Answer
$4x^2+y^2+z^2=16$
Work Step by Step
Rewrite as: $4x^2+y^2=16$
$\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$
or, $x=(y^2-2y)+(z^2-4z)+5$
We have an ellipse rotated around the x-axis whose traces are parallel to the yz plane.
So, we have:
$\dfrac{x^2}{4}+\dfrac{y^2}{16}+\dfrac{z^2}{16}=1$
or, $4x^2+y^2+z^2=16$
