Question 1138078



The standard form is {{{(x - h)^2 = 4p (y - k)}}}, where the focus is ({{{h}}}, {{{k + p}}}) and the directrix is {{{y = k - p}}}.


given: 

a focus at ({{{0}}}, {{{-3}}}) =>({{{h}}}, {{{k + p}}})=({{{0}}}, {{{-3}}}) 

so, {{{h=0}}} and {{{ k + p=-3}}}=>{{{k=-3-p}}}....eq.1

a directrix at {{{y = 3}}}=>{{{k - p=3}}}=>{{{k=3+p}}}....eq.2

from eq.1 and eq.2 we have
 
{{{-3-p=3+p}}}.....solve for {{{p}}}
{{{-3-3=p+p}}}
{{{-6=2p}}}
{{{p=-3}}}

go to {{{k=3+p}}}....eq.2, plug in {{{p}}}

{{{k=3-3}}}

{{{k=0}}}

your equation is:

{{{(x-0)^2 = 4(-3) (y - 0)}}}

{{{x^2 = -12 y }}}

{{{y= -(1/12)x^2  }}} => your answer