The parabola, a conic section, is a symmetric U-shaped curve. It appears in everyday scenarios and physics, showcasing its versatility. The parabola's mirror symmetry is evident from the trajectory of a thrown ball to the design of satellite dishes. Its ubiquity underscores its significance across diverse applications, unifying various phenomena under a single mathematical framework.
The parabola, integral to conic sections, arises from slicing a right circular cone. Defined by quadratic equations, its graph depicts points that are equal in distance from a fixed line called the directrix and point (the focus). This elegant curve embodies fundamental geometric principles, with applications ranging from projectile motion to satellite dish design.
A parabola is a curve formed by all points in a plane that are equidistant from a specified point (referred to as the focus) and a fixed line (known as the directrix).
The general equation of a parabola with axis parallel to y-axis form can be expressed as:
y = ax2 + bx + c
Where:
This equation allows us to describe various types of parabolas, including those that open upwards or downwards and may be shifted horizontally or vertically.
The equation of a parabola is simplest when the vertex is located at the origin and the axis of symmetry aligns with either the x-axis or the y-axis. The 4 possible orientations of the parabola are shown below.
These 4 equations are known as standard equations of parabolas.
Observations from these equations
The parabola latus rectum is a line segment that passes through its focus and is perpendicular to its axis. It has its endpoints on the parabola and is parallel to the directrix. The length of the latus rectum is typically denoted by 4a, where a is the distance from the vertex to the focus.
Parabola eccentricity is defined as the ratio of the distance from the focus to any point on the parabola to the perpendicular distance from that point to the directrix.
For a parabola, the eccentricity is always equal to 1. This means that the distance from any point on the parabola to the focus is always equal to the perpendicular distance from that point to the directrix. Unlike ellipses and hyperbolas, where the eccentricity can vary.
constant= e[/latex]“e” is the eccentricity, which is equal to 1.
Parametric equation of a chord joining two points “t1” and “t2” of the parabola y2 = 4ax is
2x– y (t1 + t2) + 2at1t2 = 0
A tangent to a parabola is a straight line that touches the curve at a single point.
Equation of Tangent to Parabola y2 =4ax for all m ≠ 0 at (a/m2, 2a/m) is: y = mx + a/m
Equation of Tangent to Parabola y2 =4ax at point (x1, y1) is: yy1 = 2a (x + x1)
Equation of Tangent to Parabola y2 =4ax at the point P (at2, 2at) is: ty = x + at2
The normal to a parabola is a line perpendicular to the tangent at a given point and passing through that point.
Equation of Normal to Parabola y2 =4ax for all m ≠ 0 at (am2, –2am) is: y = mx – 2am – am3.
Equation of Normal to Parabola y2 =4ax at point (x1, y1) is: y – y1 = ( –y1/2a) (x – x1)
Equation of Normal to Parabola y2 =4ax at point (at2, 2at) is: y + tx = 2at + at3.
Example 1: Find the vertex, axis, directrix, and the length of the latus rectum for the given parabola: y2 =12x
Solution: The given equation is in the form y2 =4ax, where 4a = 12, a = 3.
Vertex = (0,0)
Axis = y = 0
Directrix = x = –a So, x = –3
Length of Latus Rectum ⇒ 4a = (4 × 3) units = 12 units.
Example 2: Find the axis, vertex, tangent at the vertex, directrix, and the length of the latus rectum of the given parabola: y2 + 4y – 2x – 2 = 0
Solution: The given solution can be re-written as (y + 2)2 = 2(x + 3)
Which is in the form Y2 =4aX
where Y = y + 2, X = x + 3, 4a = 2
Hence, the axis ⇒ Y = 0 ⇒ y + 2 = 0 ⇒ y = –2
The vertex is X = 0, Y = 0, i.e. (– 3, –2).
The tangent at the vertex is:
X = 0 ⇒ x + 3 = 0 ⇒ x = – 3
The directrix is X + a = 0.
⇒ x + 3 + ½ = 0
⇒ x = – 7/2
Length of the latus rectum = 4a = 2 units.
Example 3: Determine the equation of the parabola whose focus is ( –2, 3) and a directrix given by the line x + y + 2 = 0.
Solution: Let P(x, y) be any point on the parabola. then
⇒ (x +2)2 +(y – 3)2 = (x +y + 2)2/2
⇒ 2( x2 + 4x + 4 + y2 – 6y + 9) = x2 +y2 + 4 + 4x + 4y + 2xy
⇒ x2 + y2 + 4x – 16y + 2xy + 22 = 0
Example 4: How do you find the vertex of a parabola?
Ans: The parabola vertex can be found using the formula for a parabola in standard form y=ax2 + bx + c.
(Session 2025 - 26)