Chapter 8, Problem 50RE

Solution

To find:The equation of the conic $x=4\sin t$ , $y=6\cos t$ , $0\le t\le 4\pi$ in standard form.

The equation of the conic $x=4\sin t$ , $y=6\cos t$ , $0\le t\le 4\pi$ in standard form is $\frac{x^2}{16}+\frac{y^2}{36}=1$ .

Given information:

The parametric equations of the conic are: $x=4\sin t$ , $y=6\cos t$ , $0\le t\le 4\pi$ .

Formula used:

Pythagorean Trigonometric Identity:

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

Calculation:

Isolate $\sin t$ from the equation $x=4\sin t$ and square it to obtain:

  $\begin{array}{c}x=4\sin t\\ \frac{x}{4}=\sin t\\ \sin t=\frac{x}{4}\end{array}$

Square both sides

  $\begin{array}{l}{\sin }^{2}t={\left(\frac{x}{4}\right)}^2\\ {\sin }^{2}t=\frac{x^2}{16}\end{array}$

Isolate $\cos t$ from the equation $y=6\cos t$ and square it to obtain:

  $\begin{array}{c}y=6\cos t\\ \frac{y}{6}=\cos t\\ \cos t=\frac{y}{6}\end{array}$

Square both sides:

  $\begin{array}{l}{\cos }^{2}t={\left(\frac{y}{6}\right)}^2\\ {\cos }^{2}t=\frac{y^2}{36}\end{array}$

Substitute ${\cos }^{2}t=\frac{y^2}{36}$ and ${\sin }^{2}t=\frac{x^2}{16}$ in the Pythagorean Trigonometric Identity to obtain the standard equation of the conic with the given parametric equations:

  $\begin{array}{l}{\sin }^{2}t+{\cos }^{2}t=1\\ \frac{x^2}{16}+\frac{y^2}{36}=1\end{array}$

Thus, the equation of the conic $x=4\sin t$ , $y=6\cos t$ , $0\le t\le 4\pi$ in standard form is $\frac{x^2}{16}+\frac{y^2}{36}=1$ .