I know the very basic way to find the b in this quadratic expression: $$P(x,y)=ax^2+bxy+ cy^2$$ I can first evaluate $P(0,1)=c$. Similarly, I can do $P(1,0)=a$ and then I can do $\frac{P(1,1)-P(1,-1)}{2}$ to find $b$. This took three steps.
Is there a more efficient way to do this? My Prof. gave a hint to use this fact: $$P(tx_1,...,tx_n)= t^nP(x_1,...,x_n)$$
I'm afraid I have to disagree with @Henry.
$P(1,1)=a+b+c$ and $P(-1,-1)=a+b+c$ and you cannot do anything with that to find $b$ (unless you already know $a$ and $c$).
What you want to know is $P(1,1)=a+b+c$ and $P(1,-1)=a-b+c$
Then $P(1,1)-P(1,-1)=...$ will give you a way in to finding $b$.