x=4y^2 Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.

One of the vertex form of the parabola is,
(y-k)^2=4p(x-h)   where (h,k) is the vertex and 
p is the distance between vertex and focus and also the same distance between the vertex and the directrix,
Given equation is x=4y^2
Graph of the equation is attached.
Rewrite the equation in the standard form,
y^2=1/4x
4p=1/4
=>p=1/16
(y-0)^2=4(1/16)(x-0)
Vertex is at (h,k) i.e (0,0)
Focus is at (h+p,k) i.e (1/16,0)
Axis of symmetry is the horizontal line passing through the vertex, i.e y=0
Directrix being perpendicular to the axis of symmetry is the vertical line,
Directrix is x=h-p
Directrix is x=0-1/16=-1/16
 

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