Show that $ax^2+2hxy+by^2$ is positive definite when $h^2The question asks to "show that the condition for $P(x,y)=ax^2+2hxy+by^2$ ($a$,$b$ and $h$ not all zero) to be positive definite is that $h^2<ab$, and that $P(x,y)$ has the same sign as $a$." 
Now I've seen questions similar to this before where it's a two variable quadratic and I'm not too sure how to go about it. Normally with one variable you could just show the discriminant is less than $0$ but since there's two variables I can't use the same process (since I'm not sure how the discriminant is defined for a two variable quadratic). I also tried completing the square but that doesn't seem to make the algebra anything near as simple as the answer (I got a rather complicated fraction with cubes). Is there a way to go about this type of problem? 
Thanks.

When $a,b,h$ are all positive,
$$
ax^2+2hxy+by^2 = h \bigg( \Big(\frac ab\Big)^{1/4}x + \Big(\frac ba\Big)^{1/4}y \bigg)^2 + \Big(1-\frac{h}{\sqrt{ab}}\Big)(ax^2 + by^2).
$$
If at least one of $a,b,h$ is negative, change some signs accordingly.
