Conics in $\mathbb{A}^2$

I'm trying to solve Exercise 3.1 in Hartshorne's Algebraic Geometry:

Show that any conic in $\mathbb{A}^2$ is isomorphic to $\mathbb{A}^1$ or $\mathbb{A}^1-\{0\}$.

I know from a previous exercise that the coordinate ring of a given affine conic $Q$ is either isomorphic to $A(Z(y-x^2))\cong k[y]$ or $A(Z(xy-1))\cong k[x,1/x]$. In the first case clearly $A(Q)\cong A(\mathbb{A}^1)$.

But how is $A(Q)\cong A(\mathbb{A}^1-\{0\})$ in the other case? Somehow I even am confused with the notation $A(\mathbb{A}^1-\{0\})$, since the only varieties in $\mathbb{A}^1$ are single points and $\mathbb{A}^1$ itself. Therefore $\mathbb{A}^1-\{0\}$ isn't an affine variety, right?

So how can I say, that $Q$ is isomorphic to $\mathbb{A}^1-\{0\}$? What is $A(\mathbb{A}^1-\{0\})$?

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Consider the map given by the projection on one of the two coordinate axes restricted to the hyperbola $xy=1$. –  Mariano Suárez-Alvarez Feb 13 '12 at 23:43
Take the projection map from the hyperbola $Z(XY-1)$ to $\mathbb A^1-\{0\}$ sending $(x,\frac{1}{x})$ to $x$. This is your isomorphism since the inverse map is sending $x$ to $(x,\frac{1}{x})$. The coordinate ring you're asking about is $k[X,Y]/(XY-1) \cong k[X,\frac{1}{X}]$. $\mathbb A^1-\{0\}$ is affine because the hyperbola is, but $\mathbb A^n - \{0\}$ is not affine for $n>1$.
I'm quite surprised that $\mathbb{A}^1-\{0\}$ is affine, but I think I understand now. Thank you very much for your explanation! Is there an easy way to see that this is not true for $n>1$? –  user20388 Feb 14 '12 at 15:22