Question 1085976
{{{x^2=(1/8) y }}}

there is no {{{h}}} and {{{k}}}, so vertex is at origin:
({{{0}}}, {{{0}}})

your parabola is of the form: {{{x^2=4py}}}

so, {{{4p=1/8}}}-> {{{p=1/(8*4)}}}-> {{{p=1/32}}}

directrix | {{{y = -p}}}->{{{y = -1/32}}}

focus | ({{{0}}}, {{{1/32}}})

axis of symmetry: symmetric with respect to the {{{y-axis}}}



{{{drawing( 600, 600, -1, 1, -1, 1,
circle(0,1/32,.012),locate(0.02,0.09,f(0,1/32)),
 graph( 600, 600, -1, 1, -1, 1, 8x^2,-1/32)) }}}