projection of $x^2+y^2+z^2-yz=1$ to xoy plane

Question

if the surface is projected to yoz plane, I known than the equation of the projection is $y^2+z^2-yz-1=0$. But what what about xoy plane, is the equation of project projection $x^2+y^2=1$?

Answer 1

$$ x^2+y^2+z^2-yz=1 \\ z^2-yz+\left( x^2+y^2-1 \right) =0 \\ \Delta =-4x^2-3y^2+4\ge 0 \\ x^2+\frac{3}{4}y^2\le 1 $$

Answer 2

The intersection of the plane $x^2+y^2+z^2-yz=1$ with the three coordinate planes gives you the following lines:

1. in the $xy$-plane, therefore, $z=0$: $x^{2}+y^{2}=1$,

2. in the $xz$-plane, therefore, $y=0$: $x^{2}+z^{2}=1$,

3. in the $yz$-plane, therefore, $x=0$: $y^{2}+z^{2}-yz=1$.