Answer
$$\eqalign{
& \left( {\bf{a}} \right){\text{ vertex: }}\left( {0,0} \right),{\text{ focus: }}\left( {1,0} \right),{\text{ directrix: }}x = - 1 \cr
& \left( {\bf{b}} \right){\text{ vertex: }}\left( {0,0} \right),{\text{ focus: }}\left( {0, - 2} \right),{\text{ directrix: }}y = 2 \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right){\mkern 1mu} {\mkern 1mu} {y^2} = 4x \cr
& {\text{The equation has been written in the form }}{y^2} = 4px{\text{ then,}} \cr
& \underbrace {{y^2} = 4x}_{{y^2} = 4px}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} p = 1 \cr
& {\text{The vertex of a parabola of the form }}{y^2} = 4px{\text{ is Vertex: }}\left( {0,0} \right) \cr
& {\text{The focus of a parabola of the form }}{y^2} = 4px{\text{ is }}\left( {p,0} \right),{\text{ then}} \cr
& {\text{focus: }}\left( {1,0} \right) \cr
& {\text{The directrix of a parabola of the form }}{y^2} = 4px{\text{ is }}x = - p \cr
& {\text{directrix: }}x = - 1 \cr
& {\text{In summary,}} \cr
& {\text{vertex: }}\left( {0,0} \right),{\text{ focus: }}\left( {1,0} \right),{\text{ directrix: }}x = - 1 \cr
& \cr
& \left( {\text{b}} \right){\mkern 1mu} {\mkern 1mu} {x^2} = - 8y \cr
& {\text{The equation has been written in the form }}{x^2} = 4py{\text{ then,}} \cr
& \underbrace {{x^2} = - 8y}_{{x^2} = 4py}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \Rightarrow {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 4p = - 8,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} p = - 2 \cr
& {\text{The vertex of a parabola of the form }}{x^2} = 4py{\text{ is Vertex: }}\left( {0,0} \right) \cr
& {\text{The focus of a parabola of the form }}{x^2} = 4py{\text{ is }}\left( {0,p} \right) \cr
& {\text{Focus: }}\left( {0, - 2} \right) \cr
& {\text{The directrix of a parabola of the form }}{x^2} = 4py{\text{ is }}y = - p \cr
& {\text{directrix: }}y = 2 \cr
& {\text{In summary,}} \cr
& {\text{vertex: }}\left( {0,0} \right),{\text{ focus: }}\left( {0, - 2} \right),{\text{ directrix: }}y = 2 \cr} $$
