Conic Sections is Chapter 10 of CBSE Class 11 Maths — the chapter where geometry and algebra finally shake hands. A single slice through a cone can give you a circle, a parabola, an ellipse, or a hyperbola, and each of these curves has a clean standard equation you can recognise on sight. Master this chapter and you unlock a whole family of questions in JEE, NEET-adjacent maths, and your board exam.
By the end of these notes you will be able to identify any conic from its equation, write the standard forms of the circle, parabola, ellipse and hyperbola, and pull out the focus, directrix, vertices, eccentricity, foci and latus rectum without hesitation. This is a high-weightage chapter carrying roughly 7–9 marks in boards, and the foundation for coordinate geometry, calculus, and most of analytic geometry that follows.
Table of Contents
- Key Concepts — Sections of a cone, circle, parabola, ellipse, hyperbola
- Weightage in Board & Entrance Exams
- Important Definitions
- Solved Examples
- Important Questions for Board Exams
- Quick Revision Points
Key Concepts
1. Sections of a Cone
A conic section is the curve obtained when a plane cuts a double right circular cone. Depending on the angle the cutting plane makes with the axis, you get four non-degenerate curves: the circle, ellipse, parabola and hyperbola.
[DIAGRAM: A double right circular cone with vertex V, axis, and a generator line making angle α with the axis. A cutting plane making angle β with the axis produces different conics.]
The Cutting-Angle Rule
Let the cone’s generators make a semi-vertical angle α with the axis, and let the cutting plane make angle β with the axis.
- β = 90°: the plane is perpendicular to the axis → circle.
- α < β < 90°: the plane cuts one nappe at a slant → ellipse.
- β = α: the plane is parallel to a generator → parabola.
- 0 ≤ β < α: the plane cuts both nappes → hyperbola.
Degenerate Conics
When the cutting plane passes through the vertex, you get a degenerate conic — a point, a straight line, or a pair of intersecting lines.
2. The Circle
A circle is the set of all points in a plane that are at a fixed distance (the radius r) from a fixed point (the centre).
Standard Equation
For centre (h, k) and radius r:
(x − h)² + (y − k)² = r²
If the centre is the origin (0, 0), this reduces to x² + y² = r².
General Equation
The general second-degree equation of a circle is:
x² + y² + 2gx + 2fy + c = 0
- Centre: (−g, −f)
- Radius: r = √(g² + f² − c)
- It represents a real circle only if g² + f² − c > 0; a point circle if it equals 0; and an imaginary circle if it is negative.
Note: A general second-degree equation Ax² + By² + 2Hxy + 2Gx + 2Fy + C = 0 represents a circle only when A = B and H = 0 (no xy term, equal coefficients of x² and y²).
3. The Parabola
A parabola is the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed line (the directrix).
The constant ratio of these two distances — the eccentricity e — equals exactly 1 for a parabola.
[DIAGRAM: A right-opening parabola y² = 4ax with vertex at origin, focus at (a, 0), directrix x = −a, axis along the x-axis, and the latus rectum as a vertical chord through the focus.]
Standard Forms
| Equation | Opens | Focus | Directrix | Axis |
|---|---|---|---|---|
| y² = 4ax | Right | (a, 0) | x = −a | x-axis |
| y² = −4ax | Left | (−a, 0) | x = a | x-axis |
| x² = 4ay | Up | (0, a) | y = −a | y-axis |
| x² = −4ay | Down | (0, −a) | y = a | y-axis |
Key Features (for y² = 4ax)
- Vertex: (0, 0)
- Length of latus rectum: 4a
- Latus rectum is the chord through the focus, perpendicular to the axis.
- The parabola is symmetric about its axis.
4. The Ellipse
An ellipse is the set of all points in a plane the sum of whose distances from two fixed points (the foci) is constant. That constant equals the length of the major axis, 2a.
Its eccentricity satisfies 0 < e < 1.
[DIAGRAM: A horizontal ellipse centred at origin with major axis along x-axis, vertices (±a, 0), foci (±c, 0), minor axis along y-axis with endpoints (0, ±b).]
Standard Forms
| Equation | Major axis | Vertices | Foci |
|---|---|---|---|
| x²/a² + y²/b² = 1 (a > b) | x-axis (length 2a) | (±a, 0) | (±c, 0) |
| x²/b² + y²/a² = 1 (a > b) | y-axis (length 2a) | (0, ±a) | (0, ±c) |
Key Relations
- Relation between a, b, c: c² = a² − b²
- Eccentricity: e = c/a = √(1 − b²/a²), with 0 < e < 1
- Length of major axis: 2a; minor axis: 2b
- Length of latus rectum: 2b²/a
- Sum of focal distances of any point = 2a
5. The Hyperbola
A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points (the foci) is constant (equal to 2a, the length of the transverse axis).
Its eccentricity satisfies e > 1.
[DIAGRAM: A horizontal hyperbola centred at origin with two branches opening left and right, vertices (±a, 0), foci (±c, 0), and the two asymptotes y = ±(b/a)x crossing at the centre.]
Standard Forms
| Equation | Transverse axis | Vertices | Foci | Asymptotes |
|---|---|---|---|---|
| x²/a² − y²/b² = 1 | x-axis | (±a, 0) | (±c, 0) | y = ±(b/a)x |
| y²/a² − x²/b² = 1 | y-axis | (0, ±a) | (0, ±c) | y = ±(a/b)x |
Key Relations
- Relation between a, b, c: c² = a² + b²
- Eccentricity: e = c/a = √(1 + b²/a²), with e > 1
- Length of transverse axis: 2a; conjugate axis: 2b
- Length of latus rectum: 2b²/a
- A rectangular (equilateral) hyperbola has a = b, so e = √2 and asymptotes y = ±x.
6. Eccentricity — The Unifying Idea
Every conic can be defined by a single ratio e = (distance from focus)/(distance from directrix). This eccentricity tells you instantly which conic you are looking at.
| Conic | Eccentricity (e) | c–relation |
|---|---|---|
| Circle | e = 0 | foci coincide at centre |
| Ellipse | 0 < e < 1 | c² = a² − b² |
| Parabola | e = 1 | — |
| Hyperbola | e > 1 | c² = a² + b² |
Weightage in Board & Entrance Exams
| Exam | Typical Weightage | Most-Tested Areas |
|---|---|---|
| CBSE Board (Class 11) | 7–9 marks | Circle equations, parabola focus/latus rectum, ellipse & hyperbola standard forms |
| JEE Main / Advanced | 2–4 questions | Eccentricity, foci, tangents, latus rectum, locus problems |
| Competitive (CUET etc.) | 2–3 questions | Identifying conics, finding centre/radius, standard-form conversions |
[TABLE: Question-type split — VSA (1 mark): identify the conic, state eccentricity; SA (2–3 marks): find focus/directrix/latus rectum, centre & radius of a circle; LA (5 marks): derive the equation of an ellipse/hyperbola from given foci and vertices.]
Important Definitions
| Term | Definition |
|---|---|
| Conic section | Curve formed by the intersection of a plane and a double right circular cone |
| Circle | Locus of points at a fixed distance r from a fixed centre: (x − h)² + (y − k)² = r² |
| Parabola | Locus of points equidistant from a focus and a directrix; e = 1 |
| Ellipse | Locus of points whose distances from two foci sum to a constant 2a; 0 < e < 1 |
| Hyperbola | Locus of points whose distances from two foci differ by a constant 2a; e > 1 |
| Focus | Fixed point used to define a conic |
| Directrix | Fixed line used (with the focus) to define a conic |
| Eccentricity (e) | Ratio of distance from focus to distance from directrix; fixes the conic type |
| Latus rectum | Focal chord perpendicular to the major/transverse axis; 4a (parabola) or 2b²/a (ellipse, hyperbola) |
| Asymptote | Line a hyperbola approaches but never meets: y = ±(b/a)x |
Solved Examples
Example 1
Find the centre and radius of the circle x² + y² − 6x + 4y − 12 = 0.
Answer: Here 2g = −6, 2f = 4, c = −12, so g = −3, f = 2. Centre = (−g, −f) = (3, −2). Radius = √(g² + f² − c) = √(9 + 4 + 12) = √25 = 5.
Example 2
Find the equation of a circle with centre (2, −3) and radius 4.
Answer: (x − 2)² + (y + 3)² = 4² → (x − 2)² + (y + 3)² = 16, i.e. x² + y² − 4x + 6y − 3 = 0.
Example 3
For the parabola y² = 12x, find the focus, directrix and length of the latus rectum.
Answer: Compare with y² = 4ax → 4a = 12 → a = 3. Focus = (3, 0); directrix x = −3; length of latus rectum = 4a = 12.
Example 4
Find the coordinates of the foci and the eccentricity of the ellipse x²/25 + y²/9 = 1.
Answer: a² = 25, b² = 9, so a = 5, b = 3. c² = a² − b² = 25 − 9 = 16 → c = 4. Foci = (±4, 0); eccentricity e = c/a = 4/5 = 0.8.
Example 5
Find the eccentricity, foci and length of the latus rectum of the hyperbola x²/16 − y²/9 = 1.
Answer: a² = 16, b² = 9, so a = 4, b = 3. c² = a² + b² = 25 → c = 5. Eccentricity e = c/a = 5/4 = 1.25; foci = (±5, 0); latus rectum = 2b²/a = 2(9)/4 = 4.5.
Example 6
Find the equation of the ellipse whose vertices are (±6, 0) and foci are (±4, 0).
Answer: a = 6, c = 4 → b² = a² − c² = 36 − 16 = 20. Equation: x²/36 + y²/20 = 1.
Important Questions for Board Exams
1-Mark Questions (VSA)
- What is the eccentricity of a parabola?
- Write the standard equation of a circle with centre at the origin and radius r.
- For the ellipse x²/a² + y²/b² = 1 (a > b), what is the length of the latus rectum?
- What are the asymptotes of the hyperbola x²/a² − y²/b² = 1?
- Name the conic obtained when a plane cuts a cone perpendicular to its axis.
2–3-Mark Questions (SA)
- Find the centre and radius of the circle 2x² + 2y² − 8x + 12y − 6 = 0.
- Find the focus, directrix and latus rectum of the parabola x² = −16y.
- Find the foci, vertices and eccentricity of the ellipse 9x² + 4y² = 36.
- Find the eccentricity and the length of the latus rectum of the hyperbola 9x² − 16y² = 144.
5-Mark Questions (LA)
- Define a parabola and derive its standard equation y² = 4ax taking the focus at (a, 0) and directrix x = −a.
- Find the equation of the ellipse whose foci are (±5, 0) and the length of the major axis is 18. State its eccentricity.
- Find the equation of the hyperbola with vertices (±7, 0) and one focus at (10, 0). Find its eccentricity and latus rectum.
Quick Revision Points
- Conics come from slicing a cone — circle, ellipse, parabola, hyperbola, set by the cutting angle
- Circle: (x − h)² + (y − k)² = r²; general x² + y² + 2gx + 2fy + c = 0, centre (−g, −f), r = √(g² + f² − c)
- Parabola: y² = 4ax → focus (a, 0), directrix x = −a, latus rectum 4a; e = 1
- Ellipse: x²/a² + y²/b² = 1, c² = a² − b², e = c/a (0 < e < 1), latus rectum 2b²/a
- Hyperbola: x²/a² − y²/b² = 1, c² = a² + b², e = c/a (e > 1), asymptotes y = ±(b/a)x, latus rectum 2b²/a
- Eccentricity ladder: circle 0, ellipse 0–1, parabola 1, hyperbola >1
- Latus rectum is the focal chord perpendicular to the axis
- Rectangular hyperbola: a = b, e = √2, asymptotes y = ±x
Next Chapter: Chapter 11 — Introduction to Three Dimensional Geometry
Chapter Navigation
Previous: Straight Lines Class 11 Notes
Next: Introduction to Three Dimensional Geometry Class 11 Notes
Related Chapters in Class 11 Maths
- Straight Lines Class 11 Notes
- Trigonometric Functions Class 11 Notes
- Limits and Derivatives Class 11 Notes
Practice What You Learned
Take your coordinate geometry further with our Class 12 Maths notes once you are board-ready.