Chapter 8, Problem 49RE

Solution

To find:The equation of the conic $x=5\cos t$ , $y=2\sin t$ , $0\le t\le 2\pi$ in standard form.

The equation of the conic $x=5\cos t$ , $y=2\sin t$ , $0\le t\le 2\pi$ in standard form is $\frac{x^2}{25}+\frac{y^2}{4}=1$ .

Given information:

The parametric equations of the conic are: $x=5\cos t$ , $y=2\sin t$ , $0\le t\le 2\pi$ .

Formula used:

Pythagorean Trigonometric Identity:

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

Calculation:

Isolate $\cos t$ from the equation $x=5\cos t$ and square it to obtain ${\cos }^{2}t$ :

  $\begin{array}{c}x=5\cos t\\ \frac{x}{5}=\cos t\\ \cos t=\frac{x}{5}\end{array}$

Square both sides

  $\begin{array}{l}{\cos }^{2}t={\left(\frac{x}{5}\right)}^2\\ {\cos }^{2}t=\frac{x^2}{25}\end{array}$

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

  $\begin{array}{c}y=2\sin t\\ \frac{y}{2}=\sin t\\ \sin t=\frac{y}{2}\end{array}$

Square both sides.

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

Substitute ${\cos }^{2}t=\frac{x^2}{25}$ and ${\sin }^{2}t=\frac{y^2}{4}$ in the Pythagorean Trigonometric Identity to obtain the standard equation of the conic with the given parametric equations:

  $\begin{array}{l}{\sin }^{2}t+{\cos }^{2}t=1\\ \frac{y^2}{4}+\frac{x^2}{25}=1\\ \frac{x^2}{25}+\frac{y^2}{4}=1\end{array}$

  $$

Thus, the equation of the conic $x=5\cos t$ , $y=2\sin t$ , $0\le t\le 2\pi$ in standard form is $\frac{x^2}{25}+\frac{y^2}{4}=1$ .