Question 1101490
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With the y variable squared, this parabola opens horizontally; with the coefficient on the x term positive, it opens to the right.<br>
The standard vertex form (that I use!) for a parabola that opens horizontally is<br>
{{{(x-h) = (1/(4p))(y-k)^2)}}}<br>
or<br>
{{{(y-k)^2 = 4p(x-h)}}}<br>
In this form, p is the distance from the vertex to the focus, and from the vertex to the directrix.<br>
The equation for your parabola is 
{{{(y-1)^2 = 8(x-2)}}}<br>
That means p is 2; so the vertex of the parabola is at (2,1); the focus is 2 units to the right of the vertex, at (4,1); and the directrix is the vertical line 2 units to the left of the vertex, x=0.<br><br>
{{{graph(300,300,-2,12,-10,10,x=0,y=0,y=sqrt(8(x-2))+1,y=-sqrt(8(x-2))+1))}}}<br><br>
(Sorry.... I haven't yet figured out how to do labeling on a graph....)