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.

You know that $P(x,y)$ and $a$ have the same sign, so $b$ has the same sign as $a$ (consider $P(0,1)$, so you know that $ab \geq 0$. Next:
case 1: $xy\leq 0$
$$P(x,y)=ax^2+2hxy+by^2=ax^2+by^2+2\sqrt{ab}xy-2\sqrt{ab}xy+2hxy= \\ =(\sqrt{|a|}x+\sqrt{|b|}y)^2-2\sqrt{ab}xy+2hxy=(\sqrt{|a|}x+\sqrt{|b|}y)^2+2xy(h-\sqrt{ab})$$
You know that $ab>h^2$, so $h-\sqrt{ab} \leq 0$, so $xy(h-\sqrt{ab}) \geq 0$.
case 2: $xy \geq 0$
$$P(x,y)=ax^2+2hxy+by^2=ax^2+by^2+2\sqrt{ab}xy-2\sqrt{ab}xy+2hxy= \\ =(\sqrt{|a|}x-\sqrt{|b|}y)^2+2\sqrt{ab}xy+2hxy=(\sqrt{|a|}x-\sqrt{|b|}y)^2+2xy(h+\sqrt{ab})$$
You know that $ab>h^2$, so $h+\sqrt{ab} \geq 0$, so $xy(h-\sqrt{ab}) \geq 0$.
