Answer
$r(t)=\cos \theta i+\sqrt 3|\cos \theta| j+\sin \theta k$ ; $0 \leq \theta \leq 2 \pi$
Work Step by Step
The parametric equations of a circle whose radius is $r$ are given as follows: $x=r \cos t ; y =r \sin t$
From the given problem, we have
$x^2+y^2+4z^2=4; y \geq 0$ and $x^2+z^2=1$
Re-arrange as: $(x^2+z^2)+y+3z^2=4 \implies y^2+3z^2=3$
This gives: $y^2=3-3z^2$
So, $y=\sqrt {3-3z^2}$
Let us consider that $z= \sin \theta$ and $y=\sqrt {3-3(\sin \theta)^2}=\sqrt 3|\cos \theta|$
But $x^2+z^2=1$ gives us: $x= \cos \theta$
Hence, $r(t)=\cos \theta i+\sqrt 3|\cos \theta| j+\sin \theta k$ ; $0 \leq \theta \leq 2 \pi$
