Sufficient condition placed on $f(x,y)$ to obtain isomorphism. Let $k[x,y]$ be the ring of polynomials in $x,y$ with coefficients in $k$.
Let $\mathcal{O}(W) := k[x,y]/\langle f(x,y) \rangle$, where $f(x,y)$ is the irreducible quadratic in two variables $f(x,y) = ax^2 + by^2 + cxy + dx + ey + g$.
My question is, what are the necessary and sufficient conditions we need to place on $f$ such that $\mathcal{O}(W) \cong k[x]$?
My guess is that we need to remove the cross term, so it is simply the case of setting $c=0$, since we have a simple relationship between getting $y$ from elements in $k$ and $x$.
Thanks!
 A: Let's assume that $k = \overline k$ and $k$ is of characteristic zero.
I will try to answer geometrically. Let $W$ be the zero set of your polynomial, and let's look in $\Bbb P^2$ and $X = \overline W$. We know that $W = X \backslash \{p_1, p_2\}$ for some points $p_1, p_2$ (not necessary distinct). 
Now, we now that a smooth conic is isomorphic to $\Bbb P^1$. It follows that if $p_1, p_2$ are distincts one has $Y \cong k^*$ and so $\mathcal O(W) = k[t,t^{-1}]$. This is given e.g by $f(x,y) = xy - 1$. If $p_1 = p_2$ then it follows that $W \cong \Bbb A^1$ so $\mathcal O(W) \cong k[t]$ as expected. In particular, you also have to check that your curve is smooth, else you will have $\mathcal O(W) \cong k[t] \times k[s]$ or $\mathcal O(W) \cong k[x,y)]/(xy)$ or  as shows the polynomial $f(x,y) = y(x-1)$, resp $f(x,y) = xy$ (depending if the singular point is on the line at infinity or not).
In maybe more algebraic terms, parametrizing the line at infinity $L$ will gives you a polynomial $g = f_{|L}$ of degree $2$, in one variable. The condition that $L$ is tangent at $X$ is exactly the condition that polynomial $g$ has a double root.
