2020 H2 Maths Paper 2 Q3 — Worked Solution

Part (i)

5 marks

Approach

We first find the parameter $t$ at $(14,11)$ , then differentiate parametrically to get the tangent gradient, and take the negative reciprocal for the normal.

Since $x = 14$ and $y = 11$ ,

3t^{2} + 2 = 14\quad\text{and}\quad 6t - 1 = 11,

both giving $t = 2$ (and $t = 2 \geq \frac{1}{6}$ , so this is a valid point on $C$ ).

Differentiating each equation with respect to $t$ ,

\frac{dx}{dt} = 6t,\quad\quad\frac{dy}{dt} = 6.

Using the chain rule,

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6}{6t} = \frac{1}{t}.

At $t = 2$ , the gradient of the tangent to $C$ is $\frac{1}{2}$ , so the gradient of the normal $N$ (perpendicular to the tangent) is $- 2$ .

Since $N$ passes through $(14,11)$ with gradient $- 2$ ,

y - 11 = - 2(x - 14)

y - 11 = - 2x + 28

2x + y = 39.

To check your answer: the tangent and normal gradients,

\frac{1}{2}

and

- 2

, multiply to

- 1

; and

2(14) + 11 = 39

. ✓

$\therefore$ the equation of $N$ is $2x + y = 39$ .

Part (ii)

4 marks

Approach

The region bounded by $C$ , $N$ and the $x$ -axis splits naturally at $x = 14$ : for $x$ between the start of $C$ and $14$ the upper boundary is the curve, and for $x$ between $14$ and where $N$ meets the $x$ -axis the upper boundary is the line. We find both end-points first.

Since $t \geq \frac{1}{6}$ , the curve $C$ starts where $t = \frac{1}{6}$ :

x = 3\left( \frac{1}{6} \right)^{2} + 2 = \frac{25}{12},\quad\quad y = 6\left( \frac{1}{6} \right) - 1 = 0,

i.e. $C$ starts on the $x$ -axis, at $\left( \frac{25}{12},\ 0 \right)$ .

Note: both $x$ and $y$ increase with $t$ , so $C$ rises monotonically from $\left( \frac{25}{12},0 \right)$ to $(14,11)$ .

The line $N$ meets the $x$ -axis where $y = 0$ in $2x + y = 39$ :

2x = 39

x = \frac{39}{2}.

The enclosed region runs along the $x$ -axis from $x = \frac{25}{12}$ to $x = \frac{39}{2}$ , bounded above by $C$ up to $x = 14$ and by $N$ from $x = 14$ to $x = \frac{39}{2}$ . Using $\int y\,\frac{dx}{dt}\, dt$ for the area under $C$ ,

\text{Area under }C = \int_{1/6}^{2}(6t - 1)(6t)\, dt

= \int_{1/6}^{2}\left( 36t^{2} - 6t \right)dt

= \left\lbrack 12t^{3} - 3t^{2} \right\rbrack_{1/6}^{2}

= 84 - \left( - \frac{1}{36} \right)

= \frac{3025}{36}.

The area under $N$ , from $x = 14$ to $x = \frac{39}{2}$ , is a right-angled triangle of base $\frac{39}{2} - 14 = \frac{11}{2}$ and height $11$ :

\text{Area under }N = \frac{1}{2} \cdot \frac{11}{2} \cdot 11 = \frac{121}{4}.

Adding the two areas,

\text{Total area} = \frac{3025}{36} + \frac{121}{4}

= \frac{3025}{36} + \frac{1089}{36}

= \frac{2057}{18}.

$\therefore$ the area enclosed by $C$ , $N$ and the $x$ -axis is $\frac{2057}{18} \approx 114\ \text{units}^{2}$ (3 s.f.).

Part (iii)(a)

1 mark

Since a two-way stretch with scale factor $2$ in the $x$ -direction and scale factor $3$ in the $y$ -direction scales every area by $2 \times 3 = 6$ ,

\text{Area enclosed by }D,\ M\text{ and the }x\text{-axis} = 6 \times \frac{2057}{18} = \frac{2057}{3}.

$\therefore$ the area is $\frac{2057}{3} \approx 686\ \text{units}^{2}$ (3 s.f.).

Part (iii)(b)

2 marks

Approach

We first eliminate $t$ to get the cartesian equation of $C$ , then apply the stretch $x \mapsto 2x,\ y \mapsto 3y$ directly to that equation.

From $x = 3t^{2} + 2$ , $t^{2} = \frac{x - 2}{3}$ . From $y = 6t - 1$ , $t = \frac{y + 1}{6}$ , so $t^{2} = \frac{(y + 1)^{2}}{36}$ . Equating,

\frac{(y + 1)^{2}}{36} = \frac{x - 2}{3}

(y + 1)^{2} = 12(x - 2).

Under the stretch, a point $(x,y)$ on $C$ maps to $(X,Y) = (2x,3y)$ on $D$ , so $x = \frac{X}{2}$ and $y = \frac{Y}{3}$ . Substituting,

\left( \frac{Y}{3} + 1 \right)^{2} = 12\left( \frac{X}{2} - 2 \right)

\frac{(Y + 3)^{2}}{9} = 6(X - 4)

(Y + 3)^{2} = 54(X - 4).

Note: since $y \geq 0$ on $C$ (as $t \geq \frac{1}{6}$ ), the image satisfies $Y \geq 0$ , so $D$ is the upper branch of this curve.

$\therefore$ the cartesian equation of $D$ is $(y + 3)^{2} = 54(x - 4)$ , with $y \geq 0$ .

2020 H2 Maths Paper 2, Question 3

Part (i)

Part (ii)

Part (iii)(a)

Part (iii)(b)

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