Max & Min

Let us first determine the minimum value $b$ by rewriting the constraint as $-x^2 = 3y^2 - 18$ and substituting this into the expression $3y - x^2$ to obtain $f(y) = 3y^2 + 3y - 18 = 3(y + \frac{1}{2})^2 - \frac{75}{4}$ after completing the square. The function $f(y)$ is a concave-up parabola with a minimum value of $\frac{-75}{4} \Rightarrow b = \frac{-75}{4}$ .

Now the maximum value $a$ is found when $x^2$ is at its smallest value in $3y - x^2$ , which is of course $|x^2| \ge 0$ . Substituting $x = 0$ into the constraint equation produces $y = \pm\sqrt6$ of which the positive root is larger. Hence, $a = 3\sqrt6 - 0^2 = 3\sqrt6$ .

The final calculation comes to: $a^2 - 4b = (3\sqrt6)^2 - 4(\frac{-75}{4}) = 54 + 75 = \boxed{129}$ .