Question 1011138
{{{ y= -(1/12)x^2 }}}
{{{ (1/(-1/12))y= x^2 }}}
{{{ x^2 =(12/-1)y}}}
 {{{x^2 =-12y}}}

we can see that the vertex will be at ({{{h}}},{{{k}}})=({{{0}}},{{{0}}})

 The standard form of a parabola (in this case upside down parabola) with a vertical axis of symmetry is 

{{{(x-h)^2=-4p(y-k)}}}, where {{{p}}} is the directed distance between the vertex and focus (and also between the vertex and directrix) 

since {{{h=0}}} and {{{k=0}}}, we have  

{{{(x-0)^2=-4p(y-0)}}}

{{{x^2=4p*y}}}  since you have  {{{x^2 =-12y}}}, then
{{{4p=-12}}}
{{{p=-12/4}}}
{{{p=-3}}}

focus | ({{{0}}},{{{p}}})=({{{0}}}, {{{-3}}})
p is the directed distance between  the vertex and directrix, directrix will be {{{3}}} units above x-axis :

{{{y = 3}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
green(circle(0,-3,.14)),locate(0,-3,F(0,-3)),
 graph( 600, 600, -10, 10, -10, 10, x^2/-12,3)) }}}