Question 851812
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Hi,
sketch IT (Directrix  x = 15 the clue) Parabola opening left a < 0
the vertex form of a Parabola opening right(a>0) or left(a<0), {{{x=a(y-k)^2 +h}}}
 where(h,k) is the vertex and  y = k  is the Line of Symmetry
{{{x=a(y)^2 }}}  V(0,0)   Finding value of a
p(distance of the directrix and focus left 0r right of vertex) = 15, a = 1/4p = -1/60 as a < 0
{{{x=(-1/60)(y)^2 }}}  
{{{drawing(300,300,-20,20,-20,20, grid(1), blue(line(15,20,15,-20)) ,
circle(-15,0,0.6),
graph( 300, 300,-20,20,-20,20, 0 , sqrt(-60(x))-1, -sqrt(-60(x))-1 ))}}}