jmc

Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix.

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



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

Thus, the focus $\boxed{(-3,0)}.$