I want to show this: The conic $V(y^2- x)$ (that is, the zero set of the polynomial $y^2-x$ over a field $k$) is isomorphic to the open set $\mathbb{A}^1\setminus\{0\}$, where $\mathbb{A}^1=k$.
I think that the question is wrong, and it should be $\mathbb{A}^1\setminus\{0\}=V(xy-1)$ in $\mathbb{A}^2$. Is my thinking correct? Any help is appreciated.