Answer
$x=1+t; y=2, ; z=2-2t$
Work Step by Step
We know that the parametric equations for any plane can be defined as: $x=r \cos t; y=r \sin t$
The given equations yield $4x^2+2(2)^2+z^2=16$ when $y=2$
This gives: $4x^2+z^2=8$
This can be re-arranged as follows:
$\dfrac{x^2}{2}+\dfrac{z^2}{8}=1$
Radius: $r=\sqrt 2$
Thus, $x(t)=\sqrt 2 \cos t; y(t)=2; z(t)=2 \sqrt 2 \sin t$
and $x'(t)=-\sqrt 2 \sin t; y(t)=0; z'(t)=2 \sqrt 2 \cos t$
At the point $(1,2,2)$, and $t=\dfrac{\pi}{4}$ we have
Hence, the parametric equations are: $x=1+t; y=2, ; z=2-2t$
