Is this parabola the same as this circle?

Question

Let $K$ be a field of characteristic not 2.

Is there an invertible rational function between $\mathbb{P}^1(K) = \{ [s:t] : s,t \in K, (s,t) \neq (0,0) \}$ and $V= \{ [x:y:z] : 2xy+z^2 = 0 \}$?

Here $[s:t]=[u:v]$ iff $sv=tu$ and $(s,t)\neq (0,0) \neq (u,v)$. A rational function must be homogenous to be well defined.

Answer 1

Let $\cal C\subseteq\Bbb P^2(K)$ be any non-degenerate conic with a point $P\in\Bbb P^2(K)$. Then you can always define an isomorphism $$ \Phi:\cal C\longrightarrow\Bbb P^1 $$ defined over $K$ by the following geometric construction.

First, identify $\Bbb P^1$ with the set of lines through $P$. Then $\Phi$ is defined as follows: $$ \text{if $P\neq Q\qquad$ then $\Phi(Q)=\overline{PQ}$} $$ and $\Phi(P)=$ tangent line at $P$.