Question 1089163
{{{x^2+2x-8y-31=0}}}

The "vertex" form of a parabola with its vertex at ({{{h}}},{{{ k}}}) is:

regular: {{{y = a(x-h)^2 + k}}}
sideways: {{{x = a(y -k)^2 + h }}}

The conics form of the parabola equation (the one you'll find in advanced or older texts) is:

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

where the value of {{{4p}}} is actually the same as the value of {{{1/4a}}}


{{{(x^2+2x+b^2)-b^2-31=8y}}}....since coefficients {{{a=1}}} and {{{2ab=2}}}, we have {{{2*1*b=2}}}->{{{2b=2}}}->{{{b=1}}}

{{{8y=(x^2+2x+1^2)-1^2-31}}}

{{{8y=(x+1)^2-1-31}}}

{{{y=(1/8)(x+1)^2-32/8}}}

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

the vertex is at ({{{-1}}},{{{-4}}})

The focus is "p" units from the vertex:
{{{p =1/4a=1/(4(1/8))=2}}}, 

so {{{p=2}}}

then, 
the focus is {{{2}}} unit above the vertex, at  ({{{-1}}}, {{{-2}}}), 
and the directrix is the horizontal line{{{ y = -6}}}, p or {{{two}}}{{{ units}}} below the vertex


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