Answer
$x=1+t; y=2, ; z=2-2t$
Work Step by Step
Given: $4x^2+2y^2+z^2=16$ intersects with the plane $x=1$ in a parabola.
This gives: $4x^2+2(2)^2+z^2=16$
or, $4x^2+z^2=8$
Re-arrange as: $\dfrac{x^2}{2}+\dfrac{z^2}{8}=1$
The parametric equations are given as: $x=r \cos t; y=r \sin t$
Here, we have $r=\sqrt 2$
Thus, $x=\sqrt 2 \cos t; y=2; z=2 \sqrt 2 \sin t$
and $x'=-\sqrt 2 \sin t; y=0; z'=2 \sqrt 2 \cos t$
At the point $(1,2,2)$, and $t=\dfrac{\pi}{4}$ we have
$x=1+t; y=2, ; z=2-2t$
