Chapter 10.6, Problem 41E

To determine

To graph the curve represented by the parametric equations.

Explanation of Solution

Given information:

Given parametric equation is

. $\begin{array}{l}x=4+3\cos \theta \\ y=-2+\sin \theta \end{array}$

Calculation:

The parametric equations are defined for all values of the parameter $\theta$ , So the domain for $\theta$ is $\left[0,2\pi \right]$ . Hence both $x$ and $y$ take on both positive and negative values. Hence the graph of the function is located in all the four quadrants.

Let us find the rectangular equation by eliminating the parameter $\theta$ from the two given equation.

From the first of the parametric equations, $\theta$ in terms of $x$ is given by

  $\begin{array}{c}x-4=3\cos \theta \\ \cos \theta =\frac{x-4}{3}\end{array}$

Similarly from the second equation $\theta$ in terms of $y$ is given by

  $\sin \theta =y+2$

Use the trigonometric identity ${\cos }^2\theta +{\sin }^2\theta =1$ to get the rectangular equation as

  $\begin{array}{c}{\left(\frac{x-4}{3}\right)}^2+{\left(y+2\right)}^2=1\\ \frac{{\left(x-4\right)}^2}{9}+\frac{{\left(y+2\right)}^2}{1}=1\end{array}$

This is the equation of an ellipse. It is known that equation of an ellipse having centre at $\left(h,k\right)$ and horizontal major axis is given by

  $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=\mathrm{1...........}(3)$

The vertices and foci of this ellipse are

Vertices: $\left(h\pm a,k\right)$

Foci: $\left(h\pm c,k\right)$

Comparing equation (2) with the standard equation of an ellipse given by equation (30, it is seen that the ellipse represented by equation (2) has centreat $\left(4,-2\right)$ , and $a=3,b=1$ . Thus the vertices of the ellipse are at $\left(4\pm 3,-2\right)$ or $\left(1,-2\right),\left(7,-2\right)$ .

Using graphing utility, the graph of the ellipse is drawn in the viewing window $\left[-1,8\right]\times \left[-4,1\right]$ and given below. The graph is located in the $4^{th}$ quadrant only.

  

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