Focus and Directrix of a Parabola

Last Updated : 21 Apr, 2026

A parabola is the locus of a point in a plane such that its distance from a fixed point (focus) is equal to its perpendicular distance from a fixed straight line (directrix). A parabola has only one focus, and it never lies on the directrix.

As shown in the diagram, P1M = P1S, P2M = P2S, P3M = P3S, and P4M = P4S.

👁 parabola-3

Focus of a Parabola

The focus is a fixed point that lies on the axis of a parabola and is used to define it. A parabola is the set of all points that are equidistant from the focus and a fixed line called the directrix. The position of the focus depends on the standard equation and orientation of the parabola.

👁 parabola_2

The focus for the standard equations of a parabola is defined as follows:

The focus of the parabola y² = 4ax, with the x-axis as its axis, is F(a, 0).
The focus of the parabola y² = -4ax, with the x-axis as its axis, is F(-a, 0).
The focus of the parabola x² = 4ay, with the y-axis as its axis, is F(0, a).
The focus of the parabola x² = -4ay, with the y-axis as its axis, is F(0, -a).

Locating the Focus of a Parabola

The focus of a parabola lies on its axis and is at a distance of ’a' units from the vertex. The direction of the axis depends on which variable is squared in the equation.

If x is squared, the axis is vertical (parallel to the y-axis).
If y is squared, the axis is horizontal (parallel to the x-axis).

For standard forms:

Equation	Axis	Vertex	Focus
(x - h)² = 4a(y - k)	x = h	(h, k)	(h, k + a)
(y - k)² = 4a(x - h)	y = k	(h, k)	(h + a, k)

Directrix of a Parabola

The directrix is a fixed straight line that is perpendicular to the axis of a parabola. A parabola is defined as the set of all points that are equidistant from a fixed point (focus) and a fixed line (directrix). The directrix and the focus lie on opposite sides of the vertex and are at equal distances from it.

👁 parabola_1

The directrix for the standard equations of a parabola is defined as follows:

The directrix of the parabola y² = 4ax, with the x-axis as its axis, passes through (-a, 0) and has the equation x + a = 0.
The directrix of the parabola y² = -4ax, with the x-axis as its axis, passes through (a, 0) and has the equation x - a = 0.
The directrix of the parabola x² = 4ay, with the y-axis as its axis, passes through (0, -a) and has the equation y + a = 0.
The directrix of the parabola x² = -4ay, with the y-axis as its axis, passes through (0, a) and has the equation y - a = 0.

Locating the Directrix of a Parabola

The directrix of a parabola is a line perpendicular to its axis and lies at a distance of ’a' units from the vertex. The focus and the directrix are equidistant from the vertex but lie on opposite sides. The equation of the directrix can be determined from the equation of the parabola.

If x is squared, the axis is vertical (parallel to the y-axis).
If y is squared, the axis is horizontal (parallel to the x-axis).

For standard forms:

Equation	Axis	Vertex	Directrix
(x - h)² = 4a(y - k)	x = h	(h, k)	y = k - a
(y - k)² = 4a(x - h)	y = k	(h, k)	x = h - a

Uses of the Directrix of a Parabola

The directrix helps in determining various properties of a parabola:

It is used to form the equation of a parabola.
It helps in identifying the axis of the parabola.
It is useful in finding the equations of focal chords.
It helps in determining the equation and endpoints of the latus rectum.

Sample Problems

Question 1. Find the equation of the parabola whose focus is (-4, 2) and whose directrix is x + y = 3.

Solution:

Let P (x, y) be any point on the parabola whose focus is (-4, 2) and the directrix x + y - 3 = 0.
As we already know that the distance of a point P from focus = distance of a point P from directrix
So, √(x + 4)² + (y - 2)²=
On squaring both side we get
(x + 4)² + (y - 2)²=
2((x² + 16 + 8x) + (y²+ 4 - 4y)) = x² + y² + 9 +2xy - 6x - 6y
2(x² + 20 + 8x + y²- 4y) = x² + y² + 9 +2xy - 6x - 6y
2x² + 40 + 16x + 2y²- 8y = x² + y² + 9 +2xy - 6x - 6y
x² + y² + 2xy + 10x - 2y + 31 = 0

Question 2. Find the equation of the parabola whose focus is (-4, 0) and the directrix x + 6 = 0.

Solution:

Let P (x, y) be any point on the parabola whose focus is (-4, 0) and the directrix x + 6 = 0.
As we already know that the distance of a point P from focus = distance of a point P from directrix
So, √(x + 4)² + (y )²=
On squaring both side we get
(x + 4)² + (y)²=
2x² + 32 + 16x + 2y²= x² + 36 + 12x
x² + 2y² - 4 + 14x = 0

Question 3. Find the equation of the parabola with focus (4, 0) and directrix x = 3.

Solution:

Focus = (4, 0)
Directrix = x = 3
The vertex lies midway between the focus and the directrix.
Vertex = ((4 + 3)/2, 0) = (7/2, 0)
Distance from vertex to focus:
a = 4 − 7/2 = 1/2
Since the parabola opens towards the right, its equation is:
(y − 0)² = 4a(x − 7/2)
y² = 4 × (1/2)(x − 7/2)
y² = 2(x − 7/2)
y² = 2x − 7

Question 4. Find the equation of the parabola with vertex at (0, 0) and focus at (0, 4).

Solution:

Vertex = (0, 0), Focus = (0, 4)
Since the focus lies on the y-axis, the axis of the parabola is along the y-axis.
Standard form of parabola is: x² = 4ay
Here, a = 4
So, x² = 4 × 4 × y
x² = 16y

Question 5. Find focus, directrix, and vertex of the following equation: y = x² - 2x + 3

Solution:

Complete the square:
y = x² − 2x + 3
= (x² − 2x + 1) + 2
= (x − 1)² + 2
Rewrite in standard form: (x − 1)² = y − 2
Compare with standard form: (x − h)² = 4a(y − k)
So, h = 1, k = 2, and 4a = 1 ⇒ a = 1/4
Vertex: (h, k) = (1, 2)
Focus: (h, k + a) = (1, 2 + 1/4) = (1, 9/4)
Directrix: y = k − a = 2 − 1/4 = 7/4