Answer
(a) $x = 4~sin~t$
$y = 3~cos~t$
We can see the graph below.
(b) $\frac{x^2}{16}+\frac{y^2}{9} = 1$
Work Step by Step
(a) $x = 4~sin~t$
$y = 3~cos~t$
When $t = 0$:
$x = 4~sin~0 = 0$
$y = 3~cos~0 = 3$
When $t = \frac{\pi}{6}$:
$x = 4~sin~\frac{\pi}{6} = 2$
$y = 3~cos~\frac{\pi}{6} = \frac{3\sqrt{3}}{2}$
When $t = \frac{\pi}{4}$:
$x = 4~sin~\frac{\pi}{4} = 2\sqrt{2}$
$y = 3~cos~\frac{\pi}{4} = \frac{3\sqrt{2}}{2}$
When $t = \frac{\pi}{3}$:
$x = 4~sin~\frac{\pi}{3} = 2\sqrt{3}$
$y = 3~cos~\frac{\pi}{3} = \frac{3}{2}$
When $t = \frac{\pi}{2}$:
$x = 4~sin~\frac{\pi}{2} = 4$
$y = 3~cos~\frac{\pi}{2} = 0$
When $t = \frac{2\pi}{3}$:
$x = 4~sin~\frac{2\pi}{3} = 2\sqrt{3}$
$y = 3~cos~\frac{2\pi}{3} = -\frac{3}{2}$
When $t = \pi$:
$x = 4~sin~\pi = 0$
$y = 3~cos~\pi = -3$
When $t = \frac{4\pi}{3}$:
$x = 4~sin~\frac{4\pi}{3} = -2\sqrt{3}$
$y = 3~cos~\frac{4\pi}{3} = -\frac{3}{2}$
When $t = \frac{3\pi}{2}$:
$x = 4~sin~\frac{3\pi}{2} = -4$
$y = 3~cos~\frac{3\pi}{2} = 0$
We can see the graph below.
(b) $x = 4~sin~t$
$y = 3~cos~t$
$\frac{x^2}{16}+\frac{y^2}{9} = \frac{(4~sin~t)^2}{16}+\frac{(3~cos~t)^2}{9}$
$\frac{x^2}{16}+\frac{y^2}{9} = \frac{16~sin^2~t}{16}+\frac{9~cos^2~t}{9}$
$\frac{x^2}{16}+\frac{y^2}{9} = sin^2~t+cos^2~t$
$\frac{x^2}{16}+\frac{y^2}{9} = 1$
