## Question 5:

$H: xy=c^2,\ c\neq 0;\ P\!\left(cp,\dfrac{c}{p}\right),\ p\neq 0$, lies on $H$

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $y=\dfrac{c^2}{x}=c^2x^{-1}\Rightarrow\dfrac{dy}{dx}=-c^2x^{-2}$ **or** $-\dfrac{c^2}{x^2}$; **or** $xy=c^2\Rightarrow x\dfrac{dy}{dx}+y=0$; **or** $x=ct,\ y=\dfrac{c}{t}\Rightarrow\dfrac{dy}{dx}=-\left(\dfrac{c}{t^2}\right)\!\left(\dfrac{1}{c}\right)$; at $P$: $m_T=-\dfrac{1}{p^2}$ | M1 | 2.1 |
| So $m_N=p^2$ | A1 | 2.2a |
| $y-\dfrac{c}{p}={}^{\prime\prime}p^2{}^{\prime\prime}(x-cp)$ **or** $\dfrac{c}{p}={}^{\prime\prime}p^2{}^{\prime\prime}(cp)+b\Rightarrow y={}^{\prime\prime}p^2{}^{\prime\prime}x+\text{their }b$ | M1 | 1.1b |
| Correct algebra leading to $p^3x-py+c(1-p^4)=0$ | A1* | 2.1 |
| **Total: 4 marks** | | |

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $y=\dfrac{c^2}{x}\Rightarrow p^3x-p\dfrac{c^2}{x}+c(1-p^4)=0$ **or** $x=\dfrac{c^2}{y}\Rightarrow p^3\dfrac{c^2}{y}-py+c(1-p^4)=0$ | M1 | 3.1a |
| $p^3x^2+c(1-p^4)x-c^2p=0$ **or** $py^2-c(1-p^4)y-c^2p^3=0$ | A1 | 1.1b |
| $(x-cp)(p^3x+c)=0\Rightarrow x=\ldots$ **or** $\left(y-\dfrac{c}{p}\right)(yp+cp^4)=0\Rightarrow y=\ldots$ | M1 | 3.1a |
| $x=-\dfrac{c}{p^3}$ **and** $y=-cp^3$; so $Q\!\left(-\dfrac{c}{p^3},-cp^3\right)$ | A1 | 1.1b |
| Midpoint $=\left(\dfrac{1}{2}\!\left(cp-\dfrac{c}{p^3}\right),\dfrac{1}{2}\!\left(\dfrac{c}{p}-cp^3\right)\right)$ | M1 | 1.1b |
| Correct midpoint confirmed | A1 | 1.1b |
| **Total: 6 marks** | | **Alt 1:** Let $Q=\!\left(cq,\dfrac{c}{q}\right)$; sub into normal: $p^3cq-p\dfrac{c}{q}+c(1-p^4)=0$; $p^3q^2+(1-p^4)q-p=0$; $(q-p)(p^3q+1)=0\Rightarrow q=\ldots$; giving $Q\!\left(-\dfrac{c}{p^3},-cp^3\right)$ [M1 A1 M1 A1] |

# Question (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates $xy = c^2$ to obtain $\frac{dy}{dx} = \pm kx^{-2}$; $k \neq 0$, **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 $t \equiv p$. **And** attempts to use $P\left(cp, \frac{c}{p}\right)$ to write down gradient of tangent in terms of $p$ | M1 | Starts process of establishing gradient of normal by differentiating $xy = c^2$ |
| Deduces correct normal gradient $p^2$ from their tangent gradient found using calculus | A1 | |
| Correct straight line method for equation of normal where $m_N \neq m_T$ is found by calculus. **Note:** $m_N$ must be a function of $p$ | M1 | |
| $p^3x - py + c(1-p^4) = 0$ | A1* | Correct solution only |

---

# Question (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $y = \frac{c^2}{x}$ or $x = \frac{c^2}{y}$ into printed equation to obtain equation in $x$, $c$ and $p$ only **or** in $y$, $c$ and $p$ only | M1 | |
| Obtains a 3TQ in $x$ or a 3TQ in $y$ | A1 | E.g. $p^3x^2 + cx - cp^4x = c^2p$ or $py^2 = cy - cp^4y + c^2p^3$ acceptable for 1st A mark |
| Recognises one solution already known; uses correct factorisation to give $x = \ldots$ or $y = \ldots$. Alternatively applies correct quadratic formula to solve 3TQ | M1 | |
| Correct coordinates for $Q$: $x = -\dfrac{c}{p^3}$ and $y = -cp^3$ | A1 | Can be simplified or un-simplified |
| Uses $\left(cp, \frac{c}{p}\right)$ and their $(x_Q, y_Q)$ and applies $\left(\dfrac{cp + \text{their } x_Q}{2}, \dfrac{\frac{c}{p} + \text{their } y_Q}{2}\right)$ to give $(x_M, y_M)$ where both in terms of $c$ and $p$ only | M1 | |
| Correct coordinates: $\left(\dfrac{1}{2}\!\left(cp - \dfrac{c}{p^3}\right),\ \dfrac{1}{2}\!\left(\dfrac{c}{p} - cp^3\right)\right)$ | A1 | Condone $\left(\dfrac{cp - \frac{c}{p^3}}{2},\ \dfrac{\frac{c}{p} - cp^3}{2}\right)$; also condone $x = \dfrac{1}{2}\!\left(cp - \dfrac{c}{p^3}\right)$ and $y = \dfrac{1}{2}\!\left(\dfrac{c}{p} - cp^3\right)$ for final A mark |

**Alt 1 (for first 4 marks):**

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $x = cq$ and $y = \frac{c}{q}$ into printed equation to obtain equation in $p$, $c$ and $q$ only | M1 | |
| Eliminates $c$ and obtains correct quadratic in $q$. E.g. $p^3q^2 + q - p^4q = p$ acceptable | A1 | |
| Recognises one solution already known; uses correct factorisation to give $q = \ldots$. Alternatively applies correct quadratic formula for 3TQ in $q$ | M1 | |
| Correct coordinates for $Q$: $x = -\dfrac{c}{p^3}$ and $y = -cp^3$ | A1 | Can be simplified or un-simplified |