Question 1177687
{{{4y^2 + 8y − 8x + 7 = 0}}}

{{{(4y^2 + 8y )=8x - 7 }}}

{{{4(y^2 + 2y )=8x - 7 }}}

{{{(y^2 + 2y )=2x - 7/4}}} .........complete square

{{{(y^2 + 2y +b^2)-b^2=2x - 7/4}}}.........b=2/2=1

{{{(y^2 + 2y +1^2)-1^2=2x - 7/4}}}

{{{(y+1)^2-1=2x - 7/4}}}

{{{(y+1)^2=2x - 7/4+1}}}

{{{(y+1)^2=2x - 3/4}}}

{{{(y+1)^2=2(x - (3/4)/2)}}}

{{{(y+1)^2=2(x - 3/8)}}}

so you have

{{{(y-k)^2=4p(x-h) }}}is the standard equation for a right-left facing parabola with vertex at  ({{{h}}},{{{ k }}}), 


{{{(y+1)^2=2(x-3/8)}}}-> {{{h=3/8}}}, {{{k=-1}}}, {{{4p=2 }}}->{{{p=1/2}}}

vertex: ({{{3/8}}},{{{-1}}})

focus:  ({{{h+p}}},{{{k}}}) = ({{{7/8}}},{{{-1}}})


directrix: 

parabola is symmetric around the {{{x}}}-axis and so the directrix is a line parallel to the y-axis, a distance {{{-p}}} from the vertex ( {{{3/8}}},{{{-1}}}) x-coordinate 

{{{x=3/8-p}}}.........since {{{p=1/2}}}
{{{x=3/8-1/2}}}
{{{x=3/8-4/8}}}
{{{x=-1/8}}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(3/8,-1,.12), locate(3/8,-1,V(3/8,-1)),
circle(7/8,-1,.12), locate(7/8,-0.3,F(7/8,-1)),
green(line(-1/8,-10,-1/8,10)),
 graph( 600, 600, -10, 10, -10, 10,-sqrt(2(x - 3/8))-1, sqrt(2(x - 3/8))-1)) }}}