Define $a@b = ab - b^{2}$ and $a\#b = a + b - ab^{2}$. What is $\frac {6@2}{6\#2}$?
$\text{(A)}\ - \frac {1}{2}\qquad \text{(B)}\ - \frac {1}{4}\qquad \text{(C)}\ \frac {1}{8}\qquad \text{(D)}\ \frac {1}{4}\qquad \text{(E)}\ \frac {1}{2}$
$6@2$ must be equal to $6*2-2^2$ which is 8. $6\# 2$ is equal to $6+2-6*2^2$ which is $8-24 = -16$. Therefore $\frac{6@2}{6\# 2}$ must be equal to $\frac{8}{-16} = -\frac{1}{2}$. Therefore the solution is $A$.