Question 1198612
we use the standard form 

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

for parabolas that have an axis of symmetry parallel to the {{{y}}}-axis

{{{(3x-2)^2=84y-12}}}

{{{9x^2 - 12x + 4=84y-12}}}

{{{9x^2 - 12x =84y-12-4}}}

{{{9(x^2 - (12/9)x) =84y-16}}}

{{{9(x^2 - (4/3)x+(4/6)^2) -9(4/6)^2=84y-16}}}

{{{9(x^2 - (4/3)x+(2/3)^2) -9(2/3)^2=84y-16}}}

{{{9(x - 2/3)^2 -4=84y-16}}}

{{{9(x - 2/3)^2 =84y-16+4}}}

{{{9(x - 2/3)^2 =84y-12}}}

{{{9(x - 2/3)^2 =84(y-12/84)}}}

{{{9(x - 2/3)^2 =84(y-1/7)}}}

{{{(x - 2/3)^2 =(84/9)(y-1/7)}}}

{{{(x - 2/3)^2 =(28/3)(y-1/7)}}}


compare to  {{{(x-h)^2=4p(y-k) }}}   
     

{{{4p=28/3}}}

{{{p=28/12}}}

{{{p=7/3}}}

{{{h=2/3}}}

{{{k=1/7}}}


so

 vertex is at ({{{h}}},{{{ k}}})=({{{2/3}}},{{{ 1/7}}})

focus:({{{h}}},{{{ k+p}}})=({{{2/3}}}, {{{1/7+7/3}}})=({{{2/3}}}, {{{52/21}}})

directrix: {{{y =k-p=1/7-7/3= -46/21}}}

axis of symmetry: {{{x=h}}} so {{{x=2/3}}}

latus rectum:{{{4p= 28/3}}}

 endpoints of latus rectum: ({{{h+2p}}},{{{ k+p}}}) and ({{{h-2p}}},{{{ k+p}}})

({{{h+2p}}},{{{k+p}}}) =({{{2/3+2(7/3)}}}, {{{1/7+7/3}}})=({{{16/3}}}, {{{52/21}}})
and 
({{{h-2p}}},{{{ k+p}}})=({{{2/3-2(7/3)}}},{{{ 1/7+7/3}}})=({{{-4}}}, {{{52/21}}})