Important Questions Class 11 Maths Chapter 10 Conic Sections

Conic sections are curves obtained by cutting a double-napped right circular cone with a plane. A circle, parabola, ellipse, or hyperbola forms depending on the plane’s angle with the cone.

Curves become easier when students connect their equations with focus, directrix, centre, axes, and latus rectum. Important Questions Class 11 Maths Chapter 10 help students practise circle, parabola, ellipse, hyperbola, eccentricity, latus rectum, degenerated conics, and standard equations. The 2026 NCERT chapter Conic Sections explains how plane sections of a cone create different curves used in planetary motion, telescopes, antennas, flashlights, and automobile headlights.

Key Takeaways

• Circle Formula: A circle with centre (h, k) and radius r has equation (x − h)² + (y − k)² = r².
• Parabola Formula: The standard right-opening parabola has equation y² = 4ax.
• Ellipse Relation: For an ellipse, c² = a² − b² and eccentricity is e = c/a.
• Hyperbola Relation: For a hyperbola, c² = a² + b² and eccentricity is always greater than 1.

Important Questions Class 11 Maths Chapter 10 Structure 2026

Important Questions Class 11 Maths Chapter 10 Structure 2026

Key Points For Quick Recall

• Circle: Every point on a circle stays at distance r from the centre.
• Parabola: Every point on a parabola is equidistant from the focus and directrix.
• Ellipse: Sum of distances from two foci remains constant.
• Hyperbola: Difference of distances from two foci remains constant.

Conic Sections Class 11 Chapter Overview

A cone has a vertex, axis, generator, and two nappes. A plane section creates different curves based on its position.

The angle β made by the cutting plane decides the curve. NCERT uses β = 90° for circle, α < β < 90° for ellipse, β = α for parabola, and 0 ≤ β < α for hyperbola.

Q1. What Are Conic Sections In Class 11 Maths?

Conic sections are curves obtained by intersecting a right circular cone with a plane.

The type of curve depends on the angle of the plane with the cone’s axis. The four main conics are circle, ellipse, parabola, and hyperbola.

Example: A plane parallel to a generator gives a parabola.

Q2. Why Are Important Questions Class 11 Maths Chapter 10 Useful For Exams?

Important Questions Class 11 Maths Chapter 10 are useful because the chapter has definitions, formulas, identification, and numerical problems.

Students must connect each equation with its geometry. Circle, parabola, ellipse, and hyperbola have different focus, axis, and latus rectum rules.

Example: y² = 4ax opens right when a > 0.

Q3. What Are Degenerated Conic Sections?

Degenerated conic sections occur when the cutting plane passes through the vertex of the cone.

The section may become a point, a straight line, or a pair of intersecting lines. These cases depend on the angle between the plane and cone axis.

Example: β = α gives a straight line as a degenerated parabola.

Q4. How Does A Plane Cut A Cone To Form Different Conics?

A plane forms different conics based on its angle β with the cone’s vertical axis.

1. If β = 90°, the section is a circle.
2. If α < β < 90°, the section is an ellipse.
3. If β = α, the section is a parabola.
4. If 0 ≤ β < α, the section is a hyperbola.

Final Result: The angle β decides the conic section.

Circle Class 11 Maths Important Questions

A circle is the simplest conic in coordinate form. Its equation comes directly from the distance formula.

Most circle questions ask for equation, centre, radius, or completion of squares.

Q5. What Is A Circle In Class 11 Maths?

A circle is the set of all points in a plane equidistant from a fixed point.

The fixed point is the centre. The fixed distance is the radius.

Example: If centre is (0, 0) and radius is r, equation is x² + y² = r².

Q6. What Is The Standard Equation Of Circle Class 11?

The standard equation of a circle with centre (h, k) and radius r is (x − h)² + (y − k)² = r².

It follows from the distance formula. Every point (x, y) on the circle stays r units from (h, k).

Formula:
(x − h)² + (y − k)² = r²

Q7. Find The Equation Of The Circle With Centre (−3, 2) And Radius 4.

The equation is (x + 3)² + (y − 2)² = 16.

1. Given Data:
   Centre = (−3, 2)
   Radius = 4
2. Formula Used:
   (x − h)² + (y − k)² = r²
3. Substitute values:
   (x − (−3))² + (y − 2)² = 4²
4. Simplify:
   (x + 3)² + (y − 2)² = 16

Final Result: (x + 3)² + (y − 2)² = 16

Q8. Find The Centre And Radius Of x² + y² + 8x + 10y − 8 = 0.

The centre is (−4, −5), and the radius is 7.

1. Given equation:
   x² + y² + 8x + 10y − 8 = 0
2. Rearrange:
   (x² + 8x) + (y² + 10y) = 8
3. Complete squares:
   (x² + 8x + 16) + (y² + 10y + 25) = 8 + 16 + 25
4. Simplify:
   (x + 4)² + (y + 5)² = 49
5. Compare with standard form:
   Centre = (−4, −5)
   Radius = 7

Final Result: Centre = (−4, −5), radius = 7

Q9. Does The Point (−2.5, 3.5) Lie Inside, Outside Or On x² + y² = 25?

The point lies inside the circle.

1. Given circle:
   x² + y² = 25
2. Radius squared:
   r² = 25
3. For point (−2.5, 3.5):
   x² + y² = (−2.5)² + (3.5)²
   = 6.25 + 12.25 = 18.5
4. Compare:
   18.5 < 25

Final Result: The point lies inside the circle.

Q10. Find The Equation Of The Circle With Centre (2, 2) Passing Through (4, 5).

The equation is (x − 2)² + (y − 2)² = 13.

1. Given Data:
   Centre = (2, 2)
   Point = (4, 5)
2. Find radius squared:
   r² = (4 − 2)² + (5 − 2)²
   r² = 2² + 3² = 13
3. Use standard equation:
   (x − 2)² + (y − 2)² = 13

Final Result: (x − 2)² + (y − 2)² = 13

Standard Equation Of Parabola Class 11 Questions

A parabola uses one focus and one directrix. The vertex lies halfway between them in standard position.

The squared variable tells the axis. If y² appears, the axis is along the x-axis.

Q11. What Is Parabola Class 11 Maths?

A parabola is the set of all points equidistant from a fixed point and a fixed line.

The fixed point is the focus. The fixed line is the directrix.

Example: For y² = 4ax, focus is (a, 0) and directrix is x = −a.

Q12. What Are The Four Standard Equations Of Parabola?

The four standard equations are y² = 4ax, y² = −4ax, x² = 4ay, and x² = −4ay.

Each equation has vertex at the origin. The sign and squared variable decide the direction.

1. y² = 4ax: opens right
2. y² = −4ax: opens left
3. x² = 4ay: opens upward
4. x² = −4ay: opens downward

Final Result: These are the four standard parabolas.

Q13. Find The Focus, Directrix, Axis And Latus Rectum Of y² = 8x.

The focus is (2, 0), directrix is x = −2, axis is x-axis, and latus rectum is 8.

1. Given equation:
   y² = 8x
2. Compare with:
   y² = 4ax
3. Find a:
   4a = 8
   a = 2
4. Features:
   Focus = (a, 0) = (2, 0)
   Directrix = x = −a = −2
   Axis = x-axis
   Latus rectum = 4a = 8

Final Result: Focus (2, 0), directrix x = −2, latus rectum 8

Q14. Find The Equation Of The Parabola With Focus (2, 0) And Directrix x = −2.

The equation is y² = 8x.

1. Given Data:
   Focus = (2, 0)
   Directrix = x = −2
2. Since focus lies on x-axis:
   Equation form is y² = 4ax
3. Here:
   a = 2
4. Substitute:
   y² = 4(2)x

Final Result: y² = 8x

Q15. Find The Equation Of The Parabola With Vertex (0, 0) And Focus (0, 2).

The equation is x² = 8y.

1. Given Data:
   Vertex = (0, 0)
   Focus = (0, 2)
2. Focus lies on positive y-axis.
   Equation form is x² = 4ay.
3. Here:
   a = 2
4. Substitute:
   x² = 4(2)y

Final Result: x² = 8y

Q16. Find The Equation Of A Parabola Symmetric About The y-Axis And Passing Through (2, −3).

The equation is 3x² = −4y.

1. Symmetric about y-axis means form:
   x² = 4ay or x² = −4ay
2. Point (2, −3) lies below x-axis.
   Parabola opens downward.
3. Use form:
   x² = −4ay
4. Substitute (2, −3):
   2² = −4a(−3)
   4 = 12a
   a = 1/3
5. Equation:
   x² = −4(1/3)y
6. Multiply by 3:
   3x² = −4y

Final Result: 3x² = −4y

Latus Rectum Class 11 Questions For Parabola

The latus rectum passes through the focus and stays perpendicular to the axis. Its endpoints lie on the parabola.

For y² = 4ax, the latus rectum length is 4a.

Q17. What Is Latus Rectum Of A Parabola?

The latus rectum of a parabola is a line segment through the focus and perpendicular to the axis.

Its endpoints lie on the parabola. It measures the width of the parabola at the focus.

Example: For y² = 4ax, latus rectum length = 4a.

Q18. Find The Latus Rectum Length Of y² = 12x.

The latus rectum length is 12.

1. Given equation:
   y² = 12x
2. Compare with:
   y² = 4ax
3. Find a:
   4a = 12
   a = 3
4. Latus rectum length:
   4a = 12

Final Result: 12

Q19. Find Focus And Directrix Of x² = −16y.

The focus is (0, −4), and the directrix is y = 4.

1. Given equation:
   x² = −16y
2. Compare with:
   x² = −4ay
3. Find a:
   4a = 16
   a = 4
4. Features:
   Focus = (0, −a) = (0, −4)
   Directrix = y = a = 4

Final Result: Focus (0, −4), directrix y = 4

Ellipse Class 11 Maths Important Questions

An ellipse uses two foci and a constant sum of distances. Its major axis contains the foci.

The larger denominator in the standard equation gives the major axis direction.

Q20. What Is Ellipse Class 11 Maths?

An ellipse is the set of all points whose distance sum from two fixed points is constant.

The fixed points are called foci. The midpoint of the foci is the centre.

Example: In x²/a² + y²/b² = 1, if a > b, foci lie on the x-axis.

Q21. What Is The Standard Equation Of Ellipse Class 11?

The standard equation of an ellipse with major axis on x-axis is x²/a² + y²/b² = 1.

Here, a > b. The foci are (±c, 0) and c² = a² − b².

Formula:
x²/a² + y²/b² = 1

Q22. Find Foci, Vertices, Axes Lengths, Eccentricity And Latus Rectum Of x²/25 + y²/9 = 1.

The foci are (±4, 0), vertices are (±5, 0), eccentricity is 4/5, and latus rectum is 18/5.

1. Given equation:
   x²/25 + y²/9 = 1
2. Compare with:
   x²/a² + y²/b² = 1
3. Values:
   a² = 25, so a = 5
   b² = 9, so b = 3
4. Find c:
   c² = a² − b² = 25 − 9 = 16
   c = 4
5. Features:
   Foci = (±4, 0)
   Vertices = (±5, 0)
   Major axis length = 2a = 10
   Minor axis length = 2b = 6
   Eccentricity = c/a = 4/5
   Latus rectum = 2b²/a = 2 × 9/5 = 18/5

Final Result: Foci (±4, 0), vertices (±5, 0), e = 4/5, latus rectum = 18/5

Q23. Find The Standard Form Of 9x² + 4y² = 36 And Its Foci.

The standard form is x²/4 + y²/9 = 1, and foci are (0, ±√5).

1. Given equation:
   9x² + 4y² = 36
2. Divide by 36:
   x²/4 + y²/9 = 1
3. Larger denominator is under y².
   Major axis lies on y-axis.
4. Values:
   a² = 9, so a = 3
   b² = 4, so b = 2
5. Find c:
   c² = a² − b² = 9 − 4 = 5
   c = √5

Final Result: Foci = (0, ±√5)

Q24. Find The Equation Of The Ellipse With Vertices (±13, 0) And Foci (±5, 0).

The equation is x²/169 + y²/144 = 1.

1. Vertices on x-axis mean major axis is x-axis.
2. Given:
   a = 13
   c = 5
3. Use relation:
   c² = a² − b²
4. Substitute:
   25 = 169 − b²
   b² = 144
5. Equation:
   x²/169 + y²/144 = 1

Final Result: x²/169 + y²/144 = 1

Q25. Find The Equation Of The Ellipse With Major Axis 20 And Foci (0, ±5).

The equation is x²/75 + y²/100 = 1.

1. Foci lie on y-axis.
   Major axis is y-axis.
2. Given:
   Major axis length = 20
   a = 10
   c = 5
3. Use relation:
   c² = a² − b²
4. Substitute:
   25 = 100 − b²
   b² = 75
5. Equation:
   x²/75 + y²/100 = 1

Final Result: x²/75 + y²/100 = 1

Eccentricity Class 11 Maths Questions For Ellipse

Eccentricity measures how far the foci lie from the centre compared with the vertex distance. For an ellipse, eccentricity is always less than 1.

Use e = c/a after finding c.

Q26. What Is Eccentricity Of An Ellipse?

The eccentricity of an ellipse is e = c/a.

Here, c is the distance from centre to focus. The value a is the semi-major axis.

Fact: For an ellipse, 0 < e < 1.

Q27. Find The Eccentricity Of x²/36 + y²/16 = 1.

The eccentricity is √5/3.

1. Given equation:
   x²/36 + y²/16 = 1
2. Values:
   a² = 36, so a = 6
   b² = 16, so b = 4
3. Find c:
   c² = a² − b² = 36 − 16 = 20
   c = 2√5
4. Eccentricity:
   e = c/a = 2√5/6 = √5/3

Final Result: e = √5/3

Q28. Find The Latus Rectum Of x²/49 + y²/36 = 1.

The length of latus rectum is 72/7.

1. Given equation:
   x²/49 + y²/36 = 1
2. Values:
   a² = 49, so a = 7
   b² = 36
3. Formula Used:
   Latus rectum = 2b²/a
4. Calculation:
   2b²/a = 2 × 36/7 = 72/7

Final Result: 72/7

Hyperbola Class 11 Maths Important Questions

A hyperbola uses the difference of distances from two foci. Its transverse axis contains the foci and vertices.

The positive term in the standard equation gives the transverse axis.

Q29. What Is Hyperbola Class 11 Maths?

A hyperbola is the set of all points whose distance difference from two fixed points is constant.

The fixed points are foci. The line through the foci is the transverse axis.

Example: x²/a² − y²/b² = 1 has transverse axis along x-axis.

Q30. What Is The Standard Equation Of Hyperbola Class 11?

The standard equation with transverse axis on x-axis is x²/a² − y²/b² = 1.

For y-axis transverse axis, the equation is y²/a² − x²/b² = 1. The relation is c² = a² + b².

Formula:
x²/a² − y²/b² = 1

Q31. Find Foci, Vertices, Eccentricity And Latus Rectum Of x²/9 − y²/16 = 1.

The foci are (±5, 0), vertices are (±3, 0), eccentricity is 5/3, and latus rectum is 32/3.

1. Given equation:
   x²/9 − y²/16 = 1
2. Values:
   a² = 9, so a = 3
   b² = 16, so b = 4
3. Find c:
   c² = a² + b² = 9 + 16 = 25
   c = 5
4. Features:
   Foci = (±5, 0)
   Vertices = (±3, 0)
   Eccentricity = c/a = 5/3
   Latus rectum = 2b²/a = 2 × 16/3 = 32/3

Final Result: Foci (±5, 0), vertices (±3, 0), e = 5/3, latus rectum = 32/3

Q32. Find The Standard Form And Foci Of y² − 16x² = 16.

The standard form is y²/16 − x²/1 = 1, and foci are (0, ±√17).

1. Given equation:
   y² − 16x² = 16
2. Divide by 16:
   y²/16 − x²/1 = 1
3. Values:
   a² = 16, so a = 4
   b² = 1, so b = 1
4. Find c:
   c² = a² + b² = 16 + 1 = 17
   c = √17
5. Since positive term is y²:
   Foci lie on y-axis.

Final Result: Foci = (0, ±√17)

Q33. Find The Equation Of Hyperbola With Foci (0, ±3) And Vertices (0, ±11/2).

The equation is 100y² − 44x² = 275.

1. Foci and vertices lie on y-axis.
   Equation form: y²/a² − x²/b² = 1
2. Given:
   a = 11/2
   c = 3
3. Use relation:
   b² = c² − a²
4. Calculate:
   b² = 9 − 121/4 = 36/4 − 121/4 = −85/4

This data cannot form a real hyperbola because c must be greater than a. The NCERT example has vertices (0, ±√11/2), not (0, ±11/2).

Correct NCERT-style result:
If a² = 11/4 and c = 3, then:
b² = 9 − 11/4 = 25/4

Equation:
y²/(11/4) − x²/(25/4) = 1

Multiplying gives:
100y² − 44x² = 275

Final Result: 100y² − 44x² = 275 for vertices (0, ±√11/2)

Q34. Find The Equation Of Hyperbola With Foci (0, ±12) And Latus Rectum 36.

The equation is 3y² − x² = 108.

1. Foci lie on y-axis.
   Equation form: y²/a² − x²/b² = 1
2. Given:
   c = 12
   Latus rectum = 2b²/a = 36
3. So:
   b² = 18a
4. Use relation:
   c² = a² + b²
5. Substitute:
   144 = a² + 18a
   a² + 18a − 144 = 0
6. Solve:
   a = 6, rejecting negative value.
7. Find b²:
   b² = 18 × 6 = 108
8. Equation:
   y²/36 − x²/108 = 1
9. Multiply by 108:
   3y² − x² = 108

Final Result: 3y² − x² = 108

Standard Equation Of Hyperbola Class 11 Board Pattern Questions

Hyperbola questions often test whether the positive term is x² or y². The positive term decides the transverse axis.

Use c² = a² + b², not c² = a² − b².

Q35. Find The Vertices And Foci Of 16x² − 9y² = 576.

The vertices are (±6, 0), and the foci are (±10, 0).

1. Given equation:
   16x² − 9y² = 576
2. Divide by 576:
   x²/36 − y²/64 = 1
3. Values:
   a² = 36, so a = 6
   b² = 64, so b = 8
4. Find c:
   c² = a² + b² = 36 + 64 = 100
   c = 10

Final Result: Vertices (±6, 0), foci (±10, 0)

Q36. Find The Equation Of Hyperbola With Vertices (±2, 0) And Foci (±3, 0).

The equation is x²/4 − y²/5 = 1.

1. Vertices and foci lie on x-axis.
   Equation form: x²/a² − y²/b² = 1
2. Given:
   a = 2
   c = 3
3. Use relation:
   b² = c² − a²
4. Calculate:
   b² = 9 − 4 = 5
5. Equation:
   x²/4 − y²/5 = 1

Final Result: x²/4 − y²/5 = 1

Q37. Find The Equation Of Hyperbola With Vertices (0, ±5) And Foci (0, ±8).

The equation is y²/25 − x²/39 = 1.

1. Vertices and foci lie on y-axis.
   Equation form: y²/a² − x²/b² = 1
2. Given:
   a = 5
   c = 8
3. Use relation:
   b² = c² − a²
4. Calculate:
   b² = 64 − 25 = 39
5. Equation:
   y²/25 − x²/39 = 1

Final Result: y²/25 − x²/39 = 1

NCERT Class 11 Maths Chapter 10 Questions On Applications

The application problems use conic equations after choosing a coordinate system. Most questions become simple after placing the vertex or centre at origin.

The NCERT miscellaneous examples include parabolic mirrors, deflected beams, rods, arches, and racecourse paths.

Q38. A Parabolic Mirror Has Focus 5 cm From Vertex And Depth 45 cm. Find AB.

The distance AB is 60 cm.

1. Given Data:
   Focus distance = 5 cm
   Depth = 45 cm
2. Choose equation:
   y² = 4ax
3. Here:
   a = 5
4. Equation becomes:
   y² = 20x
5. At depth 45 cm:
   x = 45
6. Substitute:
   y² = 20 × 45 = 900
   y = ±30
7. Total width:
   AB = 2 × 30 = 60 cm

Final Result: AB = 60 cm

Q39. A Beam Has Supports 12 m Apart And Centre Deflection 3 cm. Find Distance From Centre For 1 cm Deflection.

The distance is 2√6 m from the centre.

1. Use vertex at lowest point.
   Beam shape is a parabola.
2. Equation form:
   x² = 4ay
3. Half distance between supports:
   x = 6 m
4. Centre deflection:
   y = 3/100 m
5. Substitute:
   6² = 4a × 3/100
   36 = 12a/100
   a = 300
6. For 1 cm deflection:
   y = 1/100
7. Calculate:
   x² = 4 × 300 × 1/100 = 12

The NCERT example uses y = 2/100 at the required level, giving:
x² = 4 × 300 × 2/100 = 24
x = 2√6

Final Result: x = 2√6 m for the stated NCERT level

Q40. A Rod Of Length 15 cm Touches Both Axes. Point P Is 6 cm From The x-Axis End. Show Its Locus Is An Ellipse.

The locus of P is x²/81 + y²/36 = 1.

1. Let rod AB touch x-axis at A and y-axis at B.
   Let P(x, y) lie on AB.
2. Given:
   AB = 15 cm
   AP = 6 cm
   PB = 9 cm
3. From triangle near y-axis:
   cos θ = x/9
4. From triangle near x-axis:
   sin θ = y/6
5. Use identity:
   cos² θ + sin² θ = 1
6. Substitute:
   (x/9)² + (y/6)² = 1
7. Simplify:
   x²/81 + y²/36 = 1

Final Result: x²/81 + y²/36 = 1

Conic Sections Formulas Class 11 For Quick Identification

Formula recognition saves time in this chapter. The signs, squared variables, and denominators identify the conic quickly.

A circle has equal x² and y² coefficients. A hyperbola has opposite signs between squared terms.

Q41. How Can Students Identify A Circle From Its Equation?

A circle has x² and y² with equal coefficients and the same sign.

Its standard form is (x − h)² + (y − k)² = r². Completing squares gives centre and radius.

Example: x² + y² − 4x − 8y − 45 = 0 is a circle.

Q42. How Can Students Identify A Parabola From Its Equation?

A parabola has only one squared variable in its standard equation.

If y² appears, its axis is along x-axis. If x² appears, its axis is along y-axis.

Example: y² = 12x is a right-opening parabola.

Q43. How Can Students Identify An Ellipse From Its Equation?

An ellipse has x² and y² with the same sign and unequal denominators.

The larger denominator gives the major axis direction. Foci always lie on the major axis.

Example: x²/36 + y²/16 = 1 has major axis on x-axis.

Q44. How Can Students Identify A Hyperbola From Its Equation?

A hyperbola has x² and y² terms with opposite signs.

The positive term gives the transverse axis. Use c² = a² + b² for foci.

Example: y²/25 − x²/16 = 1 has transverse axis on y-axis.

Q45. Which Formula Is Often Confused Between Ellipse And Hyperbola?

The relation for c is often confused between ellipse and hyperbola.

For ellipse:
c² = a² − b²

For hyperbola:
c² = a² + b²

Example: x²/25 + y²/9 = 1 uses ellipse relation.

Important Questions Class 11 Maths Chapter-Wise