[proofplan] We first make a $\mathbb{Q}$-linear projective change of coordinates sending the given rational point $P$ to a convenient coordinate point. In those coordinates a line through $P$ is encoded by a parameter $[u:v]\in \mathbb{P}^1$, and substituting the line into the quadratic equation of $C$ factors off the known intersection at $P$. The remaining linear factor gives the residual point as homogeneous quadratic functions of $u$ and $v$. Finally, nonsingularity rules out a line component and identifies the unique exceptional parameter with the tangent line at $P$. [/proofplan]

[step:Put the conic in coordinates adapted to the rational point]Choose homogeneous coordinates $[X:Y:Z]$ on $\mathbb{P}^2_{\mathbb{Q}}$ by a projective linear change of coordinates defined over $\mathbb{Q}$ such that \begin{align*} P = [0:1:0]. \end{align*} Let \begin{align*} F: \mathbb{Q}^3 \to \mathbb{Q} \end{align*} be a nonzero homogeneous quadratic form defining $C$, so that \begin{align*} C = \{[X:Y:Z]\in \mathbb{P}^2_{\mathbb{Q}} : F(X,Y,Z)=0\}. \end{align*} Since $P\in C(\mathbb{Q})$, the coefficient of $Y^2$ in $F$ is zero. Hence there exist coefficients $A,B,C_1,D,E\in \mathbb{Q}$ such that \begin{align*} F(X,Y,Z)=A X^2+BXY+C_1XZ+DYZ+EZ^2. \end{align*} The partial derivatives at $P$ are \begin{align*} \partial_X F(P)=B,\qquad \partial_YF(P)=0,\qquad \partial_ZF(P)=D. \end{align*} Because $C$ is nonsingular at $P$, the gradient of $F$ at $P$ is nonzero, so $(B,D)\ne (0,0)$.[/step]

[guided]We use the rational point $P$ to choose coordinates compatible with the pencil of lines through $P$. Since $P$ has rational homogeneous coordinates, there is a projective linear change of coordinates over $\mathbb{Q}$ sending $P$ to $[0:1:0]$. Such a change does not alter the assertion, because rational lines, rational points, tangency, and residual intersection are preserved by projective linear transformations defined over $\mathbb{Q}$. Let \begin{align*} F: \mathbb{Q}^3 \to \mathbb{Q} \end{align*} be the homogeneous quadratic form defining the conic in the new coordinates: \begin{align*} C = \{[X:Y:Z]\in \mathbb{P}^2_{\mathbb{Q}} : F(X,Y,Z)=0\}. \end{align*} A general homogeneous quadratic form in $X,Y,Z$ has a $Y^2$ coefficient. Evaluating at $P=[0:1:0]$ isolates exactly that coefficient. Since $P\in C$, we have $F(0,1,0)=0$, so the $Y^2$ coefficient must vanish. Therefore \begin{align*} F(X,Y,Z)=A X^2+BXY+C_1XZ+DYZ+EZ^2 \end{align*} for uniquely determined coefficients $A,B,C_1,D,E\in \mathbb{Q}$. The nonsingularity hypothesis is used immediately. The partial derivatives at $P$ are \begin{align*} \partial_X F(P)=B,\qquad \partial_YF(P)=0,\qquad \partial_ZF(P)=D. \end{align*} A projective plane curve defined by one homogeneous equation is nonsingular at $P$ precisely when not all these partial derivatives vanish at $P$. Hence $(B,D)\ne(0,0)$. This nonzero linear form will later be the tangent-direction test.[/guided]

[step:Parametrize the pencil of lines through $P$] Identify the pencil of lines through $P=[0:1:0]$ with $\mathbb{P}^1_{\mathbb{Q}}$ as follows. For $[u:v]\in \mathbb{P}^1_{\mathbb{Q}}$, define the line \begin{align*} L_{[u:v]} := \{[X:Y:Z]\in \mathbb{P}^2_{\mathbb{Q}} : vX-uZ=0\}. \end{align*} A point of $L_{[u:v]}$ can be written in the form \begin{align*} [\lambda u:\mu:\lambda v], \end{align*} where $[\lambda:\mu]\in \mathbb{P}^1$ are homogeneous coordinates on the line. The value $\lambda=0$ gives $P=[0:1:0]$. Define homogeneous forms \begin{align*} H(u,v) &:= Bu+Dv,\\ G(u,v) &:= Au^2+C_1uv+Ev^2. \end{align*} Substitution into the conic equation gives \begin{align*} F(\lambda u,\mu,\lambda v) &=\lambda^2G(u,v)+\lambda\mu H(u,v)\\ &=\lambda\bigl(\lambda G(u,v)+\mu H(u,v)\bigr). \end{align*} Thus the factor $\lambda=0$ records the known intersection point $P$, and the residual intersection is determined by \begin{align*} \lambda G(u,v)+\mu H(u,v)=0. \end{align*} [/step]

[step:Construct the residual intersection map]Define \begin{align*} \varphi:\mathbb{P}^1_{\mathbb{Q}} &\to \mathbb{P}^2_{\mathbb{Q}}\\ [u:v] &\mapsto [uH(u,v):-G(u,v):vH(u,v)]. \end{align*} The three coordinates are homogeneous quadratic forms in $u$ and $v$. They do not vanish simultaneously: if $H(u,v)=0$ and $G(u,v)=0$ for some $[u:v]$, then \begin{align*} F(\lambda u,\mu,\lambda v)=0 \end{align*} for all $[\lambda:\mu]\in \mathbb{P}^1$, so the whole line $L_{[u:v]}$ would be contained in $C$. A quadratic plane curve containing a line has that line as a component, so it is reducible over an [algebraic closure](/page/Algebraic%20Closure) and is singular at the intersection point of its two linear components, or everywhere along the line in the double-line case. This contradicts the nonsingularity of $C$. For each $[u:v]$, the point $\varphi([u:v])$ lies on $L_{[u:v]}$ because its first and third coordinates are $uH(u,v)$ and $vH(u,v)$. It also lies on $C$: using the substitution formula with \begin{align*} \lambda=H(u,v),\qquad \mu=-G(u,v), \end{align*} we obtain \begin{align*} F(uH(u,v),-G(u,v),vH(u,v)) =H(u,v)\bigl(H(u,v)G(u,v)-G(u,v)H(u,v)\bigr)=0. \end{align*} Therefore $\varphi$ is a $\mathbb{Q}$-defined morphism \begin{align*} \varphi:\mathbb{P}^1_{\mathbb{Q}}\to C. \end{align*}[/step]

[guided]The residual root equation \begin{align*} \lambda G(u,v)+\mu H(u,v)=0 \end{align*} is homogeneous in $[\lambda:\mu]$. A convenient solution is \begin{align*} [\lambda:\mu]=[H(u,v):-G(u,v)]. \end{align*} Substituting this solution back into the parametrization of the line gives \begin{align*} [u:v]\mapsto [uH(u,v):-G(u,v):vH(u,v)]. \end{align*} This motivates the definition \begin{align*} \varphi:\mathbb{P}^1_{\mathbb{Q}} &\to \mathbb{P}^2_{\mathbb{Q}}\\ [u:v] &\mapsto [uH(u,v):-G(u,v):vH(u,v)]. \end{align*} We must check that this formula defines a projective morphism. First, all three coordinates are homogeneous of degree two in $u$ and $v$: $H$ has degree one, while $G$ has degree two. Second, the three coordinates must not vanish simultaneously. If they did vanish at some $[u:v]$, then $H(u,v)=0$ and $G(u,v)=0$. The substitution identity would become \begin{align*} F(\lambda u,\mu,\lambda v)=0 \end{align*} for every $[\lambda:\mu]\in \mathbb{P}^1$, meaning that the entire line $L_{[u:v]}$ lies inside $C$. A plane quadratic containing a line has that line as a component. Over an algebraic closure, such a quadratic is either a product of two linear forms or a square of one linear form; in both cases the curve is singular, either at the intersection of the two components or along the whole double line. This contradicts the hypothesis that $C$ is nonsingular. Now we verify that the image actually lies in $C$. The point $\varphi([u:v])$ has the form \begin{align*} [\lambda u:\mu:\lambda v] \end{align*} with $\lambda=H(u,v)$ and $\mu=-G(u,v)$, so it lies on the line $L_{[u:v]}$. Substituting these values into the factorization gives \begin{align*} F(uH(u,v),-G(u,v),vH(u,v)) =H(u,v)\bigl(H(u,v)G(u,v)-G(u,v)H(u,v)\bigr)=0. \end{align*} Hence $\varphi([u:v])\in C$ for every $[u:v]$, and the formula defines a morphism over $\mathbb{Q}$ because all coefficients belong to $\mathbb{Q}$.[/guided]

[step:Identify the tangent parameter with the repeated intersection at $P$] The tangent line to $C$ at $P$ is defined by the linear part of $F$ at $P$. In the chosen coordinates it is \begin{align*} T_PC=\{[X:Y:Z]\in \mathbb{P}^2_{\mathbb{Q}}: BX+DZ=0\}. \end{align*} For the line $L_{[u:v]}$, this tangent condition is exactly \begin{align*} H(u,v)=Bu+Dv=0. \end{align*} When $H(u,v)=0$, the substitution formula becomes \begin{align*} F(\lambda u,\mu,\lambda v)=\lambda^2G(u,v). \end{align*} Since $G(u,v)\ne0$ by the no-line-component argument above, the only intersection point is $\lambda=0$, namely $P$, with multiplicity two. The parametrization gives \begin{align*} \varphi([u:v])=[0:-G(u,v):0]=P. \end{align*} Thus the tangent line corresponds exactly to the parameter value returning $P$ with multiplicity two. [/step]

[step:Recover every rational point from its line through $P$]Let $Q=[x:y:z]\in C(\mathbb{Q})$ with $Q\ne P$. Then $(x,z)\ne(0,0)$, so define \begin{align*} [u:v]:=[x:z]\in \mathbb{P}^1(\mathbb{Q}). \end{align*} This parameter represents the rational line through $P$ and $Q$. Since $Q\in C$, we have \begin{align*} 0=F(x,y,z)=G(x,z)+yH(x,z). \end{align*} If $H(x,z)=0$, then this equation gives $G(x,z)=0$, contradicting the fact that $H$ and $G$ do not vanish simultaneously at a projective parameter. Therefore $H(x,z)\ne0$, and \begin{align*} y=-\frac{G(x,z)}{H(x,z)}. \end{align*} Hence \begin{align*} \varphi([x:z]) &=[xH(x,z):-G(x,z):zH(x,z)]\\ &=[x:y:z]\\ &=Q. \end{align*} Therefore every rational point of $C$ other than $P$ is obtained as the second intersection point of $C$ with the rational line through $P$ and that point.[/step]

[guided]Take any rational point \begin{align*} Q=[x:y:z]\in C(\mathbb{Q}) \end{align*} with $Q\ne P$. Since $P=[0:1:0]$, the condition $Q\ne P$ means that $x$ and $z$ are not both zero. Thus \begin{align*} [u:v]:=[x:z]\in \mathbb{P}^1(\mathbb{Q}) \end{align*} is a well-defined parameter. Geometrically, this is the line through $P$ and $Q$, because both points satisfy the equation $zX-xZ=0$. Now use the fact that $Q$ lies on the conic. Substituting $X=x$, $Y=y$, and $Z=z$ into the defining equation gives \begin{align*} 0=F(x,y,z)=G(x,z)+yH(x,z). \end{align*} We need $H(x,z)\ne0$ in order to solve for $y$. If $H(x,z)=0$, then the displayed equation forces $G(x,z)=0$. But we already proved that there is no projective parameter at which $H$ and $G$ vanish simultaneously, since such a parameter would make the corresponding line a component of the nonsingular conic. Therefore $H(x,z)\ne0$. Solving the equation gives \begin{align*} y=-\frac{G(x,z)}{H(x,z)}. \end{align*} Substituting this into the formula for $\varphi$ yields \begin{align*} \varphi([x:z]) &=[xH(x,z):-G(x,z):zH(x,z)]\\ &=[x:y:z], \end{align*} where the second equality follows by scaling the homogeneous coordinates by the nonzero scalar $H(x,z)$. Hence the point $Q$ is exactly the residual intersection of $C$ with the rational line through $P$ and $Q$.[/guided]

[step:Read the projective parametrization as the usual affine slope formula] In an affine chart containing $P$ and a chosen non-tangent affine direction, the parameter $[u:v]\in\mathbb{P}^1(\mathbb{Q})$ becomes the usual slope parameter $t\in\mathbb{Q}$ on the corresponding affine lines through $P$. The remaining projective value $[u:v]$ with denominator zero represents the vertical direction in that affine chart. If a residual point lies on the line at infinity of the chosen chart, it is still represented by the same projective formula \begin{align*} \varphi([u:v])=[uH(u,v):-G(u,v):vH(u,v)]. \end{align*} Thus the affine slope parametrization is precisely the restriction of the projective parametrization, with the vertical line and points at infinity accounted for by the projective parameter space $\mathbb{P}^1(\mathbb{Q})$. This proves the theorem. [/step]