Here is one way.

The directrix of a parabola is "p" away from the vertex along the axis of symmetry. [Also, the focus of a parabola is "p" away from the vertex.]

1) The directrix of parabola y^2 = 4(x+3).

The y is squared, so the parabola is horizontal, or the axis of symmetry is horizontal.

A standard form of the equation of a horizontal parabola, in terms of "p", is

(x- h) = [1/(4p)](y -k)^2 -------------(1)

where

(h,k) is the vertex

p is vertex to focus, or vertex to direcrix along the axis of symmetry.

If the 1/(4p) is negative, then the parabola opens to the left.

If the 1/(4p) is positive, then the parabola opens to the right.

So,

y^2 = 4(x+3) -----------------------(i)

Convert that into the form of (1),

(x+3) = [1/(4*1)](y -0)^2 ---------------(ii)

Therfore,

vertex is (-3,0)

p = 1 unit.

The 1/(4p) is 1/(4*1) is positive, so the directrix is to the left of the vertex.

The vertex (-3,0) is on the verical line x = -3. One unit to the left of that is x = -4.

Therefore, the directrix is the vertical line x = -4. -----------answer.

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2) Find the equation of the parabola whose focus is (2,3) and the directrix is x=12.

The vertex is halfway of the distance between the focus and the directrix along the axis of symmetry.

The directrix x = 12 is a vertical line, therefore the parabola is horizontal. The axis of symmetry is the horizontal line y = 3 because the focus is always on the axis of symmetry.

The focus (2,3) is on the vertical line x = 2.

So distance from focus to directrix along the axis of symmetry is 12 -2 = 10 units.

Hence the "p" is 10/2 = 5 units.

And so the vertex is (2+5,3) = (7,3)

The focus is at (2,3) and the directrix is the vertical line x=12. Therefore, the parabola opens to the left. Hence the 1/(4p) is negative.

Then, the parabola is:

(x -h) = -[1/(4p)](y -k)^2

(x -7) = -[1/(4*5)](y -3)^2

x -7 = -(1/20)(y -3)^2 --------------------answer.