Standard Equation of a Parabola

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.

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 = ax² + bx + c
(or)
x = ay² + 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.

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 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).

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 ⇒ y² = 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

y² = 4ax, y² = -4ax, x² = 4ay, and x² = -4ay are the standard equations of a parabola.

Articles Related to Parabola:

• Equation of Circle
• Equation of Ellipse
• Equation of Hyperbola
• Applications of Parabola in Real Life

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 y² = 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 x² = 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 x² - 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.