# Question 1:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = 3$, $\frac{dy}{dt} = 6t^{-2}$ | — | — |
| $\frac{dy}{dx} = \frac{6t^{-2}}{3} = 2t^{-2} = \frac{2}{t^2}$ | M1 | Their $\frac{dy}{dt}$ divided by their $\frac{dx}{dt}$ to give $\frac{dy}{dx}$ in terms of $t$, **or** their $\frac{dy}{dt}$ multiplied by their $\frac{dt}{dx}$ to give $\frac{dy}{dx}$ in terms of $t$ |
| $\frac{6t^{-2}}{3}$, simplified or un-simplified, in terms of $t$ | A1 isw | Condone $\frac{dy}{dx} = \frac{\left(\frac{6}{t^2}\right)}{3}$ for A1 |

**Special Case:** Award 1st M1 if both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are stated correctly and explicitly. **[2]**

**Way 2:** $y = 5 - \frac{18}{x+4} \Rightarrow \frac{dy}{dx} = \frac{18}{(x+4)^2} = \frac{18}{(3t)^2}$

| Answer/Working | Mark | Guidance |
|---|---|---|
| Writes $\frac{dy}{dx}$ in form $\frac{\pm\lambda}{(x+4)^2}$ **and** writes $\frac{dy}{dx}$ as function of $t$ | M1 | — |
| Correct un-simplified or simplified answer in terms of $t$ | A1 isw | — |

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## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \frac{1}{2} \Rightarrow P\left(-\frac{5}{2}, -7\right)$ | B1 | $x = -\frac{5}{2}$, $y = -7$ or $P\left(-\frac{5}{2}, -7\right)$ seen or implied |
| $\frac{dy}{dx} = \frac{2}{\left(\frac{1}{2}\right)^2} = 8$ and either: $y - (-7) = 8\left(x - \left(-\frac{5}{2}\right)\right)$ or $-7 = 8\left(-\frac{5}{2}\right) + c$ | M1 | Some attempt to substitute $t = 0.5$ into their $\frac{dy}{dx}$ which contains $t$, **and** applies $y - (\text{their } y_P) = (\text{their } m_T)(x - \text{their } x_P)$ **or** finds $c$ from $(\text{their } y_P) = (\text{their } m_T)(\text{their } x_P) + c$ |
| $T: y = 8x + 13$ | A1 cso | $y = 8x + 13$ or $y = 13 + 8x$ |

**[3]**

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## Part (c) Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \frac{x+4}{3} \Rightarrow y = 5 - \frac{6}{\left(\frac{x+4}{3}\right)}$ | M1 | An attempt to eliminate $t$ |
| Achieves correct equation in $x$ and $y$ only | A1 o.e. | — |
| $y = \frac{5(x+4)-18}{x+4} \Rightarrow y = \frac{5x+2}{x+4}$, $\{x > -4\}$ | A1 cso | $y = \frac{5x+2}{x+4}$ (or implied equation) |

**[3]**

## Part (c) Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \frac{6}{5-y} \Rightarrow x = \frac{18}{5-y} - 4$ | M1 | An attempt to eliminate $t$ |
| Achieves correct equation in $x$ and $y$ only | A1 o.e. | — |
| $(x+4)(5-y) = 18 \Rightarrow 5x - xy + 20 - 4y = 18 \Rightarrow 5x + 2 = y(x+4)$, $y = \frac{5x+2}{x+4}$, $\{x > -4\}$ | A1 cso | $y = \frac{5x+2}{x+4}$ (or implied equation) |

**[3]**

## Part (c) Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{3at - 4a + b}{3t - 4 + 4} = \frac{3at}{3t} - \frac{4a-b}{3t} = a - \frac{4a-b}{3t} \Rightarrow a = 5$ | M1 | A full method leading to the value of $a$ being found |
| $y = a - \frac{4a-b}{3t}$ and $a = 5$ | A1 | — |
| $\frac{4a-b}{3} = 6 \Rightarrow b = 4(5) - 6(3) = 2$ | A1 | Both $a = 5$ **and** $b = 2$ |

**[3] Total: 8**

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