Let $P(x,y)=Ax^2 + 2Bxy + Cy^2$, and set
$\triangle = \frac{(P_{12})^2 - P_{11}P_{22}}{4}=B^{2}-AC$
Then,
(i) If $\triangle >0$, then there are two lines through the origin such that $P(x,y)>0$ for all $(x,y)$ on one, and $P(x,y)<0$ for all $(x,y)$ on the other, with the point $(0,0)$ omitted.
To prove (i), we assume that $\triangle>0$ and compute
$P(B,-A) = AB^2 -2B^2A + CA^2= -A \triangle$
$P(C,-B) = AC^2 -2B^2C + CB^2 = -C\triangle$
Since $P$ is homogeneous of degree $2$, $P(\lambda x,\lambda y) = \lambda^{2}P(x,y)$, it will be sufficient if we can find two points at which $P$ has opposite signs, since the same will then hold on the entire line joining these to the origin.
I was wondering if someone could further elaborate on the last sentence. Let's say that we find two points at which $P$ has opposite signs, why would the same hold on the entire line joining these to the origin? And why does this follow from homogeneity?