Answer
$$1=\dfrac{(x-3)^2}{4}+\dfrac{(y-2)^2}{16} $$ $$ \text{ ellipse }$$
Work Step by Step
We have: $x=3+2 \cos t \implies \cos t=\dfrac{x-3}{2}$ and $y= 2+4\sin t \implies \sin t=\dfrac{y-2}{4}$
Now, use the identity $$\sin^2 t+\cos^2 t=1 \dfrac{(x-3)^2}{4}+\dfrac{(y-2)^2}{16}\\ 1=\dfrac{(x-3)^2}{4}+\dfrac{(y-2)^2}{16} $$
This shows an equation for an ellipse.
