## Question 6:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2/t^2}{2} \left(= -\frac{1}{t^2}\right)$ | M1* | M1 for their $(dy/dt)$/their $(dx/dt)$ in terms of $t$ with at least one term correct |
| (unsimplified form acceptable) | A1 | A1 cao (oe) – allow unsimplified even if subsequently cancelled incorrectly |
| $y - \frac{2}{t} = f(t)(x-2t)$ with any non-zero gradient as function of $t$ | M1dep* | Must have used correct point $\left(2t, \frac{2}{t}\right)$; if using $y=mx+c$ must explicitly have $c=...$ before M1 awarded |
| $y - \frac{2}{t} = -\frac{1}{t^2}(x-2t)$ | A1 | oe – need not be simplified |
| When $x=0$, $y = \frac{4}{t}$ | A1ft | Must be a function of $t$ |
| When $y=0$, $x=4t$ | A1ft | Must be a function of $t$ |
| Area of triangle $= \frac{1}{2} \times \frac{4}{t} \times 4t = 8$ (independent of $t$) | A1 | No ft – answer of 8 with no additional comment sufficient |

**OR (for first two marks):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{2}{\left(\frac{x}{2}\right)} = \frac{4}{x} \Rightarrow \frac{dy}{dx} = -\frac{4}{x^2}$ | M1* | Attempt to eliminate $t$ and correctly differentiate Cartesian equation |
| $\Rightarrow \left(\frac{dy}{dx}\right) = -\frac{4}{(2t)^2}$ | A1 | |

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