2022 H2 Maths Paper 1 Q3 — Worked Solution

Part (a)

3 marks

x = \frac{1}{2}\left( e^{3t} + 2e^{- 3t} \right),\quad y = \frac{1}{2}\left( e^{3t} - 2e^{- 3t} \right)

Differentiating with respect to $t$ ,

\frac{dx}{dt} = \frac{1}{2}\left( 3e^{3t} - 6e^{- 3t} \right) = \frac{3}{2}e^{3t} - 3e^{- 3t}

\frac{dy}{dt} = \frac{1}{2}\left( 3e^{3t} + 6e^{- 3t} \right) = \frac{3}{2}e^{3t} + 3e^{- 3t}

At $t = \frac{1}{3}\ln 2$ , $e^{3t} = e^{\ln 2} = 2$ and $e^{- 3t} = \frac{1}{2}$ , so

\frac{dx}{dt} = \frac{3}{2}(2) - 3\left( \frac{1}{2} \right) = \frac{3}{2}

\frac{dy}{dt} = \frac{3}{2}(2) + 3\left( \frac{1}{2} \right) = \frac{9}{2}

The gradient of the tangent is

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}

= \frac{9/2}{3/2}

= 3

To check your answer: at

t = \frac{1}{3}\ln 2

x = \frac{1}{2}(2 + 1) = \frac{3}{2}

and

y = \frac{1}{2}(2 - 1) = \frac{1}{2}

, both finite, and

\frac{dx}{dt} \neq 0

there, so the curve is smooth with a well-defined tangent at this point ✓

Since the tangent has gradient $3$ , the normal (perpendicular to the tangent) has gradient $- \frac{1}{3}$ .

$\therefore$ the gradient of the normal is $- \frac{1}{3}$ .

Part (b)

3 marks

Approach

the parameter appears as $e^{3t}$ and $e^{- 3t}$ in both $x$ and $y$ . Squaring $x$ and $y$ and subtracting eliminates the cross terms and the parameter together, since $x^{2}$ and $y^{2}$ each contain $e^{6t}$ and $e^{- 6t}$ with the same coefficients but opposite signs on the middle term.

x^{2} = \frac{1}{4}\left( e^{3t} + 2e^{- 3t} \right)^{2}

= \frac{1}{4}\left( e^{6t} + 4 + 4e^{- 6t} \right)

y^{2} = \frac{1}{4}\left( e^{3t} - 2e^{- 3t} \right)^{2}

= \frac{1}{4}\left( e^{6t} - 4 + 4e^{- 6t} \right)

Subtracting,

x^{2} - y^{2} = \frac{1}{4}\left\lbrack \left( e^{6t} + 4 + 4e^{- 6t} \right) - \left( e^{6t} - 4 + 4e^{- 6t} \right) \right\rbrack

= \frac{1}{4}(8)

= 2

To check your answer: at

t = \frac{1}{3}\ln 2

x = \frac{3}{2},\ y = \frac{1}{2}

, so

x^{2} - y^{2} = \frac{9}{4} - \frac{1}{4} = 2

, matching ✓

For the restriction on $x$ : since $e^{3t} > 0$ and $2e^{- 3t} > 0$ for all real $t$ , by AM–GM,

e^{3t} + 2e^{- 3t} \geq 2\sqrt{e^{3t} \cdot 2e^{- 3t}} = 2\sqrt{2}

with equality when $e^{3t} = 2e^{- 3t}$ , i.e. $e^{6t} = 2$ . Hence

x = \frac{1}{2}\left( e^{3t} + 2e^{- 3t} \right) \geq \frac{1}{2}\left( 2\sqrt{2} \right) = \sqrt{2}

To check your answer: setting

\frac{dx}{dt} = 0

gives

\frac{3}{2}e^{3t} - 3e^{- 3t} = 0 \Rightarrow e^{6t} = 2

, the same condition found from AM–GM, confirming

x = \sqrt{2}

is the minimum (turning) value of

x

✓

$\therefore$ the cartesian equation is $x^{2} - y^{2} = 2$ , with $x \geq \sqrt{2}$ .

2022 H2 Maths Paper 1, Question 3

Part (a)

Part (b)

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