Chapter 8, Problem 51RE

Solution

To find:The equation of the conic $x=-2+\cos t$ , $y=4+\sin t$ , $2\pi \le t\le 4\pi$ in standard form.

The equation of the conic $x=-2+\cos t$ , $y=4+\sin t$ , $2\pi \le t\le 4\pi$ in standard form is ${\left(x+2\right)}^2+{\left(y-4\right)}^2=1$ .

Given information:

The parametric equations of the conic are: $x=-2+\cos t$ , $y=4+\sin t$ , $2\pi \le t\le 4\pi$ .

Formula used:

Pythagorean Trigonometric Identity:

  ${\sin }^2\theta +{\cos }^2\theta =1$

Calculation:

Isolate $\cos t$ from the equation $x=-2+\cos t$ and square it to obtain ${\cos }^{2}t$ :

  $\begin{array}{l}x=-2+\cos t\\ x+2=\cos t\\ \cos t=x+2\end{array}$

Square both sides

  ${\cos }^{2}t={\left(x+2\right)}^2$

Isolate $\sin t$ from the equation $y=4+\sin t$ and square it to obtain ${\sin }^{2}t$ :

  $\begin{array}{l}y=4+\sin t\\ y-4=\sin t\\ \sin t=y-4\end{array}$

Square both sides

  ${\sin }^{2}t={\left(y-4\right)}^2$

Substitute ${\cos }^{2}t={\left(x+2\right)}^2$ and ${\sin }^{2}t={\left(y-4\right)}^2$ in the Pythagorean Trigonometric Identity to obtain the standard equation of the conic with the given parametric equations:

  $\begin{array}{l}{\sin }^{2}t+{\cos }^{2}t=1\\ {\left(y-4\right)}^2+{\left(x+2\right)}^2=1\end{array}$

Thus, the equation of the conic $x=-2+\cos t$ , $y=4+\sin t$ , $2\pi \le t\le 4\pi$ in standard form is ${\left(x+2\right)}^2+{\left(y-4\right)}^2=1$ .