The standard form of a parabola is written as #y = ax^2 +bx +c#
or in vertex form as #y =a(x-h)^2 +k# where #(h,k)# is the vertex and #1/(2a)# is the distance between the vertex and the focus as well the distance between the vertex and the directrix.
Firstly rearrange the expression to get #y# alone on the left had side.
#x^2-2x+8y+9 = 0#
#8y = -x^2 +2x -9#
#y = -1/8(x^2 - 2x +9)#
Now put it into vertex form by completing the square
#y = -1/8((x-1)^2 -1+9) = -1/8((x-1)^2+8)#
#y = -1/8(x^2-1)^2 -1#
The vertex is at #(1,-1)#.
Because #a# is negative the vertex is the maximum value of the equation - all values of #x# give values less than the vertex
#a=-1/8# so #1/(2a) = 1/(-2/8) =-4#
The focus is therefore #(1,-5)# and the directrix is at #y=3#
