2021 H2 Maths Paper 1 Q3 — Worked Solution

Part (a)

2 marks

x^{1/2} + y^{1/2} = 3

We differentiate implicitly with respect to $x$ .

\frac{1}{2}x^{- 1/2} + \frac{1}{2}y^{- 1/2}\frac{dy}{dx} = 0

\Rightarrow \frac{1}{2}y^{- 1/2}\frac{dy}{dx} = - \frac{1}{2}x^{- 1/2}

\Rightarrow \frac{dy}{dx} = - \frac{x^{- 1/2}}{y^{- 1/2}}

= - \left( \frac{y}{x} \right)^{1/2}\ \ \left( \text{shown} \right)

4 marks

Substituting $x = 1$ into the curve’s equation,

1^{1/2} + y^{1/2} = 3

\Rightarrow y^{1/2} = 2

\Rightarrow y = 4

so the point is $(1,4)$ . Using the result of (a), the gradient of the curve there is

\frac{dy}{dx} = - \left( \frac{4}{1} \right)^{1/2} = - 2

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal, $\frac{1}{2}$ . Using the point $(1,4)$ ,

y - 4 = \frac{1}{2}(x - 1)

\Rightarrow y = \frac{1}{2}x + \frac{7}{2}

To check your answer: the normal’s gradient

\frac{1}{2}

times the tangent’s gradient

- 2

gives

- 1

, confirming they are perpendicular ✓.

$\therefore$ the normal at $x = 1$ has equation $y = \frac{1}{2}x + \frac{7}{2}$ .