Parabola: Focus, Directrix and Standard Forms | NIMCET

Quick answer: A parabola is the set of points equidistant from a fixed point (the focus) and a fixed line (the directrix), so its eccentricity is $e = 1$ . For $y^2 = 4ax$ ( $a > 0$ ): focus $(a, 0)$ , directrix $x = -a$ , vertex at the origin, axis along the x-axis, latus rectum length $4a$ , and parametric point $(at^2, 2at)$ . The focal distance of a point $(x_1, y_1)$ on it is $x_1 + a$ .

The four standard forms

Equation	Focus	Directrix	Axis	Opens
$y^2 = 4ax$	$(a, 0)$	$x = -a$	x-axis	Right
$y^2 = -4ax$	$(-a, 0)$	$x = a$	x-axis	Left
$x^2 = 4ay$	$(0, a)$	$y = -a$	y-axis	Up
$x^2 = -4ay$	$(0, -a)$	$y = a$	y-axis	Down

All four have vertex $(0, 0)$ and latus rectum length $4a$ .

Key terms for y² = 4ax

Term	Value
Vertex	$(0, 0)$
Focus	$(a, 0)$
Directrix	$x = -a$
Axis	$y = 0$
Latus rectum	$x = a$ , length $4a$ , endpoints $(a, \pm 2a)$
Parametric point	$(at^2, 2at)$
Focal distance of $(x_1, y_1)$	$x_1 + a$

Shifted parabolas

$(y - k)^2 = 4a(x - h)$ has vertex $(h, k)$ , focus $(h + a, k)$ and directrix $x = h - a$ .

To handle a parabola such as $y^2 - 4y - 8x + 20 = 0$ , complete the square:

Step 1: $(y - 2)^2 = 8x - 16 = 8(x - 2)$ .

Step 2: Vertex $(2, 2)$ , $4a = 8$ so $a = 2$ , focus $(4, 2)$ , directrix $x = 0$ .

Position of a point

For $y^2 = 4ax$ , let $S_1 = y_1^2 - 4ax_1$ .

$S_1$	Point is
$< 0$	Inside
$= 0$	On the parabola
$> 0$	Outside

Worked examples

Example 1: focus and directrix

Find the focus, directrix and latus rectum of $y^2 = 12x$ .

$4a = 12$ , so $a = 3$ . Focus $(3, 0)$ , directrix $x = -3$ , latus rectum length 12.

Example 2: downward parabola

For $x^2 = -16y$ , find the focus and directrix.

$4a = 16$ , $a = 4$ . Focus $(0, -4)$ , directrix $y = 4$ .

Example 3: equation from focus and directrix

Find the parabola with focus $(2, 0)$ and directrix $x = -2$ .

The vertex is midway at the origin and $a = 2$ , so $y^2 = 8x$ .

Example 4: focal distance

Find the distance of the point on $y^2 = 8x$ with ordinate 4 from the focus.

Step 1: $16 = 8x$ , so $x = 2$ .

Step 2: Focal distance $= x + a = 2 + 2 = 4$ .

Example 5: from the definition

Find the parabola with focus $(1, 1)$ and directrix $x + y = 0$ .

Distance to focus equals distance to the directrix:

$(x - 1)^2 + (y - 1)^2 = \frac{(x + y)^2}{2}$

Simplifying gives $x^2 + y^2 - 2xy - 4x - 4y + 4 = 0$ , that is, $(x - y)^2 = 4(x + y - 1)$ .

Summary table

Item	For $y^2 = 4ax$
Eccentricity	1
Focus	$(a, 0)$
Directrix	$x = -a$
Latus rectum	$4a$
Parametric point	$(at^2, 2at)$
Focal distance	$x_1 + a$

Exam tips

Identify the form first: the squared variable tells you the axis, and the sign tells you the direction.
Complete the square for any parabola not in standard form.
Focal distance questions are fastest with $x_1 + a$ rather than the distance formula.
For a non-standard directrix, go back to the definition: distance to focus = distance to directrix.

Next, read parabola tangents, normals and focal chords, or see the NIMCET Coordinate Geometry and Vectors chapter.

Common mistakes

Taking $a = 12$ instead of 3 for $y^2 = 12x$ .
Putting the focus on the wrong axis.
Forgetting the sign of the directrix.
Treating the latus rectum length as $2a$ .

Frequently Asked Questions

What is the focus of the parabola y² = 4ax?

The focus is (a, 0), and the directrix is x = −a.

What is the length of the latus rectum of a parabola?

For y² = 4ax, it is 4a.

What are the parametric coordinates of a parabola?

For y² = 4ax, any point can be written as (at², 2at).

What is the eccentricity of a parabola?

The eccentricity of every parabola is 1.

What is the focal distance of a point on a parabola?

For y² = 4ax, the focal distance of (x₁, y₁) is x₁ + a.