Show that any smooth projective curve of genus zero over a field $K$ is isomorphic to a plane conic over $K$

I have the following question:

Show that any smooth projective curve of genus zero over a field $K$ is isomorphic to a plane conic over $K$.

Assuming that a plane conic is a conic cut by a plane, but I don't see how can I get a hyperbolic by cutting the conic. Is the definition of plane conic correct? If it does, then anyone can give me a hint, please?

Solution 1:

You can look at Liu, Algebraic Geometry and Arithmetic Curves, prop 7.4.1 p285.

The idea is use the Riemann-Roch theorem in order to show that the anticanonical divisor of the curve $C$ induces a closed immersion $C \hookrightarrow \mathbf{P}^2_K$. Thus $C$ is isomorphic to a plane curve $C'$. This curve is defined by a homogeneous polynomial of some degree $d$. But the genus formula says $0=g(C') = \dfrac{(d-1)(d-2)}{2}$. Therefore $C'$ is a plane conic.