## Calculus 8th Edition

$x=1+t\\y=2\\ z=2-2t$
Set $x=1$ in the given equation. Thus, we have $4x^2+2(2)^2+z^2=16 \implies 4x^2+z^2=8$ This gives: $\dfrac{x^2}{2}+\dfrac{z^2}{8}=1$ The parametric equations are given as: $x=r \cos t$ and $y=r \sin t$ $r=\sqrt 2$ Thus, $x=r \cos t$ and $y=r \sin t \implies x=\sqrt 2 \cos t; y=2; z=2 \sqrt 2 \sin t$ $x'=-\sqrt 2 \sin t$ and $y=0; z'=2 \sqrt 2 \cos t$ For the point $(1,2,2)$, and $t=\dfrac{\pi}{4}$ we get $x=1+t\\y=2\\z=2-2t$