Standard Equation of a Parabola (original) (raw)

Last Updated : 23 Jul, 2025

The standard form of a parabola is y = ax2 + bx + c where a, b, and c are real numbers and a is not equal to zero. A parabola is defined as the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane.

In this article, we will understand what is a Parabola, the standard equation of a Parabola, related examples, and others in detail.

What is a Parabola?

A parabola is a conic section defined as the set of all points equidistant from a point called the focus and a line called the directrix. The standard equations for a parabola depend on its orientation (opening direction) and position.

Parabola

Parabola

**Equation of a Parabola

Equation of parabola can be written in standard form or general form and both of them are added below:

General Equations of a Parabola

The general equation of a parabola is,

**y = 4a(x - h) 2 + k
****(or)**
**x = 4a(y - k) 2 + h

Where (h, k) is the vertex of a parabola.

Standard Equations of a Parabola

The standard equation of a parabola is,

**y = ax 2 + bx + c
****(or)**
**x = ay 2 + by + c

where a can never be zero.

Parts of a Parabola

Some important terms and parts of a parabola are:

A parabola has four standard equations based on the orientation of the parabola and its axis. Each parabola has a different transverse axis and conjugated axis.

Equation of Parabola Parabola Formulae of Parameters of a Parabola
**y 2 = 4ax parHorizontal Parabola Vertex = (0,0)Focus = (a, 0)The parabola opens to the right side.The equation of the axis is y = 0The equation of the directrix is x + a = 0Length of the latus rectum = 4a
**y 2 = -4ax par2Horizontal Parabola Vertex = (0,0)Focus = (-a, 0)The parabola opens to the left side.The equation of the axis is y = 0The equation of the directrix is x - a = 0Length of the latus rectum = 4a
**x 2 = 4ay par1Vertical Parabola Vertex = (0,0)Focus = (0, a)The parabola opens upwards.The equation of the axis is x = 0The equation of the directrix is y + a = 0Length of the latus rectum = 4a
**x 2 = -4ay par3Vertical Parabola Vertex = (0,0)Focus = (0, -a)The parabola opens downwards.The equation of the axis is x = 0The equation of the directrix is y - a = 0Length of the latus rectum = 4a

The following are the observations made from the standard form of equations of a parabola:

The following are the standard equations of a parabola when the axis of symmetry is either parallel to the x-axis or y-axis and the vertex is not at the origin.

Equation of Parabola Parabola Formulae of Parameters of a Parabola
(y – k)2 = 4a(x – h) par4Horizontal Parabola Vertex = (h, k)Focus = (h + a, k)The parabola opens to the right side.The equation of the axis is y = kThe equation of the directrix is x = h – aLength of the latus rectum = 4a
(y – k)2 = -4a(x – h) par5Horizontal Parabola Vertex = (h, k)Focus = (h - a, k)The parabola opens to the left side.The equation of the axis is y = kThe equation of the directrix is x = h + aLength of the latus rectum = 4a
(x – h)2 = 4a(y – k) par6Vertical Parabola Vertex = (h, k)Focus = (h, k + a)The parabola opens upwards.The equation of the axis is x = hThe equation of the directrix is y = k - aLength of the latus rectum = 4a
(x – h)2 = -4a(y – k) par7Vertical Parabola Vertex = (h, k)Focus = (h, k - a)The parabola opens downwards.The equation of the axis is x = hThe equation of the directrix is y = k + aLength of the latus rectum = 4a

**Equation of Parabola Derivation

Let P be a point on the parabola whose coordinates are (x, y). From the definition of a parabola, the distance of point P to the focus (F) is equal to the distance of the same point P to the directrix of a parabola. Now, let us consider a point X on the directrix, whose coordinates are (-a, y).

Equation of Parabola Derivation

A Horizontal Parabola

From the definition of the eccentricity of a parabola, we have

e = PF/PX = 1
⇒ PF = PX

The coordinates of the focus are (a, 0). Now, by using the coordinate distance formula, we can find the distance of point P (x, y) to the focus F (a, 0).

PF = √[(x - a)2 + (y - 0)2]
⇒ PF = √[(x - a)2 + y2] —————— (1)

The equation of the directrix is x + a = 0. To find the distance of PX, we use the perpendicular distance formula.

PX = (x + a)/√[12 + 02]
⇒ PX = x +a —————— (2)

We already know that PF = PX. So, equate equations (1) and (2).
√[(x - a)2 + y2] = (x + a)

By, squaring on both sides we get,

⇒ [(x - a)2 + y2] = (x + a)2
⇒ x2 + a2 - 2ax + y2 = x2 + a2 + 2ax
⇒ y2 - 2ax = 2ax
⇒ y2 = 2ax + 2ax ⇒ **y 2 = 4ax

Thus, we have derived the equation of a parabola. Similarly, we can derive the standard equations of the other three parabolas.

**y 2 = 4ax, y 2 = -4ax, x 2 = 4ay, and x 2 = -4ay are the standard equations of a parabola.

**Articles Related to Parabola:

**Solved Examples of Standard Equations of a Parabola

**Example 1: Find the length of the latus rectum, focus, and vertex, if the equation of the parabola is y 2 = 12x.

**Solution:

Given, equation of the parabola is y2 = 12x

By comparing the given equation with the standard form y2 = 4ax
4a = 12
⇒ a = 12/4 = 3

We know that,
Latus rectum of a parabola = 4a = 4 (3) = 12
Now, focus of the parabola = (a, 0) = (3, 0)
Vertex of the given parabola = (0, 0)

**Example 2: Find the equation of the parabola which is symmetric about the X-axis, and passes through the point (-4, 5).

**Solution:

Given, parabola is symmetric about the X-axis and has its vertex at the origin.

Thus, the equation can be of the form y2 = 4ax or y2 = -4ax, where the sign depends on whether the parabola opens towards the left side or right side.

Parabola must open left since it passes through (-4, 5) which lies in the second quadrant.

So, the equation will be: y2 = -4ax

Substituting (-4, 5) in the above equation,
⇒ (5)2 = -4a(-4)
⇒ 25 = 16a
⇒ a = 25/16

Therefore, the equation of the parabola is: y2 = -4(25/16)x (or) 4y2 = -25x.

**Example 3: Find the coordinates of the focus, the axis, the equation of the directrix, and the latus rectum of the parabola x 2 **= 16y.

**Solution:

Given, equation of the parabola is: x2 = 16y

By comparing the given equation with the standard form x2 = 4ay,
4a = 16 ⇒ a = 4

Coefficient of y is positive so the parabola opens upwards.

Also, the axis of symmetry is along the positive Y-axis.

Hence,
Focus of the parabola is (a, 0) = (4, 0).
Equation of the directrix is y = -a, i.e. y = -4 or y + 4 = 0.
Length of the latus rectum = 4a = 4(4) = 16.

**Example 4: Find the length of the latus rectum, focus, and vertex if the equation of a parabola is 2(x-2) 2 + 16 = y.

**Solution:

Given, equation of a parabola is 2(x-2)2 + 16 = y

By comparing the given equation with the general equation of a parabola y = a(x - h)2 + k, we get
a = 2
(h, k) = (2, 16)

We know that, length of latus rectum of a parabola = 4a
= 4(2) = 8

Now, focus= (a, 0) = (2, 0)
Now, Vertex = (2, 16)

**Example 5: The equation of a parabola is x 2 **- 12x + 4y - 24 = 0, then find its vertex, focus, and directrix.

**Solution:

Given,

Equation of the parabola is x2 - 12x + 4y - 24 = 0
⇒ x2 - 12x + 36 - 36 + 4y - 24 = 0
⇒ (x - 6)2 + 4y - 60 = 0
⇒ (x - 6)2 = -4(y + 15)

Obtained equation is in the form of (x – h)2 = -4a(y – k)
-4a = -4 ⇒ a = 1

So, the vertex = (h, k) = (6, - 15)
Focus = (h, k – a) = (6, -15-1) = (6, -16)

Equation of the directrix is y = k + a
⇒ y = -15 + 1 ⇒ y = -14
⇒ y + 14 = 0

Practice Problems on Standard Equation of Parabola

**Question 1: Find the focus, directrix, and vertex of the parabola y² = 8x.

**Question 2: Write the equation of the parabola with focus (3, 0) and directrix x = -3.

**Question 3: Find the equation of the tangent to the parabola y² = 12x at the point (3, 6).

**Question 4: Find the equation of the normal to the parabola x² = 4y at the point (4, 4).

**Question 5: Find the length of the latus rectum of the parabola y² = -16x.

**Question 6: Find the coordinates of the points of intersection of the line y = 2x + 1 with the parabola x² = 4y.

**Question 7: Find the equation of the parabola with vertex at (2, -1) and focus at (4, -1).

**Question 8: Find the equation of the tangent to the parabola y² = 4ax which makes an angle of 45° with the x-axis.