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) $\left(0,\dfrac{\sqrt3}{2}-1\right)$ (b) At $\theta=\dfrac\pi4$, the point is $(1,0)$ and the gradient is $0$. As $\theta\to0$, the tangent tends to the vertical line $x=-1$. (c) See sketch. (d) $\dfrac{(x-1)^2}{4}+(y+1)^2=1$

2024 ASRJC Promo Q8

2024 ASRJC Promo Q8

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