 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 $\alpha :L\to C$ 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 givesTaking the second intersection gives the explicit mapThe 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 $t\ne 0$. The full affine list istogether 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 all rational solutions.