Two non-zero [real numbers](https://artofproblemsolving.com/wiki/index.php/Real_number), $a$ and $b,$ satisfy $ab = a - b$. Which of the following is a possible value of $\frac {a}{b} + \frac {b}{a} - ab$?
$\textbf{(A)} \ - 2 \qquad \textbf{(B)} \ \frac { -1 }{2} \qquad \textbf{(C)} \ \frac {1}{3} \qquad \textbf{(D)} \ \frac {1}{2} \qquad \textbf{(E)} \ 2$
$\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \boxed{\text{E}}$. 
Another way is to solve the equation for $b,$ giving $b = \frac{a}{a+1};$ then substituting this into the expression and simplifying gives the answer of $2.$