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
**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:
- **Focus****:** Focus is the fixed point of a parabola.
- **Directrix****:** The directrix of a parabola is the line perpendicular to the axis of a parabola.
- **Focal Chord****:** The chord that passes through the focus of a parabola, cutting the parabola at two distinct points, is called the focal chord.
- **Focal Distance: The focal distance is the distance of a point (x1, y1) on the parabola from the focus.
- **Latus Rectum****:** A latus rectum is a focal chord that passes through the focus of a parabola and is perpendicular to the axis of the parabola. The length of the latus rectum is LL' = 4a.
- **Eccentricity****:** The ratio of the distance of a point from the focus to its distance from the directrix is called eccentricity (e). For a parabola, eccentricity is equal to 1, i.e., e = 1.
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 | Horizontal 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 | Horizontal 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 | Vertical 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 | Vertical 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:
- A parabola is symmetrical w.r.t its axis. For example, y2 = 4ax is symmetric w.r.t the x-axis, whereas x2 = 4ay is symmetric concerning the y-axis.
- If a parabola is symmetric about the x-axis, then the parabola opens towards the right if the x-coefficient is positive and towards the left if the x-coefficient is negative.
- If a parabola is symmetric about the y-axis, then the parabola opens upwards if the y-coefficient is positive and downwards if the y-coefficient is negative.
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) | Horizontal 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) | Horizontal 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) | Vertical 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) | Vertical 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).

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.
- y2 = -4ax
- x2 = 4ay
- x2 = -4ay
**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 = 3We 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 = 25/16Therefore, 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 = 4Coefficient 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) = 8Now, 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 = 1So, 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.
Horizontal Parabola
Horizontal Parabola
Vertical Parabola
Vertical Parabola
Horizontal Parabola
Horizontal Parabola
Vertical Parabola
Vertical Parabola