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Standard Equation of a Parabola with Examples

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The graph of a quadratic equation can be referred to as a parabola. In conic sections, a parabola is the equation of a curve, where a point on the curve is in the equidistant form of a fixed-line and a fixed point in the plane. The fixed line is known as the directrix of a parabola, and the fixed point is known as the focus of a parabola. In simple words, a parabola is referred to as the locus of a point that is equidistant from a fixed line (directrix) and a fixed point (focus). The axis of a parabola passes through the focus and is perpendicular to the directrix of a parabola. The point of intersection of the parabola with the axis is called the vertex of a parabola.

Equation 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.

Some important terms and parts of a parabola

  1. Focus: Focus is the fixed point of a parabola.
  2. Directrix:  The directrix of a parabola is the line perpendicular to the axis of a parabola. 
  3. Focal Chord: The chord that passes through the focus of a parabola, cutting the parabola at two distinct points, is called the focal chord.
  4. Focal Distance:  The focal distance is the distance of a point (x1, y1) on the parabola from the focus. 
  5. 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.
  6. 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 the Parabola

Parabola

Formulae of parameters of a parabola

y2 = 4ax

 

Vertex = (0,0)

Focus = (a, 0)

The parabola opens to the right side.

Equation of the axis is y = 0

Equation of the directrix is x + a = 0

The length of the latus rectum = 4a

y2 = -4ax

 

Vertex = (0,0)

Focus = (-a, 0)

The parabola opens to the left side.

Equation of the axis is y = 0

Equation of the directrix is x – a = 0

The length of the latus rectum = 4a

x2 = 4ay

 

Vertex = (0,0)

Focus = (0, a)

The parabola opens upwards.

Equation of the axis is x = 0

Equation of the directrix is y + a = 0

The length of the latus rectum = 4a

x2 = -4ay

 

Vertex = (0,0)

Focus = (0, -a)

The parabola opens downwards.

Equation of the axis is x = 0

Equation of the directrix is y – a = 0

The length 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 with respect to 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 the Parabola

Parabola

Formulae of parameters of a parabola

(y – k)2 = 4a(x – h)

 

Vertex = (h, k)

Focus = (h + a, k)

The parabola opens to the right side.

Equation of the axis is y = k

Equation of the directrix is x = h – a

The length of the latus rectum = 4a

(y – k)2 = -4a(x – h)

 

Vertex = (h, k)

Focus = (h – a, k)

The parabola opens to the left side.

Equation of the axis is y = k

Equation of the directrix is x = h + a

The length of the latus rectum = 4a

(x – h)2 = 4a(y – k)

 

Vertex = (h, k)

Focus = (h, k + a)

The parabola opens upwards.

Equation of the axis is x = h

Equation of the directrix is y = k – a

The length of the latus rectum = 4a

(x – h)2 = -4a(y – k)

 

Vertex = (h, k)

Focus = (h, k – a)

The parabola opens downwards.

Equation of the axis is x = h

Equation of the directrix is y = k + a

The length of the latus rectum = 4a

Derivation of equation of the parabola

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 the 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.

Solved Examples

Problem 1: Find the length of the latus rectum, focus, and vertex, if the equation of the parabola is y2 = 12x.

Solution:

Given, 

The 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,

The 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)

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

Solution:

Given,

The 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.

The 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.

Problem 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,

The equation of the parabola is: x2 = 16y

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

4a = 16 ⇒  a = 4

The 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.

Problem 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, 

The 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, 

The length of latus rectum of a parabola = 4a

= 4(2) = 8

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

Now, Vertex = (2, 16).

Problem 5: The equation of a parabola is x2 – 12x + 4y – 24 = 0, then find its vertex, focus, and directrix.

Solution:

Given, 

The 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)

The 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)

The equation of the directrix is y = k + a

⇒ y = -15 + 1 ⇒ y = -14

⇒ y + 14 = 0



Last Updated : 24 Jan, 2024
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