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
- Focus: Focus is the fixed point of a parabola.
- Directrix: Â The directrix of a parabola is the line perpendicular to the axis of a parabola.Â
- Focal Chord:Â The chord that passes through the focus of a parabola, cutting the parabola at two distinct points, is called the focal chord.
- Focal Distance: Â The focal distance is the distance of a point (x1, y1) on the parabola from the focus.Â
- 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:Â 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.
y2 = 4ax
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 |
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
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 |
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
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 |
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
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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.Â
(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).
Â
From the definition of the eccentricity of a parabola, we haveÂ
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,Â
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: Â 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.Â
Solution:
Given,Â
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,Â
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,Â
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)Â
-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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