## Question 9:

### Part (a) — Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = (t+3)^2 - 25$ | M1 | Attempts to complete the square; award for sight of $x=(t+3)^2 \pm \ldots$ where $\ldots \neq 0$ |
| $x+25=(t+3)^2 \Rightarrow (x+25)^{\frac{1}{2}}=(t+3) \Rightarrow y = \ldots$ | M1 | Rearranges to $(t+3)=\ldots$ or $(t+3)^2=\ldots$ and substitutes correctly into parametric equation for $y$ |
| $y = 3\ln(x+25)$ | A1cso | Must include "$y=$" at some point; all stages of working shown |

### Part (a) — Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 6\ln(t+3) = 3\ln(t+3)^2$ | M1 | Attempts power rule for logarithms $y=6\ln(t+3)=\ldots\ln(t+3)^2$ where $\ldots\neq6$ |
| $y=3\ln(t+3)^2 = 3\ln(t^2+6t+9) = 3\ln(x+16+9)$ | M1 | Writes as $3\ln(t+3)^2$ then multiplies out and substitutes correctly for $t$ |
| $y=3\ln(x+25)$ | A1cso | |

### Part (a) — Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=6\ln(t+3) \Rightarrow \frac{y}{6}=\ln(t+3) \Rightarrow t+3=e^{\frac{y}{6}} \Rightarrow t=e^{\frac{y}{6}}-3$ | M1 | Attempts to make $t$ the subject |
| $x=\left(e^{\frac{y}{6}}-3\right)^2+6\left(e^{\frac{y}{6}}-3\right)-16 \Rightarrow y=\ldots$ | M1 | Substitutes $t=e^{\frac{y}{6}}\pm\ldots$ correctly into $x=t^2+6t-16$ and rearranges |
| $y=3\ln(x+25)$ | A1cso | |

### Part (a) — Way 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=(t+3)^2-25$ | M1 | Completes the square |
| $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\left(\frac{6}{t+3}\right)}{2t+6} = \frac{3}{(t+3)^2} = \frac{3}{x+25} \Rightarrow y=3\ln(x+25)(+c)$ | M1 | Attempts $\frac{dy}{dx}$ where $\frac{dy}{dx}=\frac{\left(\frac{\ldots}{t+3}\right)}{at+b}$, $a,b\neq0$; uses completed square to find $\frac{dy}{dx}$ in terms of $x$; integrates |
| $t=0 \Rightarrow x=-16,\ y=6\ln3 \Rightarrow c=0$; $y=3\ln(x+25)$ | A1cso | Complete method using correct point to show $c=0$ |

### Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=0,\ y=3\ln25$ (oe e.g. $6\ln5$) | B1ft | Follow through on their Cartesian equation |
| $\frac{dy}{dx}=\frac{3}{x+\text{"25"}} \Rightarrow \frac{dy}{dx}=\frac{3}{0+\text{"25"}}\ \left(=\frac{3}{25}\right)$ or via parametric: $\frac{\frac{6}{t+3}}{2t+6}\Rightarrow \frac{\frac{6}{2+3}}{2\times2+6}=\frac{6}{50}=\frac{3}{25}$ | M1 | Correct differentiation and substitution of $x=0$ (or $t=2$) |
| $y - \text{"3}\ln25\text{"} = \frac{3}{25}(x-0)$ | dM1 | Equation of tangent using their gradient and their point; must proceed to $c=\ldots$ if using $y=mx+c$ |
| $25y - 3x = 150\ln5$ | A1 | |

# Question (b) [Tangent to Parametric Curve]:

| Working/Answer | Mark | Guidance |
|---|---|---|
| Deduces $y = 3\ln 25$ (or e.g. $y = 6\ln 5$) | **B1ft** | Follow through on Cartesian equation with $x=0$; apply isw after correct/ft value |
| Attempts $\frac{dy}{dx}$ when $x=0$, obtaining $\frac{dy}{dx} = \frac{\cdots}{x+\text{"25"}}$ and substituting $x=0$ | **M1** | Allow if using letters $A$, $B$: $\frac{dy}{dx}=\frac{\cdots}{x+B}=\frac{\cdots}{0+B}$ |
| **Or:** Finds $\frac{dy}{dx}$ when $t=2$ via $\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \dfrac{\frac{6}{(t+3)}}{{2t+6}} \Rightarrow \frac{\frac{6}{5}}{2\times2+6} = \frac{6}{50} = \frac{3}{25}$ | **M1** | Look for $\frac{dy}{dx}=\frac{\left(\frac{\cdots}{t+3}\right)}{at+b}$, $a,b\neq 0$ |
| Attempts equation of tangent; sight of $y - \text{"3}\ln 25\text{"} = \frac{3}{25}(x\{-0\})$ | **dM1** | Dependent on previous M1; if using $y=mx+c$ must reach $c=\ldots$ |
| $25y - 3x = 150\ln 5$ or any integer multiple in form $ax+by=c\ln 5$ | **A1** | |

**(3 marks)**

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