if $Ax^2+Bxy+Cy^2$ is positive definite, then $A>0$ and $B^2-4AC<0$

Solution 1:

$f(x, y) = Ax^2 + Bxy + Cy^2$ 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.

if $Ax^2+Bxy+Cy^2$ is positive definite, then $A>0$ and $B^2-4AC<0$

Solution 1:

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