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.

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

Short answer

The equation is$\left(\frac{x^2}{4}\right)+\left(\frac{y^2}{9}\right)=1.$

Step-by-step solution

Step 1. Given information.

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

Step 1. Calculation.

Graph the curve defined by the parametric equations -

$t$	$x$	$y$	$(x,y)$
$0$	role="math" localid="1648577233984" $2$	$0$	$(2,0)$
$-\mathrm{\pi }$	$-2$	$0$	$(-2,0)$
$\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$	$0$	role="math" localid="1648577098087" $3$	$(0,3)$
$\raisebox{1ex}{$-\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$	role="math" localid="1648577574094" $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$	$\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$	role="math" localid="1648577593451" $\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.,\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.\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-

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