Question 1087425
<pre>This one is similar to the other ones I did for you except that
y and x are interchanged, h and k are interchanged, graphs opening 
right and left are changed to opening up and down.

{{{(x-h)^2=4p(y-k)}}}

has vertex (h,k), and distance from vertex to both focus
and directrix is |p|.  If p is positive the parabola opens
upward with the horizontal directrix is |p| units below the 
vertex and the focus is |p| units above the vertex.  

{{{(x-0)^2=18(y-0)}}}

has vertex (0,0), and distance from vertex to both focus
and directrix is |18/4| or 9/2.  Since 9/2 is positive the 
parabola opens upward with the vertical directrix 9/2 units  
below the vertex and the focus is 9/2 above the vertex.  

So the focus is 9/2 units above vertex (0,0) which is (0,9/2),
and the directrix is a horizontal line 9/2 units below the 
vertex (0,0) which is the horizontal line y = -9/2

{{{drawing(400,400, -7,7,-7,7,

 graph(400,400, -7,7,-7,7,x^2/18),
locate(-3.3,-9/2,y=-9/2),
 
locate(.2,4.8,(matrix(1,3,0,",",9/2))),circle(0,9/2,.12),circle(0,9/2,.1),
green(line(-20,-9/2,20,-9/2)), circle(0,9/2,.15) )}}}

Edwin</pre>