Answer
(a) $x = 1+2~sin~t$
$y = 2+3~cos~t$
We can see the graph below.
(b) $\frac{(x-1)^2}{4}+\frac{(y-2)^2}{9} = 1$
Work Step by Step
(a) $x = 1+2~sin~t$
$y = 2+3~cos~t$
When $t = 0$:
$x = 1+2~sin~0 = 1$
$y = 2+3~cos~0 = 5$
When $t = \frac{\pi}{6}$:
$x = 1+2~sin~\frac{\pi}{6} = 2$
$y = 2+3~cos~\frac{\pi}{6} = 4.60$
When $t = \frac{\pi}{4}$:
$x = 1+2~sin~\frac{\pi}{4} = 2.41$
$y = 2+3~cos~\frac{\pi}{4} = 4.12$
When $t = \frac{\pi}{3}$:
$x = 1+2~sin~\frac{\pi}{3} = 2.73$
$y = 2+3~cos~\frac{\pi}{3} = 3.5$
When $t = \frac{\pi}{2}$:
$x = 1+2~sin~\frac{\pi}{2} = 3$
$y = 2+3~cos~\frac{\pi}{2} = 2$
When $t = \frac{2\pi}{3}$:
$x = 1+2~sin~\frac{2\pi}{3} = 2.73$
$y = 2+3~cos~\frac{2\pi}{3} = 0.5$
When $t = \pi$:
$x = 1+2~sin~\pi = 1$
$y = 2+3~cos~\pi = -1$
When $t = \frac{4\pi}{3}$:
$x = 1+2~sin~\frac{4\pi}{3} = -0.73$
$y = 2+3~cos~\frac{4\pi}{3} = 0.5$
When $t = \frac{3\pi}{2}$:
$x = 1+2~sin~\frac{3\pi}{2} = -1$
$y = 2+3~cos~\frac{3\pi}{2} = 2$
We can see the graph below.
(b) $x = 1+2~sin~t$
$\frac{x-1}{2} = sin~t$
$y = 2+3~cos~t$
$\frac{y-2}{3} = cos~t$
$(\frac{x-1}{2})^2+(\frac{y-2}{3})^2 =sin^2~t+cos^2~t$
$\frac{(x-1)^2}{4}+\frac{(y-2)^2}{9} = 1$
