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...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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`