Directrix of a Parabola

The directrix of a parabola is a straight line that is perpendicular to its axis of symmetry and does not intersect the parabola. This line is part of the conic section's definition, as a parabola is the set of all points that are equidistant from a line called the directrix and a fixed point called the focus.

Equation of the Directrix

If the parabola is vertical, its directrix is horizontal; if the parabola is horizontal, its directrix is vertical. The distance from the vertex to the directrix is equal to |p| (the absolute value of p), where p is the parameter that appears in the standard equation of the parabola. We can use this value to determine the equation of the directrix, since it will always be located |p| units from the vertex.

Vertex at the Origin

If the vertex is at the origin, there are two possible cases:

• The equation y² = 4px represents a horizontal parabola whose axis of symmetry is the x-axis; it opens to the right if p > 0 and to the left if p < 0. In this case, the equation of the directrix is x = -p.
• The equation x² = 4py represents a vertical parabola that opens upward if p > 0 and downward if p < 0; its axis of symmetry is the y-axis. The equation of the directrix is y = -p.

The equations are summarized in the following table.

Parabola Equation	Orientation	Opening	Equation of the Directrix
y2 = 4px	Horizontal	Opens to the right if p > 0 Opens to the left if p < 0	x = -p
x2 = 4py	Vertical	Opens upward if p > 0 Opens downward if p < 0	y = -p

Example 1

Find the equation of the directrix of the parabola \(x^2=6y\).

Solution

We begin by recognizing that the given equation matches the standard form \(x^2=4py,\) which corresponds to a vertical parabola with its vertex at the origin.

Comparing both equations, we see that \(4p=6,\) which allows us to solve for \(p\) to find the parameter value. Doing this, we obtain \(p=\dfrac{6}{4}=\dfrac{3}{2}.\)

Next, we determine the equation of the directrix. For a vertical parabola opening upward, the directrix is a horizontal line located at a distance of \(p\) below the vertex. Therefore, its equation is \(y=-p.\) Substituting the value of \(p,\) the equation of the directrix is:

$$y=-\dfrac{3}{2}$$

Example 2

Find the equation of the directrix of the horizontal parabola \(x=-10y^2\).

Solution

As a first step, we rewrite the equation in standard form to identify the key parameters. The given equation is \(x=-10y^2,\) which can be expressed as:

$$x=-10y^2 \quad → \quad y^2=-\dfrac{1}{10}x$$

This matches the form of a horizontal parabola that opens to the left, since the parameter p is negative.

Now we compare it to the standard form of a horizontal parabola:

$$y^2=4px$$

Setting the expressions equal, we have:

$$4p=-\dfrac{1}{10} \quad → \quad p=-\dfrac{1}{40}$$

The equation of the directrix for a horizontal parabola is:

$$x=-p$$

Therefore, the equation of the directrix line is:

\(x=-\left(-\dfrac{1}{40}\right)=\dfrac{1}{40}\)

Vertex at (h, k)

If the vertex is not at the origin but at a point (h, k), the directrix equation is found by adding h to the right side of the equation for horizontal parabolas, and adding k for vertical parabolas. The formulas and equations are summarized in the following table:

Parabola Equation	Orientation	Opening	Equation of the Directrix
(y - k)2 = 4p (x - h)	Horizontal	Opens to the right if p > 0 Opens to the left if p < 0	x = h - p
(x - h)2 = 4p (y - k)	Vertical	Opens upward if p > 0 Opens downward if p < 0	y = k - p

Example 1

Find the directrix of a parabola whose equation is \((y-7)^2=12(x - 4).\)

Solution

Comparing the given equation with the standard equation of a horizontal parabola \((y-k)^2=4p (x-h),\) we find that the vertex of the parabola is \((4, 7).\) Furthermore, setting \(4p\) equal to the coefficient of the linear term in \(x,\) we get \(4p=12,\) which implies \(p=3.\)

Since the equation has the form \((y-k)^2=4p (x-h),\) we know that the parabola opens to the right if p is positive and to the left if it is negative. In this case, since \(p=3>0,\) the parabola opens to the right. Therefore, the directrix is a vertical line located |p| units to the left of the vertex.

The equation of the directrix in this case is \(x=h-p.\) Substituting the known values, we obtain: \(x=4-3=1\)

Finally, we conclude that the directrix of the parabola is the vertical line given by the equation:

$$x=1$$

Example 2

If a parabola has a vertical axis of symmetry with its vertex at (1, 4) and its focus at (1, 2), find the equation of the directrix.

Solution

We notice that both the vertex and the focus have the same coordinate \(x=1,\) confirming that the axis of symmetry is vertical. Since the focus is at \((1, 2)\) and the vertex is at \((1, 4),\) the parabola opens downward because the focus lies below the vertex.

The parameter p is negative and is calculated as:

$$p=y_{\text{focus}}-y_{\text{vertex}}=2-4=-2$$

Therefore, \(p=-2.\)

For a vertical parabola that opens downward, the directrix is a horizontal line located at a distance of |p| above the vertex. Its general equation is:

$$y=y_{\text{vertex}}-p$$

Substituting the known values:

$$y=4-(-2)=4+2=6$$

Thus, the equation of the directrix is:

$$y=6$$

Example 3

Find the directrix of the parabola \((y+3)^2=-8(x+2).\)

Solution

We begin by recognizing that the expression is in the standard form of a horizontal parabola: \((y-k)^2=4p (x-h).\)

First, we identify the vertex \((h, k).\) Comparing it to the standard form, we observe that:

$$h=-2, \quad k=-3$$

Additionally, the value \(-8\) corresponds to \(4p,\) so:

$$4p=-8 \quad → \quad p=-2$$

Since p is negative, the parabola opens to the left. The directrix is a vertical line located |p| units from the vertex in the opposite direction of the opening. Its equation is:

$$x=h-p$$

Substituting the known values:

$$x=-2-(-2) \quad → \quad x=-2+2=0$$

Finally, we conclude that the directrix of the parabola is given by the equation:

$$x=0$$

Example 4

Determine the equation of the directrix line of the parabola \(x^2-4y+28=0\).

Solution

We begin by rewriting the equation in standard form. First, isolate the quadratic term and the linear terms:

$$x^2=4y-28$$

$$x^2=4 (y-7)$$

Now, comparing this equation with the standard form of a vertical parabola \((x-h)^2=4p (y-k),\) we identify that the vertex \((h, k)\) is \((0, 7).\) Setting the coefficient of the linear term in \(y\) equal to \(4p,\) we get \(4p=4,\) which implies \(p=1.\)

Since p > 0, the parabola opens upward. The directrix is a horizontal line located |p| units below the vertex, and its equation is \(y=k-p.\) Substituting the known values:

$$y=7-1$$

$$y=6$$

Finally, we conclude that the directrix of the parabola is the horizontal line given by the equation:

$$y=6$$