Solution:
Given:

Part A:
The vertex of an up-down facing parabola of the form;

Rewriting the equation given;

Therefore, the vertex is (0,0)
Part B:
A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)
Using the standard equation of a parabola;

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)
Hence,

Therefore, the focus is;

Part C:
A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)
Using the standard equation of a parabola;

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).
Hence,

Therefore, the directrix is;
