# Question 1:

## Part (a)

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $2 + 6y\frac{dy}{dx} + \left(6xy + 3x^2\frac{dy}{dx}\right) = 8x$ | M1 A1 B1 | M1: Differentiates implicitly to include either $\pm ky\frac{dy}{dx}$ or $3x^2\frac{dy}{dx}$. Ignore $\left(\frac{dy}{dx}=\right)$. A1: $(2x+3y^2)\to\left(2+6y\frac{dy}{dx}\right)$ and $(4x^2\to 8x)$. Note: If an extra "sixth" term appears award A0. B1: $6xy + 3x^2\frac{dy}{dx}$ |
| $\frac{dy}{dx} = \frac{8x-2-6xy}{6y+3x^2}$ | — | Not necessarily required |
| At $P(-1,1)$, $m(\mathbf{T}) = \frac{dy}{dx} = \frac{8(-1)-2-6(-1)(1)}{6(1)+3(-1)^2} = -\frac{4}{9}$ | dM1 A1 cso | dM1: Substituting $x=-1$ and $y=1$ into an equation involving $\frac{dy}{dx}$. A1: $-\frac{4}{9}$ or $-\frac{8}{18}$ or $-0.\dot{4}$ or awrt $-0.44$. Note: dM1 dependent on previous M1. If candidate substitutes $x=1, y=-1$, usually get $m(\mathbf{T})=-4$, award M0. |

**[5]**

## Part (b)

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $m(\mathbf{N}) = \frac{-1}{-\frac{4}{9}} = \frac{9}{4}$ | M1 | M1: Applies $m(\mathbf{N}) = -\frac{1}{\text{their } m(\mathbf{T})}$ |
| $y - 1 = \frac{9}{4}(x+1)$ | M1 | M1: Uses $y-1=(m_N)(x--1)$ or finds $c$ using $x=-1, y=1$ in $y=(m_N)x + \text{"c"}$, where $m_N = -\frac{1}{\text{their }m(\mathbf{T})}$ or $m_N = \frac{1}{\text{their }m(\mathbf{T})}$ or $m_N = -\text{their }m(\mathbf{T})$ |
| $9x - 4y + 13 = 0$ | A1 | A1: $9x-4y+13=0$ or $-9x+4y-13=0$ or $4y-9x-13=0$ or $18x-8y+26=0$ etc. Must be "$=0$". Do not allow $9x+13=4y$. Note: $m_N = -\left(\frac{6y+3x^2}{8x-2-6xy}\right)$ is M0M0 unless a numerical value is then found for $m_N$. |

**[3]**

**Total: 8 marks**