[proofplan] The proof proceeds by reducing to a normal form via classification of quadratic forms over an algebraically closed field. A non-degenerate quadratic form in three variables over $k$ (with $\operatorname{char} k \neq 2$) is, up to a linear change of coordinates, equivalent to $X_0 X_2 - X_1^2$. Irreducibility of $F$ forces non-degeneracy of the underlying symmetric bilinear form. We then verify smoothness directly on this normal form by computing partial derivatives, exhibit the explicit parametrisation $\Phi([Y_0:Y_1]) = [Y_0^2 : Y_0 Y_1 : Y_1^2]$ (the Veronese embedding of degree $2$), and verify it is an isomorphism by constructing an explicit inverse. [/proofplan]

[step:Reduce to the normal form $F = X_0 X_2 - X_1^2$ via diagonalisation of quadratic forms]Since $\operatorname{char} k \neq 2$, there is a bijection between homogeneous quadratic forms $F \in k[X_0, X_1, X_2]$ of degree $2$ and symmetric bilinear forms $B_F: k^3 \times k^3 \to k$, given by polarisation: \begin{align*} B_F(v, w) := \tfrac{1}{2}\bigl(F(v + w) - F(v) - F(w)\bigr). \end{align*} Equivalently, $F(v) = B_F(v, v)$, and $F$ is encoded by a symmetric matrix $M_F \in \operatorname{Sym}_3(k)$ via $F(v) = v^\top M_F\, v$. We claim $M_F$ is invertible. Suppose for contradiction that $\det M_F = 0$. Then $M_F$ has a non-trivial kernel: there exists $0 \neq v_0 \in k^3$ with $M_F v_0 = 0$. By a linear change of coordinates, we may assume $v_0 = (1, 0, 0)$, so that $M_F$ has zeros in the entire first row and column. The form then reads \begin{align*} F(X_0, X_1, X_2) = a X_1^2 + b X_1 X_2 + c X_2^2 = G(X_1, X_2), \end{align*} a homogeneous polynomial in $X_1, X_2$ alone. As a binary quadratic form over the algebraically closed field $k$, $G$ factors into two linear forms: $G(X_1, X_2) = (\alpha_1 X_1 + \beta_1 X_2)(\alpha_2 X_1 + \beta_2 X_2)$. Hence $F$ factors as a product of two linear forms in $k[X_0, X_1, X_2]$, contradicting the irreducibility of $F$. Therefore $\det M_F \neq 0$, so $M_F$ is invertible. Over an algebraically closed field with $\operatorname{char} k \neq 2$, every non-degenerate symmetric bilinear form on $k^3$ is equivalent (via $\mathrm{GL}_3(k)$-change of basis) to the standard hyperbolic-plus-line form $X_0 X_2 - X_1^2$ (or any other choice of non-degenerate symmetric matrix; over an algebraically closed field, all non-degenerate symmetric forms of a given rank are equivalent). Concretely, choose new coordinates $(X_0', X_1', X_2')$ such that $F$ becomes $X_0' X_2' - (X_1')^2$. Smoothness of $V(F)$ is invariant under projective changes of coordinates, as is the property of admitting an isomorphism from $\mathbb{P}^1_k$. So we may assume henceforth \begin{align*} F(X_0, X_1, X_2) = X_0 X_2 - X_1^2. \end{align*}[/step]

[guided]The classification of quadratic forms over an algebraically closed field with $\operatorname{char} k \neq 2$ states: every non-degenerate symmetric bilinear form on $k^n$ is equivalent (under $\mathrm{GL}_n(k)$) to the standard form $\sum_{i=1}^n y_i^2$, or equivalently to any other non-degenerate symmetric form. The hypothesis $\operatorname{char} k \neq 2$ is essential because we need to take square roots of arbitrary diagonal entries — over a field of characteristic $2$, the polarisation map $F \mapsto B_F$ is not invertible, and the classification fails (giving rise to the theory of quadratic forms in characteristic $2$, where additional invariants like the Arf invariant appear). Why is the standard form $X_0 X_2 - X_1^2$ rather than $X_0^2 + X_1^2 + X_2^2$? Both are non-degenerate, and they are equivalent over an algebraically closed field via the change of variables $X_0 = \tfrac{1}{2}(U_0 + U_2)$, $X_2 = \tfrac{1}{2}(U_0 - U_2)$, $X_1 = i U_1$ (where $i^2 = -1$ exists in $k$). The choice of $X_0 X_2 - X_1^2$ is convenient because it makes the rational parametrisation transparent: $X_0 X_2 = X_1^2$ is the equation $\frac{X_0}{X_1} = \frac{X_1}{X_2}$, suggesting the parameter $t = X_1/X_2 = X_0/X_1$, which gives $X_0 = t^2 X_2$, $X_1 = t X_2$. Why does irreducibility imply non-degeneracy? A degenerate quadratic form has a non-trivial kernel vector $v_0$, and after a change of basis sending $v_0$ to $e_0$, the form depends only on the remaining $n-1$ variables. In dimension $3$, this leaves a binary quadratic form, which over an algebraically closed field (with $\operatorname{char} \neq 2$) factors as a product of two linear forms — so $F$ is reducible. Conversely, every reducible $F = L_1 L_2$ has $\det M_F = 0$ (the bilinear form has rank at most $2$, so on $k^3$ it is degenerate). Hence in dimension $3$, irreducibility of $F$ is equivalent to non-degeneracy of $M_F$, given $\operatorname{char} k \neq 2$.[/guided]

[step:Verify smoothness of $V(F)$ for $F = X_0 X_2 - X_1^2$ by computing the Jacobian] The smooth locus of $V(F) \subset \mathbb{P}^2_k$ is the set of points where the gradient $\nabla F = (\partial_{X_0} F, \partial_{X_1} F, \partial_{X_2} F)$ is non-zero (as a vector in $k^3$). For $F = X_0 X_2 - X_1^2$: \begin{align*} \partial_{X_0} F &= X_2, \\ \partial_{X_1} F &= -2 X_1, \\ \partial_{X_2} F &= X_0. \end{align*} The simultaneous vanishing locus $V(\partial_{X_0} F, \partial_{X_1} F, \partial_{X_2} F)$ in $\mathbb{P}^2_k$ would require $X_0 = X_1 = X_2 = 0$ (using $\operatorname{char} k \neq 2$ for the middle equation: $-2 X_1 = 0 \implies X_1 = 0$ since $-2 \neq 0$ in $k$). But $[0:0:0]$ is not a point of $\mathbb{P}^2_k$. Hence $\nabla F$ is non-zero at every point of $\mathbb{P}^2_k$, in particular at every point of $V(F)$, so $V(F)$ is smooth. [/step]

[step:Define the parametrisation $\Phi: \mathbb{P}^1_k \to V(F)$ via the Veronese map] Define \begin{align*} \Phi: \mathbb{P}^1_k &\to \mathbb{P}^2_k \\ [Y_0 : Y_1] &\mapsto [Y_0^2 : Y_0 Y_1 : Y_1^2]. \end{align*} We verify this is well-defined and lands in $V(F)$. Well-definedness: the polynomials $Q_0 = Y_0^2$, $Q_1 = Y_0 Y_1$, $Q_2 = Y_1^2$ are all homogeneous of the same degree $2$ in $Y_0, Y_1$, so the formula respects the equivalence relation defining $\mathbb{P}^2_k$ from $k^3 \setminus \{0\}$: scaling $(Y_0, Y_1) \mapsto (\lambda Y_0, \lambda Y_1)$ scales the image by $\lambda^2$. Moreover, $(Y_0, Y_1) \neq (0, 0)$ implies $(Y_0^2, Y_0 Y_1, Y_1^2) \neq (0, 0, 0)$: if $Y_0^2 = Y_1^2 = 0$ then $Y_0 = Y_1 = 0$. So $\Phi$ is a well-defined morphism $\mathbb{P}^1_k \to \mathbb{P}^2_k$. Image lies in $V(F)$: substituting, \begin{align*} F(Y_0^2, Y_0 Y_1, Y_1^2) = Y_0^2 \cdot Y_1^2 - (Y_0 Y_1)^2 = Y_0^2 Y_1^2 - Y_0^2 Y_1^2 = 0, \end{align*} identically in $(Y_0, Y_1)$. Hence $\Phi(\mathbb{P}^1_k) \subset V(F)$. [/step]

[step:Construct the inverse morphism $\Psi: V(F) \to \mathbb{P}^1_k$ on each affine chart] We define $\Psi: V(F) \to \mathbb{P}^1_k$ piecewise on the open cover $V(F) = U \cup U'$, where \begin{align*} U &:= V(F) \cap \{X_0 \neq 0\}, \\ U' &:= V(F) \cap \{X_2 \neq 0\}. \end{align*} These two opens cover $V(F)$: a point $[X_0 : X_1 : X_2] \in V(F)$ with $X_0 = X_2 = 0$ would satisfy $0 \cdot 0 - X_1^2 = -X_1^2 = 0$, so $X_1 = 0$ as well, contradicting that homogeneous coordinates cannot all be zero. Hence $X_0 \neq 0$ or $X_2 \neq 0$ at every point, and $U \cup U' = V(F)$. On $U$, define \begin{align*} \Psi_U: U &\to \mathbb{P}^1_k \\ [X_0 : X_1 : X_2] &\mapsto [X_0 : X_1]. \end{align*} This is well-defined: if $X_0 \neq 0$, then $(X_0, X_1) \neq (0, 0)$ (the first coordinate is non-zero), so the image lies in $\mathbb{P}^1_k$; and the formula respects scaling $(X_0, X_1, X_2) \mapsto (\lambda X_0, \lambda X_1, \lambda X_2)$. On $U'$, define \begin{align*} \Psi_{U'}: U' &\to \mathbb{P}^1_k \\ [X_0 : X_1 : X_2] &\mapsto [X_1 : X_2]. \end{align*} Well-defined by the same argument: $X_2 \neq 0$ gives $(X_1, X_2) \neq (0, 0)$. We verify the two definitions agree on the overlap $U \cap U' = V(F) \cap \{X_0 \neq 0, X_2 \neq 0\}$. On this overlap, the equation $X_0 X_2 = X_1^2$ holds and $X_0, X_2 \neq 0$. Then \begin{align*} [X_0 : X_1] = [X_0 X_2 : X_1 X_2] = [X_1^2 : X_1 X_2] = [X_1 : X_2] \end{align*} in $\mathbb{P}^1_k$, where the second equality uses $X_0 X_2 = X_1^2$ and the third divides by $X_1$ (which is non-zero on $U \cap U'$: if $X_1 = 0$ and $X_0 X_2 = X_1^2 = 0$, then $X_0 = 0$ or $X_2 = 0$, contradicting $X_0, X_2 \neq 0$). Hence $\Psi_U = \Psi_{U'}$ on $U \cap U'$, and they glue to a global morphism $\Psi: V(F) \to \mathbb{P}^1_k$. [/step]

[step:Verify $\Phi$ and $\Psi$ are mutually inverse]We verify $\Psi \circ \Phi = \operatorname{id}_{\mathbb{P}^1_k}$ and $\Phi \circ \Psi = \operatorname{id}_{V(F)}$. For $\Psi \circ \Phi$: take $[Y_0 : Y_1] \in \mathbb{P}^1_k$. Then $\Phi([Y_0 : Y_1]) = [Y_0^2 : Y_0 Y_1 : Y_1^2]$. \begin{itemize} \item If $Y_0 \neq 0$, the image lies in $U$ (first coordinate $Y_0^2 \neq 0$), and $\Psi_U([Y_0^2 : Y_0 Y_1 : Y_1^2]) = [Y_0^2 : Y_0 Y_1] = [Y_0 : Y_1]$ (cancelling the common factor $Y_0$, which is non-zero). \item If $Y_1 \neq 0$, the image lies in $U'$ (third coordinate $Y_1^2 \neq 0$), and $\Psi_{U'}([Y_0^2 : Y_0 Y_1 : Y_1^2]) = [Y_0 Y_1 : Y_1^2] = [Y_0 : Y_1]$ (cancelling the common factor $Y_1$). \end{itemize} At least one of $Y_0, Y_1$ is non-zero (homogeneous coordinates), so $\Psi(\Phi([Y_0 : Y_1])) = [Y_0 : Y_1]$ in all cases. For $\Phi \circ \Psi$: take $[X_0 : X_1 : X_2] \in V(F)$, so $X_0 X_2 = X_1^2$. \begin{itemize} \item If $X_0 \neq 0$ (point in $U$), then $\Psi([X_0 : X_1 : X_2]) = [X_0 : X_1]$, so \begin{align*} \Phi([X_0 : X_1]) = [X_0^2 : X_0 X_1 : X_1^2] = [X_0^2 : X_0 X_1 : X_0 X_2] = [X_0 : X_1 : X_2], \end{align*} using $X_1^2 = X_0 X_2$ in the second equality and cancelling $X_0$ in the third (valid since $X_0 \neq 0$). \item If $X_2 \neq 0$ (point in $U'$), then $\Psi([X_0 : X_1 : X_2]) = [X_1 : X_2]$, so \begin{align*} \Phi([X_1 : X_2]) = [X_1^2 : X_1 X_2 : X_2^2] = [X_0 X_2 : X_1 X_2 : X_2^2] = [X_0 : X_1 : X_2], \end{align*} using $X_1^2 = X_0 X_2$ and cancelling $X_2 \neq 0$. \end{itemize} Hence $\Phi \circ \Psi = \operatorname{id}_{V(F)}$.[/step]

[guided]The pair $(\Phi, \Psi)$ is the prototype of a rational parametrisation of a smooth conic. The geometric content: pick a point $p_0 \in V(F)$ — say $p_0 = [1:0:0]$, which lies on $V(X_0 X_2 - X_1^2)$ since $1 \cdot 0 - 0 = 0$. The pencil of lines through $p_0$ is parametrised by $\mathbb{P}^1_k$ (each line is determined by its slope, plus the direction "along $X_0 = 0$"). Each line meets the conic in $p_0$ and one other point (by [Bézout for Lines](/theorems/2162), since the conic has degree $2$ and a line has degree $1$, and the line is not contained in the irreducible conic). Sending the line to its other intersection point with $V(F)$ is the parametrisation $\Phi$. The map $\Psi$ goes the other direction: given $p \in V(F)$, draw the line through $p$ and $p_0$, and record its slope. The two charts $U = \{X_0 \neq 0\}$ and $U' = \{X_2 \neq 0\}$ correspond to the two ways of writing the slope: in $U$ the slope is $X_1/X_0$; in $U'$, where the line might be vertical relative to the $X_0$-axis, we instead use $X_2/X_1$. The fact that we need two charts to define $\Psi$ is not a technical inconvenience but reflects an intrinsic feature: morphisms between projective varieties are typically defined locally, glued along open covers. A single global formula (e.g.\ $[X_0 : X_1 : X_2] \mapsto [X_0 : X_1]$) would fail at the point $[0:0:1]$ (where both coordinates vanish), but this point is recovered by the second chart. The verification that $\Psi_U$ and $\Psi_{U'}$ agree on the overlap is the crucial gluing check. The equation $X_0 X_2 = X_1^2$ of the conic is exactly what makes the two formulas coincide — without this constraint, the maps $[X_0 : X_1 : X_2] \mapsto [X_0 : X_1]$ and $[X_0 : X_1 : X_2] \mapsto [X_1 : X_2]$ would in general give different points of $\mathbb{P}^1_k$.[/guided]

[step:Conclude smoothness and the parametrisation in original coordinates] By Step 2, $V(F)$ is smooth in the normal-form coordinates. Smoothness is preserved under projective changes of coordinates (smoothness of $V(F)$ at a point $p$ is the condition that the local ring $\mathcal{O}_{V(F), p}$ is a regular local ring, which is intrinsic to the variety and not the embedding), so the original irreducible conic is also smooth. By Steps 3-5, in the normal-form coordinates the map $\Phi([Y_0 : Y_1]) = [Y_0^2 : Y_0 Y_1 : Y_1^2]$ is an isomorphism $\mathbb{P}^1_k \to V(X_0 X_2 - X_1^2)$. Composing with the inverse of the linear change of coordinates from Step 1 produces an isomorphism $\mathbb{P}^1_k \to V(F)$ given by $[Y_0 : Y_1] \mapsto [Q_0(Y_0, Y_1) : Q_1(Y_0, Y_1) : Q_2(Y_0, Y_1)]$, where $(Q_0, Q_1, Q_2)$ is the image of $(Y_0^2, Y_0 Y_1, Y_1^2)$ under the inverse linear change of coordinates: each $Q_i$ is a $k$-linear combination of $Y_0^2$, $Y_0 Y_1$, $Y_1^2$, hence a quadratic form in $Y_0, Y_1$. This completes the proof. [/step]