2024 H2 Mathematics Paper 1 Question 9

Differentiation I: Tangents and Normals, Parametric Curves

Answers

$k = 1.$

$(9 + 2 \sqrt{3}, 3 + 6 \sqrt{3}) .$

Out of syllabus.

(a)

\begin{aligned} x & = 3 t^{2} + 2 t \\ \frac{d x}{d t} & = 6 t + 2 \\ y & = t^{2} + 2 t^{3} \\ \frac{d y}{d t} & = 2 t + 6 t^{2} \end{aligned}

\begin{aligned} \frac{d y}{d x} & = \frac{d x}{d t} \div \frac{d y}{d t} \\ = \frac{2 t + 6 t^{2}}{6 t + 2} \\ = \frac{2 t (1 + 3 t)}{2 (3 t + 1)} \\ = t ∎ \end{aligned}

(b)

Since the tangent makes an angle of $60^{\circ}$ with the $x -axis,$

\begin{aligned} \frac{d y}{d x} & = \tan 60^{\circ} \\ t & = \sqrt{3} \end{aligned}

\begin{aligned} x & = 3 t^{2} + 2 t \\ = 9 + 2 \sqrt{3} \\ y & = t^{2} + 2 t^{3} \\ = 3 + 6 \sqrt{3} \end{aligned}

Coordinates of $P$ :

(9 + 2 \sqrt{3}, 3 + 6 \sqrt{3}) ∎

(c)

Out of syllabus.