Question 41896
First re-arrange the given equation to compare with standard form.
{{{y=(1/2)(x-8)^2+2}}}
or {{{(y-2)=4*(1/8)*(x-8)^2}}}


Let us transform the equation to the new co-ordinate system (X,Y) such that X=x-8 and Y=y-2.
Then the given equation becomes {{{X^2 = 4(1/8)Y}}}.
This is a parabola with the Y-axis as its axis.
Comparing with the standard equation {{{X^2 = 4aY}}}, a={{{1/8}}}.


AXIS OF SYMMETRY AND DIRECTION OF OPENING
Here the axis of symmetry is x = 8 and as {{{a = 1/8}}} i.e. positive so the parabola opens upwards.


VERTEX
The vertex of the parabola {{{X^2 = 4aY}}} is the point (X=0,Y=0) with respect to the new co-ordinate system.
Now, x=8+X and y=2+Y.
So when X=0 and Y=0, x=8 and y=2.
So the co-ordinates of the vertex w.r.t. the old co-ordinate system are (x=8,y=2)


FOCUS
The co-ordinates of focus of the parabola {{{X^2 = 4aY}}} are (X=0,Y=a) in new co-ordinate system and so (x=8,y=2+a) in old co-ordinate system.
As here a=2 so the co-ordinates of the focus are (x=8,y={{{17/8}}}).


DIRECTRIX
The equation of directrix of the parabola {{{X^2 = 4aY}}} is Y+a=0 in new co-ordinate system. Transferring to old co-ordinate system the equation is {{{(y-2)+1/8=0}}} or {{{y=15/8}}}.


{{{graph(400,300,-1,19,-1,20,0.5*(x-8)^2+2)}}}