A curve $C$ has parametric equations \[x = 1 - 2 \cos 2\theta, \quad y = \sin 2\theta - 1, \quad \

Question

A curve $C$ has parametric equations $$x = 1 - 2 \cos 2\theta, \quad y = \sin 2\theta - 1, \quad \text{for}\hspace{0.5em} 0 \le \theta \le \frac{3\pi}{8}.$$ Find the coordinates of the point where $C$ cuts the $y$-axis, giving your answer in exact form.[2] Show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2} \cot 2\theta$. Find the coordinates of the points and the gradient of $C$ where $\theta = \frac{\pi}{4}$. What can be said about the tangent to $C$ as $\theta \to 0$?[5] Hence sketch $C$, labelling the exact coordinates of axial intercepts and endpoints, and showing clearly the features of the curve at the points where $\theta = 0$ and $\theta = \frac{\pi}{4}$.[3] Find a cartesian equation of $C$.[2]

Accepted Answer

(a) The coordinate of the point where $C$ cuts the $y$-axis is $\left( 0,\frac{\sqrt{3}}{2}-1 ight)$. (b) The tangent of $C$ as $	heta 	o 0$ will get steeper and steeper until it tends towards a vertical line. (c) Endpoints: $\left( -1,-1 ight)$, $\left( 1+\sqrt{2},\ \frac{\sqrt{2}}{2}-1 ight)$; axial intercepts: $\left( 1,0 ight)$, $\left( 0,\ \frac{\sqrt{3}}{2}-1 ight)$. (d) ${{\left( \frac{1-x}{2} ight)}^{2}}+{{\left( y+1 ight)}^{2}}=1$

2024 ASRJC Promo Q8

2024 ASRJC Promo Q8