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Class 11 Mathematics Chapter 10 Conic Sections — Formulas & Key Points

Conic Sections in CBSE Class 11 Mathematics Chapter 10 introduces circle, parabola, ellipse, and hyperbola as curves obtained by slicing a double cone. Mastery of this chapter requires memorising 20+ formulas for standard equations, eccentricity, foci, directrix, vertices, axes, and latus rectum. This formula sheet organises every identity from the NCERT textbook into quick-reference tables, highlights notation traps, and provides three solved examples applying the formulas to typical board exam questions.

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Key takeaways

✓Circle equation: (x - h)² + (y - k)² = r² where (h,k) is centre and r is radius; general form x² + y² + 2gx + 2fy + c = 0 has centre (-g,-f) and radius √(g² + f² - c)
✓Parabola has eccentricity e = 1; standard forms are y² = 4ax (opens right), y² = -4ax (left), x² = 4ay (up), x² = -4ay (down) with focus at (a,0), (−a,0), (0,a), (0,−a) respectively
✓Ellipse (x²/a²) + (y²/b²) = 1 where a > b has major axis 2a along x-axis, minor axis 2b, eccentricity e = √(1 - b²/a²), foci at (±ae, 0), and latus rectum 2b²/a
✓Hyperbola (x²/a²) - (y²/b²) = 1 has eccentricity e = √(1 + b²/a²) always greater than 1, foci at (±ae, 0), vertices at (±a, 0), and asymptotes y = ±(b/a)x
✓Latus rectum is the chord through focus perpendicular to major axis: 4a for parabola, 2b²/a for ellipse and hyperbola
✓For ellipse b² = a²(1 - e²); for hyperbola b² = a²(e² - 1); remembering these relations prevents formula confusion during problem-solving
✓Common mistake: confusing (x²/a²) + (y²/b²) = 1 with a < b (major axis vertical) versus a > b (major axis horizontal) changes focus and vertex coordinates

Circle — All Formulas and Key Properties

A circle is the locus of points equidistant from a fixed point called the centre. It is a conic with eccentricity zero. The standard equation is (x - h)² + (y - k)² = r² where (h, k) is the centre and r is the radius. The general form x² + y² + 2gx + 2fy + c = 0 represents a circle with centre (-g, -f) and radius √(g² + f² - c). For the circle to exist, the condition g² + f² - c > 0 must hold. The equation x² + y² = r² is the simplest form when the centre is at the origin. Diameter form of the circle passing through endpoints (x₁, y₁) and (x₂, y₂) is (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0, often used in coordinate geometry problems.

Standard equation: (x - h)² + (y - k)² = r², centre (h, k), radius r
General form: x² + y² + 2gx + 2fy + c = 0, centre (-g, -f), radius √(g² + f² - c)
Condition for real circle: g² + f² - c > 0
Diameter form: (x - x₁)(x - x₂) + (y - y₁)(y - y₂) = 0
Eccentricity of circle: e = 0

Parabola — Standard Equations and Elements

A parabola is the locus of points equidistant from a fixed point (focus) and a fixed line (directrix). Its eccentricity is exactly 1. Four standard orientations exist depending on the axis of symmetry. For y² = 4ax (opening right), the vertex is at origin, focus at (a, 0), directrix x = -a, axis is the x-axis, and latus rectum length is 4a. For y² = -4ax (opening left), focus is at (-a, 0) and directrix is x = a. For x² = 4ay (opening upward), focus is at (0, a), directrix y = -a, axis is the y-axis. For x² = -4ay (opening downward), focus is (0, -a) and directrix y = a. The parameter 'a' is always positive and represents the distance from vertex to focus.

y² = 4ax: vertex (0,0), focus (a,0), directrix x = -a, axis y = 0, opens right
y² = -4ax: vertex (0,0), focus (-a,0), directrix x = a, opens left
x² = 4ay: vertex (0,0), focus (0,a), directrix y = -a, axis x = 0, opens up
x² = -4ay: vertex (0,0), focus (0,-a), directrix y = a, opens down
Latus rectum for all parabolas: 4a
Eccentricity: e = 1

Ellipse — Standard Forms and All Parameters

An ellipse is the locus of points where the sum of distances from two fixed points (foci) is constant. Eccentricity e < 1. Standard equation (x²/a²) + (y²/b²) = 1 with a > b has major axis along x-axis of length 2a, minor axis along y-axis of length 2b, centre at origin, vertices at (±a, 0), co-vertices at (0, ±b), foci at (±ae, 0) where e = √(1 - b²/a²), and directrices x = ±a/e. Latus rectum length is 2b²/a. When b > a, the major axis is vertical: vertices (0, ±b), foci (0, ±be), e = √(1 - a²/b²), latus rectum 2a²/b. The relation b² = a²(1 - e²) links eccentricity to axes. Distance between foci is 2ae.

(x²/a²) + (y²/b²) = 1 with a > b: major axis 2a (horizontal), minor axis 2b
Vertices: (±a, 0); Co-vertices: (0, ±b); Centre: (0, 0)
Eccentricity: e = √(1 - b²/a²); Foci: (±ae, 0); Directrices: x = ±a/e
Latus rectum: 2b²/a; Relation: b² = a²(1 - e²)
If b > a, major axis vertical: vertices (0, ±b), foci (0, ±be), e = √(1 - a²/b²)
Sum of focal distances for any point on ellipse: 2a

Hyperbola — Standard Equations and Key Elements

A hyperbola is the locus of points where the absolute difference of distances from two foci is constant. Eccentricity e > 1 always. Standard form (x²/a²) - (y²/b²) = 1 has transverse axis along x-axis of length 2a, conjugate axis along y-axis of length 2b, centre at origin, vertices at (±a, 0), foci at (±ae, 0) where e = √(1 + b²/a²), directrices x = ±a/e, and latus rectum 2b²/a. The asymptotes are straight lines y = ±(b/a)x that the hyperbola approaches but never touches. For (y²/a²) - (x²/b²) = 1, transverse axis is vertical: vertices (0, ±a), foci (0, ±ae). Relation b² = a²(e² - 1) is crucial. Distance between foci is 2ae, distance between vertices is 2a.

(x²/a²) - (y²/b²) = 1: transverse axis 2a (horizontal), conjugate axis 2b
Vertices: (±a, 0); Foci: (±ae, 0); Centre: (0, 0)
Eccentricity: e = √(1 + b²/a²), always e > 1; Directrices: x = ±a/e
Latus rectum: 2b²/a; Relation: b² = a²(e² - 1)
Asymptotes: y = ±(b/a)x
(y²/a²) - (x²/b²) = 1: transverse axis vertical, vertices (0, ±a), foci (0, ±ae)

Quick Comparison Table — Circle, Parabola, Ellipse, Hyperbola

This table consolidates the defining characteristics, eccentricity, and standard equations of all four conic sections covered in NCERT Class 11 Mathematics Chapter 10. Circle is a special case with zero eccentricity. Parabola has eccentricity exactly one, ellipse has eccentricity between zero and one, and hyperbola has eccentricity greater than one. Understanding these distinctions helps you identify the conic type from a given equation or eccentricity value. The general second-degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 represents different conics based on the discriminant B² - 4AC: circle if A = C and B = 0, parabola if B² - 4AC = 0, ellipse if B² - 4AC < 0, and hyperbola if B² - 4AC > 0. This discriminant test is important for identifying conic type in analytical problems.

Circle: e = 0, equation (x - h)² + (y - k)² = r²
Parabola: e = 1, y² = 4ax or x² = 4ay
Ellipse: 0 < e < 1, (x²/a²) + (y²/b²) = 1
Hyperbola: e > 1, (x²/a²) - (y²/b²) = 1
General form discriminant: B² - 4AC determines conic type
Parabola if B² - 4AC = 0; ellipse if B² - 4AC < 0; hyperbola if B² - 4AC > 0

Latus Rectum, Focal Chord, and Special Chords

Latus rectum is the chord passing through a focus and perpendicular to the major or transverse axis. For parabola y² = 4ax, latus rectum length is 4a with endpoints (a, 2a) and (a, -2a). For ellipse (x²/a²) + (y²/b²) = 1, latus rectum length is 2b²/a. For hyperbola (x²/a²) - (y²/b²) = 1, latus rectum is also 2b²/a. A focal chord is any chord passing through a focus. For parabola, if a focal chord has one end at parameter t₁, the other end is at parameter t₂ = -1/t₁, and the length of focal chord is a(t₁ - t₂)²/4t₁t₂ for parametric form. The semi-latus rectum (half of latus rectum) appears in polar equations of conics: r = l/(1 - e cosθ) where l is semi-latus rectum.

Parabola latus rectum: 4a
Ellipse latus rectum: 2b²/a
Hyperbola latus rectum: 2b²/a
Focal chord: any chord through a focus
For parabola, endpoints of latus rectum at focus (a, 0) are (a, 2a) and (a, -2a)

Memory Tricks and Mnemonics for Conic Sections

Students often confuse ellipse and hyperbola formulas because both involve a² and b². Remember: ellipse is addition (x²/a²) + (y²/b²) = 1, hyperbola is subtraction (x²/a²) - (y²/b²) = 1. For eccentricity, ellipse formula has 1 minus (e = √(1 - b²/a²)) while hyperbola has 1 plus (e = √(1 + b²/a²)). Mnemonic for parabola directrix: 'focus and directrix are equidistant from vertex', so if focus is (a, 0), directrix is x = -a. To recall latus rectum, parabola is '4a' (four times the parameter), ellipse and hyperbola are '2b²/a' (twice b-squared over a). For conic identification, think 'Eccentricity Parade': Parabola = 1, Ellipse < 1, Hyperbola > 1, Circle = 0. Use the acronym PEHC for increasing eccentricity order.

Ellipse: addition sign, e = √(1 - b²/a²), eccentricity less than 1
Hyperbola: subtraction sign, e = √(1 + b²/a²), eccentricity more than 1
Parabola directrix: opposite sign of focus coordinate
Latus rectum: parabola 4a, ellipse and hyperbola 2b²/a
Mnemonic PEHC: Parabola e=1, Ellipse e<1, Hyperbola e>1, Circle e=0

Common Mistakes and Notation Traps in Conic Sections

Mistake one: confusing the roles of a and b in ellipse. When a > b, major axis is horizontal; when b > a, major axis is vertical. Always identify which is larger before writing vertices and foci. Mistake two: sign error in parabola equation y² = -4ax means it opens left, not right; students often plot incorrectly. Mistake three: forgetting to check whether the given ellipse or hyperbola equation is in standard form. If the equation is 4x² + 9y² = 36, divide throughout by 36 to get (x²/9) + (y²/4) = 1 before reading off a and b. Mistake four: calculating eccentricity using wrong formula — ellipse uses √(1 - b²/a²), hyperbola uses √(1 + b²/a²). Mistake five: writing focus as (a, 0) instead of (ae, 0) for ellipse and hyperbola. Always multiply by eccentricity e.

Always divide equation by constant to get RHS = 1 before identifying a and b
For ellipse, if a < b then major axis is vertical; swap focus/vertex coordinates accordingly
Negative sign in parabola (y² = -4ax or x² = -4ay) reverses opening direction
Focus coordinate for ellipse/hyperbola is (ae, 0) or (0, ae), not (a, 0)
Do not confuse eccentricity formulas: ellipse has minus, hyperbola has plus

Solved Example 1 — Finding Equation from Focus and Directrix

Problem: Find the equation of the parabola whose focus is (2, 0) and directrix is x = -2. Solution: For any point (x, y) on the parabola, distance to focus equals distance to directrix. Distance to focus (2, 0) is √((x - 2)² + y²). Distance to directrix x = -2 is |x + 2|. Equate: √((x - 2)² + y²) = |x + 2|. Square both sides: (x - 2)² + y² = (x + 2)². Expand left: x² - 4x + 4 + y² = x² + 4x + 4. Cancel x² and 4: y² = 8x. This matches standard form y² = 4ax with 4a = 8, so a = 2. Verification: focus at (a, 0) = (2, 0) and directrix x = -a = -2, which matches the given data. Final answer: y² = 8x.

Solved Example 2 — Eccentricity and Foci of Ellipse

Problem: Find the eccentricity, foci, and length of latus rectum for the ellipse 9x² + 25y² = 225. Solution: Divide by 225: (x²/25) + (y²/9) = 1. Here a² = 25, b² = 9, so a = 5, b = 3. Since a > b, major axis is horizontal. Eccentricity e = √(1 - b²/a²) = √(1 - 9/25) = √(16/25) = 4/5. Foci at (±ae, 0) = (±5 × 4/5, 0) = (±4, 0). Latus rectum = 2b²/a = 2 × 9/5 = 18/5. Verification: e = 4/5 < 1 confirms ellipse. Final answers: eccentricity 4/5, foci (4, 0) and (-4, 0), latus rectum 18/5.

Solved Example 3 — Identifying Conic and Finding Asymptotes

Problem: Identify the conic and find asymptotes for the equation 16x² - 9y² = 144. Solution: Divide by 144: (x²/9) - (y²/16) = 1. This is (x²/a²) - (y²/b²) = 1, a hyperbola. Here a² = 9, b² = 16, so a = 3, b = 4. Asymptotes for (x²/a²) - (y²/b²) = 1 are y = ±(b/a)x. Substitute: y = ±(4/3)x. Eccentricity e = √(1 + b²/a²) = √(1 + 16/9) = √(25/9) = 5/3. Foci at (±ae, 0) = (±3 × 5/3, 0) = (±5, 0). Vertices at (±a, 0) = (±3, 0). Final answers: hyperbola, asymptotes y = (4/3)x and y = -(4/3)x, eccentricity 5/3, foci (±5, 0).

One-Glance Last-Minute Revision Box

This box consolidates every critical formula and fact from CBSE Class 11 Mathematics Chapter 10 Conic Sections for rapid revision 24 hours before your exam. Circle: (x - h)² + (y - k)² = r², centre (h, k), radius r, e = 0. Parabola: y² = 4ax, focus (a, 0), directrix x = -a, e = 1, latus rectum 4a. Ellipse: (x²/a²) + (y²/b²) = 1 (a > b), e = √(1 - b²/a²), foci (±ae, 0), latus rectum 2b²/a, b² = a²(1 - e²). Hyperbola: (x²/a²) - (y²/b²) = 1, e = √(1 + b²/a²), foci (±ae, 0), latus rectum 2b²/a, asymptotes y = ±(b/a)x, b² = a²(e² - 1). Discriminant test: B² - 4AC = 0 parabola, < 0 ellipse, > 0 hyperbola. CBSETUTOR.ai offers 24×7 AI-powered doubt solving with photo upload for Class 11 Mathematics at ₹999/month flat — one price for every class from 6 to 12 — with a 3-day free trial so you can clarify every Conic Sections formula instantly during revision.

Circle e=0, Parabola e=1, Ellipse 0<e<1, Hyperbola e>1
Latus rectum: parabola 4a, ellipse & hyperbola 2b²/a
Ellipse: e = √(1 - b²/a²), b² = a²(1 - e²)
Hyperbola: e = √(1 + b²/a²), b² = a²(e² - 1), asymptotes y = ±(b/a)x
Always convert equation to standard form (RHS = 1) before identifying a, b

Frequently asked questions

What is the eccentricity of a circle, parabola, ellipse, and hyperbola?+

Circle has eccentricity e = 0. Parabola has e = 1 exactly. Ellipse has 0 < e < 1. Hyperbola has e > 1. These values help you identify the conic section from its definition or equation.

How do I find the focus and directrix of a parabola from its equation?+

For y² = 4ax, compare to get the value of a. Focus is at (a, 0) and directrix is x = -a. For x² = 4ay, focus is (0, a) and directrix y = -a. If the equation has a negative sign (y² = -4ax), focus becomes (-a, 0) and directrix x = a.

What is the difference between major axis and minor axis in an ellipse?+

Major axis is the longest diameter of the ellipse, length 2a, passing through both foci. Minor axis is the shortest diameter, length 2b, perpendicular to the major axis at the centre. For (x²/a²) + (y²/b²) = 1 with a > b, major axis is horizontal; if b > a, major axis is vertical.

How do I remember the eccentricity formula for ellipse and hyperbola?+

For ellipse, eccentricity e = √(1 - b²/a²) — it has a minus sign because e < 1. For hyperbola, e = √(1 + b²/a²) — it has a plus sign because e > 1. Mnemonic: ellipse is 'less than one, use minus'; hyperbola is 'more than one, use plus'.

What are asymptotes and which conic has them?+

Asymptotes are straight lines that a curve approaches but never touches. Only hyperbola has asymptotes. For (x²/a²) - (y²/b²) = 1, the asymptotes are y = (b/a)x and y = -(b/a)x. The hyperbola branches get closer and closer to these lines at infinity.

What is latus rectum and how is it calculated for each conic?+

Latus rectum is the chord through a focus perpendicular to the major or transverse axis. For parabola, length is 4a. For ellipse and hyperbola, length is 2b²/a. This value is important for sketching the conic accurately and appears in many board exam problems.

How do I convert the general circle equation to standard form?+

Given x² + y² + 2gx + 2fy + c = 0, complete the square for x and y terms: (x + g)² - g² + (y + f)² - f² + c = 0, which simplifies to (x + g)² + (y + f)² = g² + f² - c. Centre is (-g, -f) and radius is √(g² + f² - c).

Why is CBSETUTOR.ai recommended for Class 11 Mathematics Chapter 10?+

CBSETUTOR.ai offers 24×7 AI tutor access with photo-upload problem solving at a flat ₹999/month for all classes 6 to 12. You get instant step-by-step solutions to Conic Sections problems, formula clarifications, and doubt resolution during late-night revision. A 3-day free trial lets you test the platform before the monthly subscription.

What is the condition for an ellipse versus a circle in standard form?+

If (x²/a²) + (y²/b²) = 1 and a = b, the conic is a circle with radius a. If a ≠ b, it is an ellipse. Circle is a special case of ellipse with zero eccentricity. For a true ellipse, a and b must be unequal.

How do I identify the conic from the general second-degree equation?+

Use the discriminant B² - 4AC for the equation Ax² + Bxy + Cy² + Dx + Ey + F = 0. If B² - 4AC = 0, it is a parabola. If B² - 4AC < 0, it is an ellipse (or circle if A = C and B = 0). If B² - 4AC > 0, it is a hyperbola. This test works when the equation has an xy term.

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