[proofplan] Translate coordinates so that the known rational point is the origin, then write the quadratic equation of the conic as a linear part plus a homogeneous quadratic part. A line through the origin with rational direction intersects the conic in the known point and, unless the direction is exceptional, one further point obtained by an explicit rational formula. Conversely, any rational point different from the known point determines a rational direction, so the construction recovers it. The exceptional directions are exactly the finitely many directions where the denominator in the formula vanishes or where the direction is treated by the point at infinity of the pencil. [/proofplan] [step:Translate the conic so that the rational point is the origin] Let $P=(a,b) \in C(\mathbb{Q})$. Define the translated polynomial \begin{align*} G: \mathbb{A}^2_{\mathbb{Q}} &\to \mathbb{A}^1_{\mathbb{Q}} \\ (u,v) &\mapsto F(a+u,b+v). \end{align*} Since $F(a,b)=0$ and $F$ has degree $2$, there are uniquely determined forms \begin{align*} \ell: \mathbb{A}^2_{\mathbb{Q}} &\to \mathbb{A}^1_{\mathbb{Q}},\\ q: \mathbb{A}^2_{\mathbb{Q}} &\to \mathbb{A}^1_{\mathbb{Q}} \end{align*} with $\ell$ homogeneous of degree $1$ and $q$ homogeneous of degree $2$ such that \begin{align*} G(u,v)=\ell(u,v)+q(u,v). \end{align*} The nonsingularity of $P$ means that the linear form $\ell$ is not the zero form. [guided] We move the known rational point to the origin because every line through $P$ then becomes a one-dimensional subspace through $(0,0)$ in the translated coordinates. Define \begin{align*} G: \mathbb{A}^2_{\mathbb{Q}} &\to \mathbb{A}^1_{\mathbb{Q}} \\ (u,v) &\mapsto F(a+u,b+v). \end{align*} The coefficients of $G$ are rational because $F$ has coefficients in $\mathbb{Q}$ and $a,b \in \mathbb{Q}$. Also $G(0,0)=F(a,b)=0$, so $G$ has no constant term. Since $F$ has total degree $2$, the translated polynomial $G$ has total degree at most $2$, and therefore decomposes uniquely into its homogeneous degree-$1$ and degree-$2$ parts: \begin{align*} G(u,v)=\ell(u,v)+q(u,v). \end{align*} Here $\ell: \mathbb{A}^2_{\mathbb{Q}} \to \mathbb{A}^1_{\mathbb{Q}}$ is a rational linear form and $q: \mathbb{A}^2_{\mathbb{Q}} \to \mathbb{A}^1_{\mathbb{Q}}$ is a rational homogeneous quadratic form. The condition that $P$ is nonsingular is exactly the condition that the first-order part of $F$ at $P$ is nonzero, so $\ell \neq 0$. [/guided] [/step] [step:Compute the second intersection with a line of rational direction] Let $d=(m,n) \in \mathbb{Q}^2 \setminus \{(0,0)\}$ be a rational direction. Define the affine line through the translated origin in direction $d$ by \begin{align*} L_d: \mathbb{A}^1_{\mathbb{Q}} &\to \mathbb{A}^2_{\mathbb{Q}} \\ r &\mapsto (rm,rn). \end{align*} Substituting this line into $G$ gives \begin{align*} G(rm,rn)=r\ell(m,n)+r^2q(m,n) =r\bigl(\ell(m,n)+r q(m,n)\bigr). \end{align*} The root $r=0$ is the known point $P$. If $q(m,n)\neq 0$, the second root is \begin{align*} r_d=-\frac{\ell(m,n)}{q(m,n)} \in \mathbb{Q}. \end{align*} Thus the corresponding second point on $C$ is \begin{align*} \varphi([m:n]) = \left(a+r_d m,\ b+r_d n\right) = \left(a-\frac{\ell(m,n)m}{q(m,n)},\ b-\frac{\ell(m,n)n}{q(m,n)}\right). \end{align*} [guided] A direction through the origin is represented by a nonzero vector $d=(m,n) \in \mathbb{Q}^2$. The same geometric direction is unchanged if we multiply $(m,n)$ by a nonzero rational scalar, so the correct parameter space of directions is $\mathbb{P}^1_{\mathbb{Q}}$. For such a direction define the line \begin{align*} L_d: \mathbb{A}^1_{\mathbb{Q}} &\to \mathbb{A}^2_{\mathbb{Q}} \\ r &\mapsto (rm,rn). \end{align*} Substituting this parametrized line into the translated conic gives \begin{align*} G(rm,rn) &=\ell(rm,rn)+q(rm,rn)\\ &=r\ell(m,n)+r^2q(m,n)\\ &=r\bigl(\ell(m,n)+r q(m,n)\bigr), \end{align*} where we used the degree-$1$ homogeneity of $\ell$ and the degree-$2$ homogeneity of $q$. The factor $r$ corresponds to the known point $(0,0)$, hence to $P$ in the original coordinates. When $q(m,n)\neq 0$, the other root is obtained by solving the remaining linear equation \begin{align*} \ell(m,n)+r q(m,n)=0. \end{align*} Thus \begin{align*} r_d=-\frac{\ell(m,n)}{q(m,n)} \in \mathbb{Q}. \end{align*} Because $\ell(m,n)$ and $q(m,n)$ are rational numbers, this second intersection parameter is rational. Translating back to the original coordinates gives a rational point of $C$: \begin{align*} \varphi([m:n]) = (a+r_d m,\ b+r_d n). \end{align*} This formula depends only on the projective direction $[m:n]$, because replacing $(m,n)$ by $\lambda(m,n)$ changes $r_d$ to $r_d/\lambda$ and leaves $(r_d m,r_d n)$ unchanged. [/guided] [/step] [step:Exclude only finitely many exceptional directions] The homogeneous quadratic form $q$ is nonzero because translation preserves the degree-$2$ part of the irreducible conic equation $F$. Hence the condition $q(m,n)=0$ cuts out a homogeneous quadratic equation on $\mathbb{P}^1_{\mathbb{Q}}$, and therefore has at most two geometric solutions. The chosen affine slope chart omits only the vertical direction $[0:1]$. Thus the directions where the displayed affine formula is not available form a finite set. The hypothesis that no line through $P$ is contained in $C$ rules out a direction $d=(m,n)$ with $\ell(m,n)=q(m,n)=0$: in that case $G(rm,rn)=0$ for every $r \in \mathbb{A}^1_{\mathbb{Q}}$, so the whole line $L_d(\mathbb{A}^1_{\mathbb{Q}})$ would lie in the translated conic, and translating back would give a line through $P$ contained in $C$. If $\ell(m,n)=0$ and $q(m,n)\neq 0$, then $r_d=0$, so the line is tangent at $P$ and the second intersection coincides with $P$ with multiplicity. This accounts for the tangent direction among the exceptional directions when one wants a parametrization of $C \setminus \{P\}$. [/step] [step:Recover every rational point away from the exceptional set from its direction] Let $Q=(x,y) \in C(\mathbb{Q})$ with $Q \neq P$. Define the translated vector \begin{align*} w := (x-a,\ y-b) \in \mathbb{Q}^2 \setminus \{(0,0)\}. \end{align*} Then $Q$ lies on the line through $P$ with projective direction $[w] \in \mathbb{P}^1_{\mathbb{Q}}$. Since $Q \in C(\mathbb{Q})$, we have \begin{align*} 0=G(w)=\ell(w)+q(w). \end{align*} If $q(w)\neq 0$, then the second-intersection formula gives \begin{align*} r_w=-\frac{\ell(w)}{q(w)}=1, \end{align*} so $\varphi([w])=Q$. Thus every rational point different from $P$ and not lying in an exceptional direction is recovered from its rational direction. [guided] Now take an arbitrary rational point $Q=(x,y) \in C(\mathbb{Q})$ with $Q \neq P$. The line through $P$ and $Q$ has rational direction because both endpoints have rational coordinates. In translated coordinates this direction is represented by \begin{align*} w := (x-a,\ y-b) \in \mathbb{Q}^2 \setminus \{(0,0)\}. \end{align*} Since $Q$ lies on $C$, the translated point $w$ lies on the translated conic, so \begin{align*} 0=G(w)=\ell(w)+q(w). \end{align*} If this direction is not exceptional, then $q(w)\neq 0$. The second-intersection parameter attached to the direction $[w]$ is therefore \begin{align*} r_w=-\frac{\ell(w)}{q(w)}. \end{align*} But the equation $\ell(w)+q(w)=0$ gives $-\ell(w)/q(w)=1$. Hence the construction sends the direction $[w]$ to the translated point $w$, and after translating back it sends $[w]$ to $Q$. This proves that the parametrization recovers every rational point away from $P$ and the finitely many exceptional directions. [/guided] [/step] [step:Conclude that projection from the rational point gives a birational parametrization] Define the open subset of directions \begin{align*} U := \{[m:n] \in \mathbb{P}^1_{\mathbb{Q}} : q(m,n) \neq 0 \text{ and } \ell(m,n) \neq 0\}. \end{align*} On $U$, the assignment \begin{align*} [m:n]\mapsto (a+r_d m,\ b+r_d n) \end{align*} is given by rational functions in $m$ and $n$, so it defines a morphism $\varphi:U \to C$. Define the open subset of the conic \begin{align*} V := \{(x,y) \in C : q(x-a,y-b) \neq 0\}. \end{align*} For $[m:n] \in U$, the translated image vector is $r_d(m,n)$, and since $r_d=-\ell(m,n)/q(m,n)$ with $\ell(m,n)\neq 0$, it is nonzero and satisfies $q(r_d m,r_d n)=r_d^2q(m,n)\neq 0$. Hence $\varphi(U)\subseteq V$. Define projection from $P$ on $V$ by \begin{align*} \pi:V &\to U \\ (x,y) &\mapsto [x-a:y-b]. \end{align*} This is a morphism because $q(x-a,y-b)\neq 0$ implies $(x-a,y-b)\neq (0,0)$, and the defining equation $\ell(x-a,y-b)+q(x-a,y-b)=0$ implies $\ell(x-a,y-b)\neq 0$. For $[m:n] \in U$, homogeneity gives $\pi(\varphi([m:n]))=[r_d m:r_d n]=[m:n]$. Conversely, for $(x,y)\in V$, set $w=(x-a,y-b)$. Since $G(w)=0$ and $q(w)\neq 0$, the construction gives $r_w=-\ell(w)/q(w)=1$, so $\varphi(\pi(x,y))=(x,y)$. Thus $\varphi:U\to V$ and $\pi:V\to U$ are inverse morphisms on nonempty Zariski-open subsets, proving that the rational map $\mathbb{P}^1_{\mathbb{Q}}\dashrightarrow C$ is a rational parametrization. The rational-point statement follows from the same inverse calculation, with the excluded points exactly $P$ and the finitely many exceptional directions described above. [guided] We now make the birational statement precise by naming the actual open sets. Define \begin{align*} U := \{[m:n] \in \mathbb{P}^1_{\mathbb{Q}} : q(m,n) \neq 0 \text{ and } \ell(m,n) \neq 0\}. \end{align*} This is a Zariski-open subset because it is the complement of the zero loci of the homogeneous forms $q$ and $\ell$. On this set the formula \begin{align*} [m:n]\mapsto (a+r_d m,\ b+r_d n), \qquad r_d=-\frac{\ell(m,n)}{q(m,n)}, \end{align*} is regular in projective coordinates on the open where $q\neq 0$, and the condition $\ell\neq 0$ removes the tangent direction that would map back to $P$. Define the corresponding open subset of the conic by \begin{align*} V := \{(x,y) \in C : q(x-a,y-b) \neq 0\}. \end{align*} If $[m:n]\in U$, the translated image vector is $r_d(m,n)$. Since $r_d=-\ell(m,n)/q(m,n)$ and $\ell(m,n)\neq 0$, we have $r_d\neq 0$. Homogeneity of $q$ gives \begin{align*} q(r_d m,r_d n)=r_d^2q(m,n)\neq 0, \end{align*} so the image lies in $V$. The inverse map is projection from the known point. Define \begin{align*} \pi:V &\to U \\ (x,y) &\mapsto [x-a:y-b]. \end{align*} This is well-defined because $q(x-a,y-b)\neq 0$ implies $(x-a,y-b)\neq (0,0)$. Moreover, if $w=(x-a,y-b)$, then $G(w)=0$ gives $\ell(w)+q(w)=0$; since $q(w)\neq 0$, also $\ell(w)\neq 0$, so $[w]\in U$. Finally we verify the two compositions. For $[m:n]\in U$, the translated image is $r_d(m,n)$ with $r_d\neq 0$, hence \begin{align*} \pi(\varphi([m:n]))=[r_d m:r_d n]=[m:n]. \end{align*} For $(x,y)\in V$, let $w=(x-a,y-b)$. Since $G(w)=0$ and $q(w)\neq 0$, the second-intersection parameter attached to $[w]$ is \begin{align*} r_w=-\frac{\ell(w)}{q(w)}=1. \end{align*} Therefore $\varphi(\pi(x,y))=(x,y)$. The maps $\varphi:U\to V$ and $\pi:V\to U$ are inverse morphisms on Zariski-open subsets, so the rational map $\mathbb{P}^1_{\mathbb{Q}}\dashrightarrow C$ is a rational parametrization. On rational points, the same computation recovers every point except the known point $P$ and the finitely many excluded directions. [/guided] [/step]