2017 H2 Maths Paper 2 Q1 — Worked Solution

Part (i)

3 marks

Since $A$ and $B$ lie on both $C$ and the line $y = 2x$ , we substitute the parametric coordinates into the line’s equation.

2t = 2\left( \frac{3}{t} \right)

\Rightarrow 2t = \frac{6}{t}

\Rightarrow t^{2} = 3

\Rightarrow t = \pm \sqrt{3}

So $A$ and $B$ correspond to $t = \sqrt{3}$ and $t = - \sqrt{3}$ :

A = \left( \sqrt{3},\ 2\sqrt{3} \right),\quad B = \left( - \sqrt{3},\ - 2\sqrt{3} \right)

By the distance formula,

AB^{2} = \left( 2\sqrt{3} - \left( - 2\sqrt{3} \right) \right)^{2} + \left( \sqrt{3} - \left( - \sqrt{3} \right) \right)^{2}

= \left( 4\sqrt{3} \right)^{2} + \left( 2\sqrt{3} \right)^{2}

= 48 + 12

= 60

To check your answer:

A

and

B

both satisfy

xy = 6

, the cartesian equation of

C

\sqrt{3} \cdot 2\sqrt{3} = 6

and

\left( - \sqrt{3} \right)\left( - 2\sqrt{3} \right) = 6

. ✓

$\therefore$ $AB = \sqrt{60} = 2\sqrt{15}$ .

Part (ii)

5 marks

Approach

We find the gradient of $C$ at a general parameter $t$ , write down the tangent at $P$ , and find where it meets each axis. Rather than guessing the shape of the locus, we compute $F$ ’s coordinates in terms of $p$ and eliminate $p$ directly.

Since $x = \frac{3}{t}$ and $y = 2t$ ,

\frac{dx}{dt} = - \frac{3}{t^{2}},\quad\frac{dy}{dt} = 2

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = 2 \div \left( - \frac{3}{t^{2}} \right) = - \frac{2t^{2}}{3}

At $P\left( \frac{3}{p},2p \right)$ , the gradient is $- \frac{2p^{2}}{3}$ , so the tangent at $P$ is

y - 2p = - \frac{2p^{2}}{3}\left( x - \frac{3}{p} \right)

Setting $y = 0$ to find $D$ :

- 2p = - \frac{2p^{2}}{3}\left( x - \frac{3}{p} \right)

\Rightarrow x - \frac{3}{p} = \frac{3}{p}

\Rightarrow x = \frac{6}{p}

So $D = \left( \frac{6}{p},\ 0 \right)$ .

Setting $x = 0$ to find $E$ :

y - 2p = - \frac{2p^{2}}{3}\left( 0 - \frac{3}{p} \right)

\Rightarrow y - 2p = 2p

\Rightarrow y = 4p

So $E = (0,\ 4p)$ .

$F$ is the midpoint of $D$ and $E$ :

F = \left( \frac{1}{2}\left( \frac{6}{p} + 0 \right),\ \frac{1}{2}(0 + 4p) \right) = \left( \frac{3}{p},\ 2p \right)

So $F$ coincides with $P$ itself for every value of $p$ : for this curve, the midpoint of the two intercepts cut off by a tangent is always the point of tangency. Writing $F = (x,y)$ , so $x = \frac{3}{p}$ and $y = 2p$ , we eliminate $p$ by multiplying:

xy = \frac{3}{p} \cdot 2p = 6

To check your answer: at

p = 1

P = (3,2)

D = (6,0)

E = (0,4)

, and the midpoint of

D

and

E

(3,2) = P

. ✓

$\therefore$ the curve traced by $F$ has cartesian equation $xy = 6$ (with $x \neq 0,\ y \neq 0$ ) — the same curve as $C$ .

2017 H2 Maths Paper 2, Question 1

Part (i)

Part (ii)

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