Let $p_1,\ p_2,\ p_3,\ p_4$ be any 4 different points on $\mathbb{CP}^1$ and $x_1,\ x_2,\ x_3,\ x_4$ are 4 different points on $\mathbb{CP}^2$.

How can I show that there is unique conic $Q$ passing through $x_i$ and isomorphism $f: Q\to\mathbb{CP}^1$ such that $f(x_i)=p_i$ for all $i$?


I'd start with the second part, the isomorphism. For that I'd use cross ratios, since these are about the most fundamental concept which stays invariant under projective transformations. So you can define


Here the left hand side is a cross ratio of four points on a conic, which is defined as the cross ratio of the four lines connecting these points to some other point on the conic. Personally, I'd consider $x_1$ and $p_1$ as “$\infty$”, $x_2$ resp. $p_2$ as “$0$” and $x_3$ resp. $p_3$ as “$1$”. Then the left hand side means that you compute the cross ratio of $q$ with respect to the basis formed by the $x_i$, and the right hand side indicates that the image of $q$ is simply the value you just computed, but marked with respect to the projective basis of the $p_i$.

In all of this, I didn't use $x_4$ and $p_4$. Furthermore, there is a one-parameter family of conics passing through four given points. So from that family, you'd have to chose that member which satisfies $f(x_4)=p_4$. So start with the right hand side, i.e. compute


All points on a conic see four given points under a fixed cross ratio, so this gives you a definition of the conic:


The notation here indicates a cross ratio as seen from a given point, which again is the cross ratio of the lines through that point. Writing determinants as brackets, you can compute this as


This is a quadratic equation in $q$, hence a conic. If you take it all together, you get $Q$ defined as the set of points $q$ which satsifies

$$ [p_1,p_3,q][p_2,p_4,q][x_1,x_4][x_2,x_3]= [p_1,p_4,q][p_2,p_3,q][x_1,x_3][x_2,x_4] $$


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