The curve C is defined by the parametric equations x=cos⁡t+12cos⁡5t,y=sin⁡t+12sin⁡5tx = \cos t + \dfrac{1}{2} \cos 5t, \quad y = \sin t + \dfrac{1}{2} \sin 5t \quad for 0≤t≤π20 \le t \le \dfrac{\pi}{2}. Another curve D has the following parametric equations. x=2+kcos⁡θ,y=32sin⁡θx = 2 + \sqrt{k} \cos \theta, \quad y = \dfrac{3}{2} \sin \theta \quad for 0≤θ≤2π0 \le \theta \le 2\pi and k>0k > 0.

  1. (a)

    Find the coordinates of the point on the curve C corresponding to t=0t = 0.

  2. (b)

    Find the cartesian equation for curve D.

  3. (c)

    Sketch the curves C and D on the same diagram.

  4. (d)

    Hence determine the range of values of k such that the equation

    (cos⁡t+0.5cos⁡5t−2)2k+4(sin⁡t+0.5sin⁡5t)29=1\dfrac{\left(\cos t + 0.5\cos 5t - 2\right)^2}{k} + \dfrac{4\left(\sin t + 0.5\sin 5t\right)^2}{9} = 1

    has no real solutions.

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Worked solution

4(a)

1. Substitute $t = 0$ into the parametric equations of curve C to find the corresponding $x$ and $y$ values.

Substitute t=0t = 0 into the equation for xx:
x=cos⁡(0)+12cos⁡(0)x = \cos(0) + \frac{1}{2} \cos(0)
x=1+12(1)=32x = 1 + \frac{1}{2}(1) = \frac{3}{2}

Substitute t=0t = 0 into the equation for yy:
y=sin⁡(0)+12sin⁡(0)y = \sin(0) + \frac{1}{2} \sin(0)
y=0+12(0)=0y = 0 + \frac{1}{2}(0) = 0

The coordinates are (32,0)\left( \frac{3}{2}, 0 \right).

Answer

\left( \frac{3}{2}, 0 \right)

4(b)

1. Express $\cos \theta$ and $\sin \theta$ in terms of $x$ and $y$ respectively.

x=2+kcos⁡θ  ⟹  cos⁡θ=x−2kx = 2 + \sqrt{k} \cos \theta \implies \cos \theta = \frac{x - 2}{\sqrt{k}}

y=32sin⁡θ  ⟹  sin⁡θ=2y3y = \frac{3}{2} \sin \theta \implies \sin \theta = \frac{2y}{3}

2. Use the Pythagorean identity to eliminate the parameter $\theta$ and obtain the Cartesian equation.

Using the identity cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1:

(x−2k)2+(2y3)2=1\left( \frac{x - 2}{\sqrt{k}} \right)^2 + \left( \frac{2y}{3} \right)^2 = 1

(x−2)2k+4y29=1\frac{(x - 2)^2}{k} + \frac{4y^2}{9} = 1

Answer

(x−2)2k+4y29=1\frac{(x - 2)^2}{k} + \frac{4y^2}{9} = 1

4(c)

1. Sketch curve C by interpreting the given parametric equations using a graphing calculator.

Using a graphing calculator, plot the parametric equations for curve C: x=cos⁡t+12cos⁡5tx = \cos t + \frac{1}{2}\cos 5t, y=sin⁡t+12sin⁡5ty = \sin t + \frac{1}{2}\sin 5t for 0≤t≤π20 \le t \le \frac{\pi}{2}.
The calculator will display a self-intersecting (looped) curve strictly within the first quadrant. The starting point at t=0t=0 is (1.5,0)(1.5, 0) and the ending point at t=π2t=\frac{\pi}{2} is (0,1.5)(0, 1.5). Mark these coordinates on the axes.

2. Sketch curve D using the standard properties of its cartesian equation.

From part (b), the cartesian equation for curve D is (x−2)2k+y29/4=1\frac{(x-2)^2}{k} + \frac{y^2}{9/4} = 1.
This represents an ellipse centered at (2,0)(2, 0).
The vertical semi-axis length is 94=1.5\sqrt{\frac{9}{4}} = 1.5, giving the highest point (2,1.5)(2, 1.5) and lowest point (2,−1.5)(2, -1.5). Note that the highest point aligns horizontally with the endpoint (0,1.5)(0, 1.5) of curve C.
The horizontal semi-axis length is k\sqrt{k}, giving the xx-intercepts (2−k,0)(2-\sqrt{k}, 0) and (2+k,0)(2+\sqrt{k}, 0).
Draw a representative full ellipse for an arbitrary k>0k > 0, labelling these key points.

Answer

A sketch containing:

  • Curve C: A looped curve in the first quadrant starting at (1.5,0)(1.5, 0) and ending at (0,1.5)(0, 1.5).
  • Curve D: A full ellipse centered at (2,0)(2, 0), passing through (2,1.5)(2, 1.5) and (2,−1.5)(2, -1.5), with horizontal extremities at (2−k,0)(2-\sqrt{k}, 0) and (2+k,0)(2+\sqrt{k}, 0).

4(d)

1. Link the given equation to the intersection of Curve C and Curve D.

Substitute the parametric equations of Curve C, x=cos⁡t+0.5cos⁡5tx = \cos t + 0.5\cos 5t and y=sin⁡t+0.5sin⁡5ty = \sin t + 0.5\sin 5t, into the given equation:
(x−2)2k+y29/4=1\dfrac{\left(x - 2\right)^2}{k} + \dfrac{y^2}{9/4} = 1
This matches the cartesian equation of Curve D found in part (b). Thus, the given equation represents points where Curve C intersects Curve D. For the equation to have no real solutions, the two curves must not intersect.

2. Determine the maximum $x$-coordinate of Curve C.

From part (a), the starting point of Curve C (at t=0t = 0) is (1.5,0)(1.5, 0).
Since cos⁡t≤1\cos t \le 1 and cos⁡5t≤1\cos 5t \le 1, the maximum possible value for xx on Curve C is:
xmax=1+0.5(1)=1.5x_{max} = 1 + 0.5(1) = 1.5
Thus, Curve C lies entirely in the region x≤1.5x \le 1.5, connecting (1.5,0)(1.5, 0) to (0,1.5)(0, 1.5).

3. Determine the minimum $x$-coordinate of Curve D.

Curve D is an ellipse centered at (2,0)(2, 0) with a horizontal semi-axis of k\sqrt{k} and a vertical semi-axis of 32\frac{3}{2}.
The left-most point (minimum xx-coordinate) of the ellipse is xmin=2−kx_{min} = 2 - \sqrt{k}.

4. Establish the geometric condition for no intersection and solve for $k$.

For the curves to not intersect, Curve D must lie strictly to the right of Curve C. We require the minimum xx of Curve D to be strictly greater than the maximum xx of Curve C:
2−k>1.52 - \sqrt{k} > 1.5
k<0.5\sqrt{k} < 0.5
k<0.25k < 0.25
(Note: If k≥0.25k \ge 0.25, the point (1.5,0)(1.5, 0) of Curve C lies inside or on the ellipse. Since the highest point of Curve C, (0,1.5)(0, 1.5), is always outside the ellipse — as DD's maximum yy is 1.51.5 occurring exclusively at x=2x=2 — Curve C will inevitably cross Curve D, guaranteeing real solutions.)

Given the condition k>0k > 0, the range of values is:
0<k<0.250 < k < 0.25

Answer

0<k<140 < k < \frac{1}{4}

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