if $Ax^2+Bxy+Cy^2$ is positive definite, then $A>0$ and $B^2-4AC<0$ if $Ax^2+Bxy+Cy^2$ is positive definite, then $A>0$ and $B^2-4AC<0$.
$Ax^2+Bxy+Cy^2=A(x+\frac{B}{2A}y)^2-\frac{B^2-4AC}{4A}y^2$, hence under the condition $A>0$ and $B^2-4AC<0$, the quadratic form is positive definition. But I am wondering why it is necessary for $A>0$ and $B^2-4AC<0$ if the quadratic form is positive definite.

$f(x, y) = Ax^2 + Bxy + Cy^2$ must be positive for $(x, y) \ne (0, 0)$, in particular:

*

*$f(1, 0) = A > 0$,

*$f(B, -2A) = A B^2 -2AB^2 +4A^2C = A ( 4AC - B^2) > 0$.

Note that in the second line $(x, y)$ is chosen such that the quadratic term $(x+\frac{B}{2A}y)^2$ vanishes.
Alternatively you can argue that the matrix
$$
 \begin{pmatrix} A & B/2 \\ B/2 & C \end{pmatrix}
$$
is positive definite if and only if all leading principal minors are positive, that is Sylvester's criterion.
