[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 $(B,D)\ne (0,0)$.
[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]
[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*}
[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]
[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.
[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]
[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]
