Question 1204729





{{{y^2 - 12x = 0}}}

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

vertex form of your parabola is: 

{{{(y - k)^2 = 4a(x - h)}}} where ({{{h}}}, {{{k}}}) is vertex and ({{{h + a}}}, {{{k}}}) is focus

your parabola in vertex form will be:

{{{(y - 0)^2 = 12(x - 0)}}} 

as you can see, {{{h=0}}}, {{{k=0}}}, {{{4a=12}}} =>{{{ a=3}}}

vertex is at origin ({{{0}}},{{{0}}})

focus is at ({{{0 +a}}}, {{{0}}})= ({{{0 +3}}}, {{{0}}})=({{{3}}},{{{0}}})

 the axis of symmetry is {{{y=k }}}=>{{{ y=0}}}

The focus and the directrix lie on either sides of vertex of the parabola and are equidistant from the vertex.

so, {{{x=-a }}}and the equations of the directrix is{{{ x=-3 }}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(3,0,.12), locate(3,0.5,V(3,0)),
green(line(-3,10,-3,-10)),

 graph( 600, 600, -10, 10, -10, 10, sqrt(12x), -sqrt(12x))) }}}