Question 1170439
given:
the parabola that has its vertex at the origin => ({{{h}}},{{{k}}})=({{{0}}},{{{0}}})
and that directrix {{{y = 1/6}}}

The standard form is {{{(x - h)^2 = 4p (y - k)}}}, where the focus is ({{{h}}}, {{{k + p}}}) and the directrix is {{{y = k - p}}}. If the parabola is rotated so that its vertex is ({{{h}}},{{{k}}}) and its axis of symmetry is parallel to the {{{x}}}-axis, it has an equation of {{{(y - k)^2 = 4p (x - h)}}}, where the focus is ({{{h + p}}}, {{{k}}}) and the directrix is {{{x = h - p}}}}

so. use {{{(x - h)^2 = 4p (y - k)}}}

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

{{{x^2 = 4p *y }}}

{{{y=x^2 /4p }}}

since directrix {{{y = 1/6}}}, the directrix is {{{y = k - p}}}, and {{{k=0}}}, we have

{{{1/6 = 0 - p}}}

{{{p =  - 1/6}}}


{{{y=x^2 /(4(- 1/6)) }}}

{{{y=x^2 /(-4/6) }}}

{{{y=x^2 /(-2/3) }}}

{{{y=(-3/2) x^2 }}}-> your answer


the focus is ({{{h}}}, {{{k + p}}}) =({{{0}}}, {{{- 1/6}}})


{{{ drawing ( 600, 600, -5, 5, -5, 5,
circle(0,-1/6,.05),locate(0.2,-1/6,F(0,-1/6)),
graph( 600, 600, -5, 5, -5, 5, (-3/2) x^2 , 1/6) )}}}