= Solution
For every $q\in L$, the line $pq$ meets the nonsingular <conic> in two points counted with multiplicity by the <Bézout theorem>. One is $p$; define $\alpha(q)$ as the second, using $\alpha(q)=p$ for the unique tangent line at $p$. Conversely, for $r\in C\setminus\{p\}$, intersect the line $pr$ with $L$, and send $p$ itself to the intersection of its tangent with $L$. Since $p\notin L$, each intersection with $L$ is unique. These constructions are inverse, so \b[$\alpha:L\to C$ is a bijection] preserving the stated collinearity. This is <parametrization of a conic by projection>.
For the particular <conic> take $p=[1:0:1]$ and $L=\{x=0\}$. If $q=[0:s:t]$, points on $pq$ have coordinates $[a:bs:a+bt]$. Substitution into the <conic> gives
$$
b\bigl[-a(s+2t)+b(s^2-t^2)\bigr]=0.
$$
Taking the second intersection gives the explicit map
$$
\boxed{\alpha([0:s:t])=[s^2-t^2:s(s+2t):s^2+st+t^2].}
$$
The three coordinates do not vanish simultaneously. If $s+2t=0$, the formula gives $p$, as required for the tangent direction.
For rational points, put $r=s/t$ when $t\ne0$. The full affine list is
$$
\boxed{x=\frac{r^2-1}{r^2+r+1},\qquad y=\frac{r(r+2)}{r^2+r+1},\qquad r\in\mathbb Q,}
$$
together with the parameter $r=\infty$, which gives $(x,y)=(1,1)$. The denominator is $(r+1/2)^2+3/4>0$ for rational $r$. Conversely, every rational point other than $p$ determines a rational joining line and hence a rational point of $L$; the point $p$ is obtained at $r=-2$. Thus this lists \b[all rational solutions].
\Image[/past-exam-of-the-geometry-course-of-the-university-of-oxford/2009-conic-projection.png]
{title=Projection from a rational conic point to a line}
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