2025 ASRJC P2 Q4

2025 ASRJC P2 Q4

6 marks

The curve \(C\) is defined by the parametric equations \(x = \cos t + \frac{1}{2}\cos 5t, \quad y = \sin t + \frac{1}{2}\sin 5t \quad \text{for}\hspace{0.5em} 0 \le t \le \frac{\pi}{2}.\)

  1. Find the coordinates of the point on the curve \(C\) corresponding to \(t=0\).[1]

Another curve \(D\) has the following parametric equations. \(x = 2 + \sqrt{k}\cos\theta, \quad y = \frac{3}{2}\sin\theta \quad \text{for}\hspace{0.5em} 0 \le \theta \le 2\pi \text{ and}\hspace{0.5em} k > 0.\)

  1. Find the cartesian equation for curve \(D\).[2]
  2. Sketch the curves \(C\) and \(D\) on the same diagram.[2]
  3. Hence determine the range of values of \(k\) such that the equation \(\frac{(\cos t + 0.5\cos 5t - 2)^2}{k} + \frac{4(\sin t + 0.5\sin 5t)^2}{9} = 1\) has no real solutions.[1]

Solution:

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Answer:(a) \(\left(\frac32,0\right)\). (b) \(\frac{(x-2)^2}{k}+\frac{4y^2}{9}=1\). (c) Sketch shown with the stated parameter interval. (d) \(0<k<\frac14\).

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