Answer
(a) $x = \sqrt{5}~sin~t$
$y = \sqrt{3}~cos~t$
We can see the graph below.
(b) $\frac{x^2}{5}+\frac{y^2}{3} = 1$
Work Step by Step
(a) $x = \sqrt{5}~sin~t$
$y = \sqrt{3}~cos~t$
When $t = 0$:
$x = \sqrt{5}~sin~0 = 0$
$y = \sqrt{3}~cos~0 = \sqrt{3}$
When $t = \frac{\pi}{6}$:
$x = \sqrt{5}~sin~\frac{\pi}{6} = \frac{\sqrt{5}}{2}$
$y = \sqrt{3}~cos~\frac{\pi}{6} = \frac{3}{2}$
When $t = \frac{\pi}{4}$:
$x = \sqrt{5}~sin~\frac{\pi}{4} = \frac{\sqrt{10}}{2}$
$y = \sqrt{3}~cos~\frac{\pi}{4} = \frac{\sqrt{6}}{2}$
When $t = \frac{\pi}{3}$:
$x = \sqrt{5}~sin~\frac{\pi}{3} = \frac{\sqrt{15}}{2}$
$y = \sqrt{3}~cos~\frac{\pi}{3} = \frac{\sqrt{3}}{2}$
When $t = \frac{\pi}{2}$:
$x = \sqrt{5}~sin~\frac{\pi}{2} = \sqrt{5}$
$y = \sqrt{3}~cos~\frac{\pi}{2} = 0$
When $t = \frac{2\pi}{3}$:
$x = \sqrt{5}~sin~\frac{2\pi}{3} = \frac{\sqrt{15}}{2}$
$y = \sqrt{3}~cos~\frac{2\pi}{3} = -\frac{\sqrt{3}}{2}$
When $t = \pi$:
$x = \sqrt{5}~sin~\pi = 0$
$y = \sqrt{3}~cos~\pi = -\sqrt{3}$
When $t = \frac{4\pi}{3}$:
$x = \sqrt{5}~sin~\frac{4\pi}{3} = -\frac{\sqrt{15}}{2}$
$y = \sqrt{3}~cos~\frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$
When $t = \frac{3\pi}{2}$:
$x = \sqrt{5}~sin~\frac{3\pi}{2} = -\sqrt{5}$
$y = \sqrt{3}~cos~\frac{3\pi}{2} = 0$
We can see the graph below.
(b) $x = \sqrt{5}~sin~t$
$y = \sqrt{3}~cos~t$
$\frac{x^2}{5}+\frac{y^2}{3} = \frac{( \sqrt{5}~sin~t)^2}{5}+\frac{(\sqrt{3}~cos~t)^2}{3}$
$\frac{x^2}{5}+\frac{y^2}{3} = \frac{5~sin^2~t}{5}+\frac{3~cos^2~t}{3}$
$\frac{x^2}{5}+\frac{y^2}{3} = sin^2~t+cos^2~t$
$\frac{x^2}{5}+\frac{y^2}{3} = 1$
