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How to Find Equation of Directrix of a Parabola?

The equation of the directrix of a parabola can be determined in different ways depending on the given information about the parabola. These can be classified into three basic cases:
  1. You know the coordinates of vertex and focus
  2. You know the equation of the parabola
The method to determine the equation of directrix for each of these cases is described below:

1. You know the coordinates of vertex and focus

Suppose you know the coordinates of vertex and focus of a parabola. In order to find the equation of its directrix, we will use two properties of the directrix to find its equation:
  1. The directrix is perpendicular to the axis of the parabola
  2. The distance of any point on the parabola, and thus its vertex, from the directrix is equal to its distance from the focus
From the first property:

Since we already know the coordinates of focus and vertex of the parabola, and we know that the axis of a parabola passes through the focus and vertex, hence we can calculate the slope of the axis of the parabola from the given information.
Suppose the coordinates of vertex are `(1, 2)` and those of the focus are `(2, 2)`. Then, the slope of the axis is, by the formula for slope of a straight line,
`"Slope of axis" = (y_2 - y_1)/(x_2 - x_1) = (2 - 2)/(2 - 1) = 0`
Since the slope of the axis of the parabola is zero, this means that the axis of the parabola is parallel to the x-axis (all lines having slope zero are parallel to the x-axis). It follows that since the directrix is perpendicular to the axis of the parabola, hence the directrix is perpendicular to the x-axis, or parallel to the y-axis. We know that the equation of any straight line parallel to the y-axis is given by `x = a`,  where `a` is its intercept on the x-axis. Thus we assume that the equation of the directrix of the given parabola is `y = a`.

From the second property

Now we will apply the second property of directrices: the distance of any point on the parabola, and particularly the vertex, from the directrix is equal to its distance from the focus. This means that the vertex is the midpoint between the focus and the point where the directrix of the parabola meets its axis. Thus, if `(x, y)` is the point where the axis and directrix intersect each other, then the vertex `(1, 2)` is the midpoint of `(x, y)` and `(2, 2)`. By using midpoint formula, we have
`(1, 2) = ((x + 1)/2, (y + 2)/2)`
Solving the above two equations (one for `x` and one for `y`):
`(x + 2)/2 = 1`
`x + 2 = 2`
`x = 2 - 2`
`x = 0`
and
`(y + 2)/2 = 2`
`y + 2 = 4`
`y = 4 - 2`
`y = 2`
Thus, the coordinates of the point of intersection of the directrix and axis are `(0, 2)`. This can be graphically represented as follows:
As you can see in the above image, the vertex `(1, 2)` is the midpoint of the focus `(2, 2)` and the point of intersection of the directrix and the axis `(0, 2)`.

Now we know two things about the directrix. One is that it is a vertical line (since it is perpendicular to the axis) and second is that it passes through the point `(0, 2)`. The equation of the directrix is `x = a` where `a` is equal to 0 since it passes through `(0, 2)`.

Thus the equation of the directrix is `x = 0`

2. You know the equation of the parabola

The equation of the directrix can be determined from the equation of the parabola by comparing the equation of the parabola with one of the four standard forms. Since two equations can only be compared when they are written in the same way, hence if the equation of the parabola is not given in the standard form already, we will have to change it into the standard form by following the method described here. It involves mainly the completing the square method.

For example, you have the equation of the parabola `y^2 = 8x` We already know that the directrix of the standard parabola `y^2 = 4ax` is given by `x = -a`. Thus, comparing `y^2 = 8x` with `y^2 = 4ax`, we get
`4a = 8`, that is
`a = 2`
Therefore the equation of the directrix for the parabola `y^2 = 8x` is `x = -2`.
Similarly you can find the equation of the directrix for all equations that are directly comparable with one of the four standard forms.

But if the equation of the parabola is not in the standard form already, you can convert it into one of the four standard forms as described in this post. For example, if you have the equation `y = x^2 + 3x + 2`, you can convert it into the standard form by following the procedure described here. The steps of converting the equation into standard form are not given in this post for its length. However, after following the steps as described in the post linked above, you will get the following equation for `y = x^2 + 3x + 2`:
`(x + 3/2)^2 = (y + 1/4)`
We will now substitute `X` in place of `x + 3/2` and `Y` in place of `y + 1/4`, so the equation becomes
`X^2 = Y`
On comparing the above equation with
`x^2 = 4ay`
we get
`4a = 1` 
`a = 1/4` 
Since we know that the parabola `x^2 = 4ay` has directrix at `y = -a`, therefore this parbola with `a = 1/4` has directrix at
`X = -1/4`
Now we need to convert the capital `X` back to `x`. Since we had substituted `x + 3/2` in place of `X`, therefore we can write,
`x + 3/2 = X`
The equation of the directrix is `X = -1/4`. Thus substituting `X = -1/4` into the above equation we get
`x + 3/2 = -1/4`
`x = -1/4 - 3/2`
`x = -7/4`
Therefore the parabola `y = x^2 + 3x + 2` has the directrix at `x = -7/4`.

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