Question

In problem graph the curve whose parametric equations are given and show its orientation. Find the rectangular equation of each curve. Verify your graph using a graphing utility.

$x=2\cos t,y=3\sin t;0\le t\le 2\mathrm{\pi }.$

Short answer

The answer is$\frac{x^2}{4}+\frac{y^2}{9}=1.$

Step-by-step solution

Step 1. Given information.

$x=2\cos t,y=3\sin t$is given.

Step 2. Calculation.

Graph the curve defined by the parametric equations -

$t$	$x$	$y$	$(x,y)$
$0$	$2$	$0$	$\left(2,0\right)$
$\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$	$0$	$3$	$\left(0,3\right)$
$\mathrm{\pi }$	$-2$	$0$	$\left(-2,0\right)$
$\raisebox{1ex}{$3\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$	$0$	$-3$	$\left(0,-3\right)$

We find the rectangular equation by eliminating the parameter t from the parametric equations.

$$\begin{gathered} x=2\cos t \\ \cos t=\frac{x}{2} \\ y=3\sin t \\ \sin t=\frac{y}{3} \end{gathered}$$

Now put these value in identity-

role="math" localid="1648406214121" $$\begin{gathered} {\cos }^{2}t+{\sin }^{2}t=1 \\ {\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{3}\right)}^2=1 \\ \frac{x^2}{4}+\frac{y^2}{9}=1 \end{gathered}$$