I'm trying to solve the above exercise in Ghatmann's notes on algebraic geometry.
I tried to consider the real case and I suspect that X1 should correspond to a plane that intersects with exactly one of the cones, whereas X2 is the case in which both cones are intersected (not at the origin).
Next i tried to look at the general case of a quadratic polynomial in two variables to see what happens if I act on it with an affine change of coordinates(i.e. linear transformation followed by translation), this quickly got very complicated.
My guess is that I need to figure out how the irreducibility condition affects the general equation, and maybe extend the intuition I gained for the real case to arbitrary fields(of characteristic $\ne$ 2).
However, right now I'm kind of stuck, so any help would be appreciated. Thanks.
Note that the field $K$ is assumed to be algebraically closed.
Assume $K$ is algebraically closed, with characteristic not $2$.
Suppose your irreducible conic is given by $ax^2+bxy+cy^2+dx+ey+f=0$ (up to a constant factor).
Consider the quadratic form $ax^2+bxy+cy^2$. Since $K$ is algebraically closed, this factors into a product of homogeneous linear forms, $a\ell_1\ell_2$, with $\ell_1$ and $\ell_2$ monic in $x$. If $\ell_1=\ell_2$, then let $dx+ey=d\ell_3$. If $\ell_3=\ell_1$ as well, then we could rewrite our equation as $a\ell_3^2 + d\ell_3 + f=0$, which is reducible, since $K$ is algebraically closed. Otherwise, $\ell_3 \ne \ell_1$, so let $x'=\sqrt{-a}\ell_1$, $y'=d\ell_3 + f$ to get that our original conic is given by $y' - x'^2$ in our new variables.
Otherwise, if $\ell_1\ne \ell_2$, take $x_1=a\ell_1$, $y_1=\ell_2$, and rewrite our conic in terms of $x_1$ and $y_1$, so that it is of the form $$x_1y_1+Ax_1 + By_1+C =0.$$ Let $x_2=x_1+B$, $y_2=y_1+A$, so that the conic can be rewritten as $$x_2y_2 + (C-AB)=0.$$ If $C-AB=0$, then the conic is reducible, so finally let $x'=x_2$, $y'=\frac{-1}{C-AB}y_2$, so that $$x'y' -1=0,$$ as desired.