Answer
$\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$
See graph
Work Step by Step
We are given the curve in parametric form:
$\begin{cases}
x=3\sin t\\
y=4\cos t
\end{cases}$
Build a table of values:
$t=0; x=0; y=4; (x,y)=(0,4)$
$t=\dfrac{\pi}{2}; x=3; y=0; (x,y)=(3,0)$
$t=\pi; x=0; y=-4; (x,y)=(0,-4)$
$t=\dfrac{3\pi}{2}; x=-3; y=0; (x,y)=(-3,0)$
$t=2\pi; x=0; y=4; (x,y)=(0,4)$
Plot the points and join them to graph the curve:
Determine the rectangular form of the equation:
$\begin{cases}
\sin t=\dfrac{x}{3}\\
\cos t=\dfrac{y}{4}
\end{cases}$
$\sin^2 t+\cos^2 =1$
$\left(\dfrac{x}{3}\right)^2+\left(\dfrac{y}{4}\right)^2=1$
$\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$
