## Question 5:

### Way 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Either $y = \frac{c^2}{x} = c^2x^{-1} \Rightarrow \frac{dy}{dx} = -c^2x^{-2}$ or $-\frac{c^2}{x^2}$; or $xy=c^2 \Rightarrow x\frac{dy}{dx}+y=0$; or $x=cp,\ y=\frac{c}{p} \Rightarrow \frac{dy}{dx}=\frac{dy}{dp}\cdot\frac{dp}{dx} = -\left(\frac{c}{p^2}\right)\left(\frac{1}{c}\right)$; and so at $P\!\left(ct,\frac{c}{t}\right)$, $m_T = -\frac{1}{t^2}$ | M1 | 3.1a |
| $y - \frac{c}{t} = {-\frac{1}{t^2}}(x-ct)$ | M1 | 1.1b |
| or $\frac{c}{t} = {-\frac{1}{t^2}}(ct)+b \Rightarrow y = -\frac{1}{t^2}x + \frac{2c}{t}$ | A1 | 1.1b |
| $y=0 \Rightarrow x = 2ct\ \{\Rightarrow x_A = 2ct\}$, $\ x=0 \Rightarrow y = \frac{2c}{t}\ \{\Rightarrow y_B = \frac{2c}{t}\}$ | M1, A1 | 1.1b |
| $\{OB = 2OA \Rightarrow\}\ \frac{2c}{t} = 2(2ct) \Rightarrow t = \ldots$ | M1 | 2.1 |
| $\left\{t^2 = \frac{1}{2}\Rightarrow\right\}\ t = \frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$ or awrt 0.707 | A1 | 1.1b |
| $\{\text{Area }(OAB) = 32 \Rightarrow\}\ \frac{1}{2}(2ct)\!\left(\frac{2c}{t}\right) = 32 \Rightarrow c = \ldots\ \{\Rightarrow c=4\}$ | M1 | 2.1 |
| Deduces the **numerical** value $x_p$ and $y_p$ using their values of $t$ and $c$ | M1 | 2.2a |
| $P(2\sqrt{2},\ 4\sqrt{2})$ or $P$(awrt 2.83, awrt 5.66) or $x=2\sqrt{2}$ and $y=4\sqrt{2}$ | A1 | 1.1b |

### Way 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Same requirement as 1st M mark in Way 1 | M1 | 3.1a |
| e.g. $\left\{t=\frac{1}{\sqrt{2}}\Rightarrow P\!\left(\frac{c}{\sqrt{2}},\sqrt{2}\,c\right)\Rightarrow\right\}\ y-\sqrt{2}\,c = -2\!\left(x-\frac{c}{\sqrt{2}}\right)$ using $m_T=-2$ and their $P$ found by correct method | M1, A1 | 1.1b |
| $y=0\Rightarrow x=\sqrt{2}\,c\ \{\Rightarrow x_A=\sqrt{2}\,c\}$, $\ x=0\Rightarrow y=2\sqrt{2}\,c\ \{\Rightarrow y_B=2\sqrt{2}\,c\}$ | M1, A1 | 1.1b |
| $\{OB=2OA\Rightarrow\}\ m_T=-2$ and their $m_T=-\frac{1}{t^2}=-2\Rightarrow t=\ldots$ | M1 | 2.1 |
| $\left\{t^2=\frac{1}{2}\Rightarrow\right\}\ t=\frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$ or awrt 0.707 $\left\{\Rightarrow P\!\left(\frac{c}{\sqrt{2}},\sqrt{2}\,c\right)\right\}$ | A1 | 1.1b |
| $\{\text{Area }(OAB)=32\Rightarrow\}\ \frac{1}{2}\sqrt{2}\,c\!\left(2\sqrt{2}\,c\right)=32\Rightarrow c=\ldots\ \{\Rightarrow c=4\}$ | M1 | 2.1 |
| Deduces the **numerical** value $x_p$ and $y_p$ using their values of $t$ and $c$ | M1 | 2.2a |
| $P(2\sqrt{2},\ 4\sqrt{2})$ or $P$(awrt 2.83, awrt 5.66) or $x=2\sqrt{2}$ and $y=4\sqrt{2}$ | A1 | 1.1b |

### Way 3:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Same requirement as 1st M mark in Way 1 | M1 | 3.1a |
| e.g. $y-8\sqrt{2}=-2(x-0)$ or $y-0=-2(x-4\sqrt{2})$ using $m_T=-2$ and either $A(4\sqrt{2},0)$ or $B(0,8\sqrt{2})$ found by correct method | M1, A1 | 1.1b |
| $\{\text{Area }(OAB)=32,\ OB=2OA\Rightarrow\}\ \frac{1}{2}(x)(2x)=32\Rightarrow x=\ldots$ | M1 | 2.1 |
| $x=4\sqrt{2}\ \{\Rightarrow x_A=4\sqrt{2}\}$ or $y=8\sqrt{2}\ \{\Rightarrow y_B=8\sqrt{2}\}$ | A1 | 1.1b |
| $\{OB=2OA\Rightarrow\}\ m_T=-2$ and $m_T=-\frac{1}{t^2}=-2\Rightarrow t=\ldots$ | M1 | 2.1 |
| $\left\{t^2=\frac{1}{2}\Rightarrow\right\}\ t=\frac{1}{\sqrt{2}}$ or $\frac{\sqrt{2}}{2}$ or awrt 0.707 $\left\{\Rightarrow P\!\left(\frac{c}{\sqrt{2}},\sqrt{2}\,c\right)\right\}$ | A1 | 1.1b |
| $\sqrt{2}\,c - 8\sqrt{2} = -2\!\left(\frac{c}{\sqrt{2}}-0\right)\Rightarrow c=\ldots\ \{\Rightarrow c=4\}$ | M1 | 1.1b |
| Deduces the **numerical** value $x_p$ and $y_p$ using their values of $t$ and $c$ | M1 | 2.2a |
| $P(2\sqrt{2},\ 4\sqrt{2})$ or $P$(awrt 2.83, awrt 5.66) or $x=2\sqrt{2}$ and $y=4\sqrt{2}$ | A1 | 1.1b |

### Way 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete process substituting $y-8\sqrt{2}=-2(x-0)$ or $y-0=-2(x-4\sqrt{2})$ into $xy=c^2$ and applying $b^2-4ac=0$ to their resulting $2x^2-8\sqrt{2}\,x+c^2=0$ | M1 | 3.1a |
| e.g. $y-8\sqrt{2}=-2(x-0)$ or $y-0=-2(x-4\sqrt{2})$ using $m_T=-2$ and either $A(4\sqrt{2},0)$ or $B(0,8\sqrt{2})$ found by correct method | M1, A1 | 1.1b |
| $\{\text{Area }(OAB)=32,\ OB=2OA\Rightarrow\}\ \frac{1}{2}(x)(2x)=32\Rightarrow x=\ldots$ | M1 | 2.1 |
| $x=4\sqrt{2}\ \{\Rightarrow x_A=4\sqrt{2}\}$ or $y=8\sqrt{2}\ \{\Rightarrow y_B=8\sqrt{2}\}$ | A1 | 1.1b |
| dependent on 2nd M mark: $\{xy=c^2\Rightarrow\}\ x(-2x+8\sqrt{2})=c^2\ \{\Rightarrow 2x^2-8\sqrt{2}\,x+c^2=0\}$ or $\{xy=c^2\Rightarrow\}\ \frac{1}{2}(8\sqrt{2}-y)y=c^2\ \{\Rightarrow y^2-8\sqrt{2}\,y+2c^2=0\}$ | dM1 | 2.1 |
| $\{b^2-4ac=0\Rightarrow\}\ (8\sqrt{2})^2-4(2)(c^2)=0\Rightarrow c=\ldots\ \{\Rightarrow c=4\}$ | A1, M1 | 1.1b |
| Deduces the **numerical** value $x_p$ and $y_p$ using their value of $c$ | M1 | 2.2a |
| $P(2\sqrt{2},\ 4\sqrt{2})$ or $P$(awrt 2.83, awrt 5.66) or $x=2\sqrt{2}$ and $y=4\sqrt{2}$ | A1 | 1.1b |
| Note (all Ways): For final M1, allow for correct method giving $x_p=2\sqrt{2}$ or $y_p=4\sqrt{2}$ or $x_p=$ awrt 2.83 or $y_p=$ awrt 5.66 o.e. | | |

# Question 5:

## Way 1

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates $xy = c^2$ to give $\frac{dy}{dx} = \pm kx^{-2}$, $k \neq 0$; or by product rule to give $\pm x\frac{dy}{dx} \pm y$; or by parametric differentiation to give $\left(\text{their } \frac{dy}{dt}\right) \times \frac{1}{\left(\text{their } \frac{dx}{dt}\right)}$, condoning $p \equiv t$. **And** attempts to use $P\left(ct, \frac{c}{t}\right)$ to write gradient of tangent in terms of $t$ | M1 | |
| Correct straight line method for equation of tangent where $m_T \neq m_N$ found by calculus. **Note:** $m_T$ must be a function of $t$ | M1 | |
| Correct equation of tangent (simplified or un-simplified) | A1 | |
| Attempts to find either $x$-coordinate of $A$ or $y$-coordinate of $B$ | M1 | |
| **Both** $\{x\text{-coordinate of } A\text{ is}\}\ 2ct$ **and** $\{y\text{-coordinate of } B\text{ is}\}\ \frac{2c}{t}$ | A1 | |
| See scheme | M1 | |
| See scheme | A1 | |
| See scheme | M1 | |
| See scheme | M1 | |
| See scheme | A1 | |

## Way 2

| Answer/Working | Mark | Guidance |
|---|---|---|
| Same as 1st M1 in Way 1 | M1 | |
| See scheme | M1 | |
| Correct equation of tangent (simplified or un-simplified) | A1 | |
| Attempts to find either $x$-coordinate of $A$ or $y$-coordinate of $B$ | M1 | |
| **Both** $\{x\text{-coordinate of } A\text{ is}\}\ \sqrt{2}c$ **and** $\{y\text{-coordinate of } B\text{ is}\}\ 2\sqrt{2}c$ | A1 | |
| Recognising gradient of tangent is $-2$ and puts this equal to their $\frac{dy}{dx}$ and finds $t = \ldots$ | M1 | |
| See scheme | A1 | |
| See scheme | M1 | |
| See scheme | M1 | |
| See scheme | A1 | |

## Way 3

| Answer/Working | Mark | Guidance |
|---|---|---|
| Same as 1st M1 in Way 1 | M1 | |
| See scheme | M1 | |
| Correct equation of tangent (simplified or un-simplified) | A1 | |
| Uses $y = 2x$ and Area $(OAB) = 32$ to find either $x_A$ or $y_B$ | M1 | |
| Either $\{x\text{-coordinate of } A\text{ is}\}\ 4\sqrt{2}$ or $\{y\text{-coordinate of } B\text{ is}\}\ 8\sqrt{2}$ | A1 | |
| Recognising gradient of tangent is $-2$ and puts this equal to their $\frac{dy}{dx}$ and finds $t = \ldots$ | M1 | |
| See scheme | A1 | |
| Substitutes their $P$ (in terms of $c$, from correct method) into equation of tangent and finds $c = \ldots$ | M1 | |
| See scheme | M1 | |
| See scheme | A1 | |

## Way 4

| Answer/Working | Mark | Guidance |
|---|---|---|
| See scheme | M1 | |
| See scheme | M1 | |
| Correct equation of tangent (simplified or un-simplified) | A1 | |
| Uses $y = 2x$ and Area $(OAB) = 32$ to find either $x_A$ or $y_B$ | M1 | |
| Either $\{x\text{-coordinate of } A\text{ is}\}\ 4\sqrt{2}$ or $\{y\text{-coordinate of } B\text{ is}\}\ 8\sqrt{2}$ | A1 | |
| See scheme | M1 | |
| See scheme | A1 | |
| See scheme | M1 | |
| See scheme | M1 | |
| See scheme | A1 | |