Question 1190798
the equation of the parabola with latus rectum {{{ 6}}} 
length of latus rectum is {{{ 6}}}, so the leading coefficient is ±{{{1/6}}}
vertex on the line {{{y = 2}}}=>{{{k=2}}}

so far we have:

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

passing through ({{{2}}}, {{{8}}})

use it to calculate {{{h}}}  



{{{8 = (1/6)(2-h)^2+2}}}

{{{8 = h^2/6 - (2 h)/3 + 8/3}}} .......multiply by common denominator {{{6}}}

{{{48 = h^2 - 4 h + 16}}}

{{{ h^2 - 4 h + 16-48=0}}}

{{{ h^2 - 4 h -32=0}}}

{{{(h + 4) (h - 8)=0}}}

=>{{{h=-4}}} or {{{h=8}}}


and, your equation is:

{{{y = (1/6)(x+4)^2+2}}}
or
{{{y = (1/6)(x-8)^2+2}}}


graph: {{{y = (1/6)(x+4)^2+2}}}

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,8,.12), locate(2,8,p(2,8)),
circle(-4,2,.12), locate(-4,2,V(-4,2)),locate(3,2.5,y=2),
 graph( 600, 600, -10, 10, -10, 10, (1/6)(x+4)^2+2,2)) }}} 


graph: {{{y = (1/6)(x-8)^2+2}}}

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(2,8,.12), locate(2,8,p(2,8)),
circle(8,2,.12), locate(8,2,V(8,2)),locate(3,2.5,y=2),
 graph( 600, 600, -10, 10, -10, 10, (1/6)(x-8)^2+2,2)) }}}