C4 June 2017 Q1
1. The curve \(C\) has parametric equations \[x = 3t - 4, \quad y = 5 - \frac{6}{t}, \qquad t > 0\]
(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) (2)
The point \(P\) lies on \(C\) where \(t = \dfrac{1}{2}\)
(b) Find the equation of the tangent to \(C\) at the point \(P\). Give your answer in the form \(y = px + q\), where \(p\) and \(q\) are integers to be determined. (3)
(c) Show that the cartesian equation for \(C\) can be written in the form \[y = \frac{ax + b}{x + 4}, \qquad x > -4\] where \(a\) and \(b\) are integers to be determined. (3)
| Scheme | Marks |
|---|---|
| \(x = 3t - 4,\ y = 5 - \dfrac{6}{t},\ t > 0\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 3,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = 6t^{-2}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{6t^{-2}}{3}\ \left\{= \dfrac{6}{3t^2} = 2t^{-2} = \dfrac{2}{t^2}\right\}\) their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) divided by their \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) to give \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) or their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) multiplied by their \(\dfrac{\mathrm{d}t}{\mathrm{d}x}\) to give \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) \(\dfrac{6t^{-2}}{3}\), simplified or un-simplified, in terms of \(t\). See note. | M1 A1 isw |
| Award Special Case 1st M1 if both \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) are stated correctly and explicitly. | |
| (2) |
Notes
Note: You can recover the work for part (a) in part (b).
Note: Condone \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\left(\frac{6}{t^2}\right)}{3}\) for A1
Note: You can ignore subsequent working following on from a correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\).
Way 2
| Scheme | Marks |
|---|---|
| \(y = 5 - \dfrac{18}{x + 4} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{18}{(x + 4)^2} = \dfrac{18}{(3t)^2}\) Writes \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in the form \(\dfrac{\pm\lambda}{(x + 4)^2}\), and writes \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) as a function of \(t\). Correct un-simplified or simplified answer, in terms of \(t\). See note. | M1 A1 isw |
| (2) |
| Scheme | Marks |
|---|---|
| \(\left\{t = \dfrac{1}{2} \Rightarrow\right\}\ P\left(-\dfrac{5}{2}, -7\right)\) \(x = -\dfrac{5}{2},\ y = -7\) or \(P\left(-\dfrac{5}{2}, -7\right)\) seen or implied. | B1 |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2}{\left(\frac{1}{2}\right)^2}\) and either \(\bullet\ y - \text{"}{-7}\text{"} = \text{"}8\text{"}\left(x - \text{"}{-\tfrac{5}{2}}\text{"}\right)\) \(\bullet\ \text{"}{-7}\text{"} = (\text{"}8\text{"})(\text{"}{-\tfrac{5}{2}}\text{"}) + c\) So, \(y = (\text{their } m_T)x + \text{"}c\text{"}\) Some attempt to substitute \(t = 0.5\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which contains \(t\) in order to find \(m_T\) and either applies \(y - (\text{their } y_P) = (\text{their } m_T)(x - \text{their } x_P)\) or finds \(c\) from \((\text{their } y_P) = (\text{their } m_T)(\text{their } x_P) + c\) and uses their numerical \(c\) in \(y = (\text{their } m_T)x + c\) | M1 |
| T: \(y = 8x + 13\) \(y = 8x + 13\) or \(y = 13 + 8x\) | A1 cso |
| Note: their \(x_P\), their \(y_P\) and their \(m_T\) must be numerical values in order to award M1 | |
| (3) |
Notes
Note: Using a changed gradient (i.e. applying \(\dfrac{-1}{\text{their } \frac{\mathrm{d}y}{\mathrm{d}x}}\) or \(\dfrac{1}{\text{their } \frac{\mathrm{d}y}{\mathrm{d}x}}\) or \(-\left(\text{their } \frac{\mathrm{d}y}{\mathrm{d}x}\right)\)) is M0.
Note: Final A1: A correct solution is required from a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
Note: Final A1: You can ignore subsequent working following on from a correct solution.
| Scheme | Marks |
|---|---|
| Way 1 | |
| \(\left\{t = \dfrac{x + 4}{3} \Rightarrow\right\}\ y = 5 - \dfrac{6}{\left(\frac{x + 4}{3}\right)}\) An attempt to eliminate \(t\). See notes. Achieves a correct equation in \(x\) and \(y\) only | M1 A1 o.e. |
| \(\Rightarrow y = 5 - \dfrac{18}{x + 4} \Rightarrow y = \dfrac{5(x + 4) - 18}{x + 4}\) | |
| So, \(y = \dfrac{5x + 2}{x + 4},\ \{x > -4\}\) \(y = \dfrac{5x + 2}{x + 4}\) (or implied equation) | A1 cso |
| (3) | |
| Way 2 | |
| \(\left\{t = \dfrac{6}{5 - y} \Rightarrow\right\}\ x = \dfrac{18}{5 - y} - 4\) An attempt to eliminate \(t\). See notes. Achieves a correct equation in \(x\) and \(y\) only | M1 A1 o.e. |
| \(\Rightarrow (x + 4)(5 - y) = 18 \Rightarrow 5x - xy + 20 - 4y = 18\) | |
| \(\{\Rightarrow 5x + 2 = y(x + 4)\}\) So, \(y = \dfrac{5x + 2}{x + 4},\ \{x > -4\}\) \(y = \dfrac{5x + 2}{x + 4}\) (or implied equation) | A1 cso |
| (3) | |
| Way 3 | |
| \(y = \dfrac{3at - 4a + b}{3t - 4 + 4} = \dfrac{3at}{3t} - \dfrac{4a - b}{3t} = a - \dfrac{4a - b}{3t} \Rightarrow a = 5\) A full method leading to the value of \(a\) being found \(y = a - \dfrac{4a - b}{3t}\) and \(a = 5\) | M1 A1 |
| \(\dfrac{4a - b}{3} = 6 \Rightarrow b = 4(5) - 6(3) = 2\) Both \(a = 5\) and \(b = 2\) | A1 |
| (3) | |
| Note: Some or all of the work for part (c) can be recovered in part (a) or part (b) | |
| (8 marks) |
Notes
Note: 1st M1: A full attempt to eliminate \(t\) is defined as either
- rearranging one of the parametric equations to make \(t\) the subject and substituting for \(t\) in the other parametric equation (only the RHS of the equation required for M mark)
- rearranging both parametric equations to make \(t\) the subject and putting the results equal to each other.
Note: Award M1A1 for \(\dfrac{6}{5 - y} = \dfrac{x + 4}{3}\) or equivalent.