Prove the sign and zeroes of $Ax^2 + 2Bxy + Cy^2$ (without using the second derivative test)

Question

Let $$P(x,y) = Ax^2 + 2Bxy + Cy^2 \\A \neq 0 \quad x,y \in \mathbb R.$$ Without using the second derivative test, prove that

1. If $AC - B^2 > 0$, then
   (i) $P$ has no zeroes outside the origin and
   (ii) $\operatorname{sgn}(P) = \operatorname{sgn}(A)$ for all $x,y$
2. If $AC - B^2 = 0$, then
   (i) $P$ has zeroes outside the origin and
   (ii) outside these zeroes, $\operatorname{sgn}(P) = \operatorname{sgn}(A)$
3. If $AC - B^2 < 0$, then
   (i) $P$ has zeroes outside the origin
   (ii) there exist $x,y$ such that $\operatorname{sgn}(P) = \operatorname{sgn}(A)$ and
   (iii) there also exist $x,y$ such that $\operatorname{sgn}(P) = -\operatorname{sgn}(A)$

Please note:

1. The second derivative test of calculus is not available, as this question is intended as a lemma for proving the second derivative test
2. This post is related but distinct

My proof is below: I request verification, critique, or alternate proofs.

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Let $$f(x,y) = (x -\frac B A y)^2 \\g(y) = \left(\frac y A \right)^2.$$ Then $$P(x,y) = A \cdot [f(x,y) + (AC-B^2)g(y)]\\ \operatorname{sgn}(P(x,y)) = \operatorname{sgn}(A) \cdot \operatorname{sgn}[f(x,y) + (AC-B^2)g(y)].$$

Observe that the range of both $f$ and $g$ is $[0, \infty)$.

Case 1: If $AC - B^2 > 0$, then $\operatorname{sgn}(P) = \operatorname{sgn}(A)$ unless both $f$ and $g$ are zero. But $g(y) = 0$ implies $y = 0$, and $f(x,0) = 0$ implies $x = 0$, so this can only happen at the origin.

Case 2: If $AC - B^2 = 0$, then $\operatorname{sgn}(P) = \operatorname{sgn}(A)$ unless $f$ is zero, which happens for infinite values of $x, y$.

Case 3: If $AC - B^2 < 0$, then for $|x| \gg |y|$, the first term dominates, and $\operatorname{sgn}(P) = \operatorname{sgn}(A)$, and for $|y| \gg |x|$, the second term dominates, and $\operatorname{sgn}(P) = -\operatorname{sgn}(A)$. Furthermore, for any fixed $y$, then varying $x$ causes $f$ to take on its entire range, so there exist infinite values of $x,y$ which make $P$ zero.

Answer 1

As an alternative, we could proceed as follows, for $y\neq 0$ by $t=\frac x y$

$$P(x,y)=Ax^2 + 2Bxy + Cy^2=y^2\left(At^2+2Bt+C\right)=y^2p(t)$$

then the sign of $P(x,y)$ is equal to the sign of $p(t)$ which can be studied by its discriminant $4B^2-4AC$.

The case $y=0$ can be considered a part.