac{81}{4}\left(\tfrac{81}{4} - 8\right)^2\) | ddM1 A1 |
| \(x = \tfrac{441}{8} = 55.125\) (or awrt 55.1) Hence \(B\left(\tfrac{441}{8}, \tfrac{81}{4}\right)\) | A1 |
| (6) |
M1: Forming an equation in terms of \(y\) only.
dM1: A realisation that \((y - 1)\) is a factor. A1: Correct factorisation
Correct y-coordinate (see below!)
ddM1: Candidate uses their \(y\)-coordinate to find their \(x\)-coordinate. Decide to award A1 here for correct y-coordinate. A1: Correct \(x\)-coordinate
Aliter 7. (c) Way 3
| Scheme | Marks |
|---|---|
| \(t = \sqrt{y}\) | |
| So \(x = \left(\sqrt{y}\right)^3 - 8\left(\sqrt{y}\right)\) | |
| \(2x - 5y - 9 = 0\) yields | |
| \(2\left(\sqrt{y}\right)^3 - 16\left(\sqrt{y}\right) - 5y - 9 = 0\) | M1 |
| \(\Rightarrow 2\left(\sqrt{y}\right)^3 - 5y - 16\left(\sqrt{y}\right) - 9 = 0\) | |
| \(\left(\sqrt{y} + 1\right)\left\{\left(2y - 7\sqrt{y} - 9\right) = 0\right\}\) | dM1 |
| \(\left(\sqrt{y} + 1\right)\left\{\left(\sqrt{y} + 1\right)\left(2\sqrt{y} - 9\right) = 0\right\}\) | A1 |
| \(y = \tfrac{81}{4} = 20.25\) (or awrt 20.3) | |
| \(x = \left(\sqrt{\tfrac{81}{4}}\right)^3 - 8\left(\sqrt{\tfrac{81}{4}}\right)\) | ddM1 A1 |
| \(x = \tfrac{441}{8} = 55.125\) (or awrt 55.1) Hence \(B\left(\tfrac{441}{8}, \tfrac{81}{4}\right)\) | A1 |
| (6) |
M1: Forming an equation in terms of \(y\) only.
dM1: A realisation that \(\left(\sqrt{y} + 1\right)\) is a factor. A1: Correct factorisation.
Correct y-coordinate (see below!)
ddM1: Candidate uses their \(y\)-coordinate to find their \(x\)-coordinate. Decide to award A1 here for correct y-coordinate. A1: Correct \(x\)-coordinate