A variety $X$ is called normal if for any point $P$\in$X$ ,the local ring $\mathbb{O}_P$ is an integrally closed ring. Show that any conic in $\mathbb{P}^2$ is normal.
Here is my attempt: Use the the fact that any conic in $\mathbb{P}^2$ is isomorphic to $\mathbb{P}^1$. So now I need to show for any point $P$\in$\mathbb{P}^1$ ,the local ring $\mathbb{O}_P$ is an integrally closed ring. Over $\mathbb{P}^1$, the local ring $\mathbb{O}_P$ is isomorphic to $k[x_{0},x_{1}]_{(m_{P})}$ which is the zero degee piece of the localization at $m_{p}$=[$f$\in$k[x_{0},x_{1}]$|$f$ homogeneous and $f(P)$=$0$]. Thus, I have to show the local ring $k[x_{0},x_{1}]_{(m_{P})}$ is integrally closed. I know UFD is integrally closed. But how can I show that this local ring is a UFD? Or is there another way to prove the conclusion?