Verified answer

1) complete squares

x^2 - 2x = x^2 - 2x + 1 - 1 = (x-1)^2 - 1

=>

x^2-2x+8y+9= (x-1)^2 -1 + 8y + 9 =

= (x-1)^2 + 8 (y+1)= 0

The vertex is located where both parenthesis are 0, then the vertex is the point (1,-1).

The focus is located at the same abscisa of the vertex, but below it, and the directrix is an horizontal line located at the same distance above:

F = (1,-1-a)

dir: y = -1+a

The parabola is the locus of point equidistant from the focus and the directrix:

Given a point (x,y) it distance from the focus is

d( (x,y), (1,-1-a)) = sqrt((x-1)^2 + (y+1+a)^2)

Its distance from the directrix is y+1-a

=>

(x-1)^2 + (y+1+a)^2 = (y+1-a)^2

=>

(x-1)^2 + (y+1)^2 + a^2 + 2a(y+1) = (y+1)^2 + a^2 - 2a(y+1)

=>

(x-1)^2 + 4a(y+1) = 0

so a=2.

F = (1, -3)

dir: y = 1

For a quadratic in regularly happening type, ax²+bx+c, the vertex is at -b/2a, which on your case is 0/-6=0. At x=0, y=10 so vertex is at (0, 10). The parabola opens downwards because of the fact a=-3 (it is unfavourable), so the vertex is a optimal. Vertex: (0,10); optimal; axis of symmetry: x=0, opens down