I read this in Reid's book Undergraduate Algebraic Geometry. I don't even know whether my question is worded correctly. It says "In a suitable coordinate system, any conic in $P_{\mathbb{R}}^2$ is one of the following":
$$(1)\quad \text{nondegenerate conic: }X^2+Y^2-Z^2=0$$ $$(2)\quad \text{empty set: }X^2+Y^2+Z^2=0$$ $$(3)\quad \text{line pair: }X^2-Y^2=0$$ $$(4)\quad \text{one point $(0,0,1)$: }X^2+Y^2=0$$ $$(5)\quad \text{double line: }X^2=0$$
It then asks to compare with them the cases in $\mathbb{R}^2$. I am pretty sure that the ellipse, parabola, and hyperbola are corresponding to case (1).
$x^2+y^2=0$ corresponds to $X^2+Y^2=0, Z=1$, so case (4).
$x^2+y^2=-1$ corresponds to $X^2+Y^2+Z^2=0$, so case (2).
For $x^2=-1$, I can write it as $X^2+Z^2=0$. What does that correspond to? It seems like the Y axis in the plane $Z=0$. Is it the one point case? It is then a point at infinity, is it right?
For $0=-1$, is it $Z=0$ or $Z^2=0$? So double line, case (5)? Is it the line contains all points at infinity? Thank you very much for your help!
$x^2=-1$ is a conjugate complex pair of lines, namely $x=\pm i$. They intersect at the point at infinity in the $y$ direction, i.e. have the real point $(0,1,0)$ in common. Projectively speaking, $X^2+Z^2=0$ is no different from $X^2+Y^2=0$, so they both are $(4)$. But in $\mathbb R^2$ the two situations are distinct since you don't “see” the solution at infinity there. In $\mathbb R^2$ would “look” like an empty set again.
$0=-1$ is certainly the empty set, but the projective formulation $Z^2=0$ describes the double line at infinity, so that's case $(5)$ again.