10.4 Parabola – Class 11 Mathematics

Introduction

The parabola is one of the most important conic sections. It appears not only in mathematics but also in physics, astronomy, and engineering. A parabola is formed when a plane cuts a right circular cone parallel to a generator (slant edge) of the cone.

Analytically, a parabola is defined as the locus of a point which moves in a plane such that its distance from a fixed point (focus) is equal to its perpendicular distance from a fixed line (directrix).

Parabolas are found in the paths of projectiles, the shapes of satellite dishes, car headlights, solar cookers, and even suspension bridges. For JEE, KCET, and COMEDK, parabola questions are highly scoring since they often rely on well-structured formulas and properties.

Standard Definition

A parabola is the set of all points P(x, y) such that:

PF = PD

Where,

• F = Focus (a, 0)
• D = Directrix x = –a
• Vertex = (0, 0)
• Axis = x-axis

Derivation of Standard Equation of a Parabola

Let point P(x, y) be on parabola.

• Distance PF = √((x – a)² + y²)
• Distance PD = |x + a|

Condition: PF = PD
⇒ √((x – a)² + y²) = |x + a|

Squaring: (x – a)² + y² = (x + a)²
⇒ x² – 2ax + a² + y² = x² + 2ax + a²
⇒ y² = 4ax

Thus, standard equation of parabola is:

y² = 4ax

This parabola opens towards the positive x-axis.

Standard Forms of Parabola

Depending on orientation of axis:

1. y² = 4ax (opens right)
2. y² = –4ax (opens left)
3. x² = 4ay (opens upward)
4. x² = –4ay (opens downward)

Key Elements of Parabola

For parabola y² = 4ax:

• Vertex: (0, 0)
• Focus: (a, 0)
• Directrix: x = –a
• Axis: x-axis
• Latus Rectum: Line through focus perpendicular to axis, length = 4a

General Equation of a Parabola

If vertex is (h, k) and axis is parallel to x-axis, equation becomes:
(y – k)² = 4a(x – h)

Similarly for other orientations.

Properties of a Parabola

1. Symmetry: Parabola is symmetric about its axis.
2. Focus–Directrix property: Distance from focus = distance from directrix.
3. Tangent at a point: Slope form and point form.
4. Normal at a point: Perpendicular to tangent.
5. Length of latus rectum: 4a.

Equation of Tangent to a Parabola

At a Point (x₁, y₁) on y² = 4ax

Equation: yy₁ = 2a(x + x₁)

Parametric Form

If point is P(at², 2at), tangent equation is:
yt = x + at²

Equation of Normal

At point P(at², 2at):
Equation of normal: y + xt = 2at + at³

Parametric Representation of Parabola

For parabola y² = 4ax, point P can be written as:
(x, y) = (at², 2at)

This parameter t simplifies tangent and normal equations, making them exam favorites.

Focal Chord

A chord passing through the focus of parabola.

If P(at₁², 2at₁) and Q(at₂², 2at₂) are endpoints,
then t₁t₂ = –1

Latus Rectum

Equation: x = a
Length = 4a
Endpoints: (a, 2a), (a, –2a)

Solved Examples

Example 1: Equation of Tangent

Find tangent at point (4, 4) on parabola y² = 16x.

Solution:
Equation: yy₁ = 2a(x + x₁), with a = 4, (x₁, y₁) = (4, 4)
⇒ y(4) = 8(x + 4)
⇒ 4y = 8x + 32
⇒ y = 2x + 8

Example 2: Normal Equation

Find normal at point (at², 2at) for parabola y² = 4ax.

Solution:
Equation: y + xt = 2at + at³

Example 3: Focal Chord

If one end of focal chord is (at₁², 2at₁), other end is (at₂², 2at₂) with t₁t₂ = –1.

Proof: Using property of parabola and focus.

Example 4: Directrix Condition (JEE style)

Find equation of parabola with focus (2, 0) and directrix x = –2.

Solution:
Vertex is (0, 0), so the equation is y² = 8x.

Example 5: Latus Rectum

Find endpoints of the latus rectum for y² = 4x.

Solution:
a = 1, so endpoints are (1, 2), (1, –2).

Applications of Parabola

1. Physics:
   • Path of projectile under gravity.
   • Reflectors in headlights and solar cookers.
2. Astronomy:
   • Telescopes and satellite dishes use parabolic mirrors.
3. Engineering:
   • Suspension bridges, satellite antennas.
4. Exams:
   • JEE: Tangents, normals, focal chords.
   • KCET: Direct questions on equation and properties.

Common Mistakes

• Forgetting correct form (x² vs y²).
• Wrong tangent equation (confusing with ellipse form).
• Missing factor of 2a in tangent/normal equations.
• Confusing directrix orientation.

Exam Weightage

• CBSE Boards: 6–8 marks (equation-based).
• JEE Main: 1–2 questions (~8 marks).
• JEE Advanced: 1 conceptual problem (~4 marks).
• KCET/COMEDK: 2–3 direct problems (2–3 marks each).

Practice Problems

1. Find equation of parabola with focus (3, 0) and directrix x = –3.
2. Find equation of tangent to parabola y² = 12x at (3, 6).
3. Find normal to parabola y² = 4x at (1, 2).
4. Show that focal chord satisfies t₁t₂ = –1.
5. Find coordinates of vertex, focus, and directrix of x² = –8y.
6. Find the length of the latus rectum of y² = 20x.
7. Find slope of tangent at P(at², 2at).
8. Prove that product of slopes of focal chord endpoints = –1.
9. If parabola passes through (1, 2) with an axis along the x-axis, find the equation.
10. Write parametric coordinates for y² = 8x.

FAQs

Q1. Which conic has eccentricity e = 1?
Parabola.

Q2. What is the parametric form of parabola y² = 4ax?
(x, y) = (at², 2at).

Q3. What is the latus rectum length of y² = 4ax?
4a.

Q4. Which property makes parabolas useful in reflectors?
All rays parallel to the axis reflect through focus.

Q5. Are parabolas common in JEE Advanced?
Yes, usually in tangent-normal or locus-based problems.

Conclusion

The parabola is one of the most fascinating conic sections, with applications ranging from projectile motion to satellite communication. Defined by its focus-directrix property, the parabola combines simple geometry with rich analytical formulas.

We studied its standard equations, tangents, normals, parametric representation, focal chords, and latus rectum. For JEE, KCET, and COMEDK, problems often involve finding tangent/normal equations, using parametric coordinates, and applying focal chord properties.

Mastering parabola ensures not just exam success but also builds a strong base for advanced coordinate geometry and real-world problem solving.