jmc

Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix. Completing the square on $x,$ we get $$y=\frac{1}{12}(x-3{)}^2-\frac{1}{3}.$$To make the algebra a bit easier, we can find the directrix of the parabola $y=\frac{1}{12}{x}^2,$ shift the parabola right by 3 units to get $y=\frac{1}{12}(x-3{)}^2$ (which does not change the directrix), and then shift it downward $\frac{1}{3}$ units to find the directrix of the parabola $y=\frac{1}{12}(x-3{)}^2-\frac{1}{3}.$

Since the parabola $y=\frac{1}{12}{x}^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$ Let $y=d$ be the equation of the directrix.



Let $\left(x,\frac{1}{12}{x}^2\right)$ be a point on the parabola $y=\frac{1}{12}{x}^2.$ Then $$P{F}^2=x^2+{\left(\frac{1}{12}{x}^2-f\right)}^2$$and $P{Q}^2={\left(\frac{1}{12}{x}^2-d\right)}^2.$ Thus, $$x^2+{\left(\frac{1}{12}{x}^2-f\right)}^2={\left(\frac{1}{12}{x}^2-d\right)}^2.$$Expanding, we get $$x^2+\frac{1}{144}{x}^4-\frac{f}{6}{x}^2+f^2=\frac{1}{144}{x}^4-\frac{d}{6}{x}^2+d^2.$$Matching coefficients, we get $$\begin{array}{rl}1-\frac{f}{6}& =-\frac{d}{6},\\ f^2& =d^2.\end{array}$$From the first equation, $f-d=6.$ Since $f^2=d^2,$ $f=d$ or $f=-d.$ We cannot have $f=d,$ so $f=-d.$ Then $-2d=6,$ so $d=-3.$

Thus, the equation of the directrix of $y=\frac{1}{12}{x}^2$ is $y=-3,$ so the equation of the directrix of $y=\frac{1}{12}(x-3{)}^2-\frac{1}{3}$ is $\boxed{y=-\frac{10}{3}}.$