NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections in English and Hindi Medium updated for CBSE term exams 2023-24. Get the modified solutions of chapter 10 class 11th mathematics based on rationalised textbooks and revised syllabus for academic year 2023-24.

**NCERT Solutions for Class 11 Maths Chapter 10 in English Medium**

Class 11 Maths Exercise 10.1 in English

Class 11 Maths Exercise 10.2 in English

Class 11 Maths Exercise 10.3 in English

Class 11 Maths Exercise 10.4 in English

Class 11 Maths Misc. Exercise 10 in English

**NCERT Solutions for Class 11 Maths Chapter 10 in Hindi Medium**

Class 11 Maths Exercise 10.1 in Hindi

Class 11 Maths Exercise 10.2 in Hindi

Class 11 Maths Exercise 10.3 in Hindi

Class 11 Maths Exercise 10.4 in Hindi

Class 11 Maths Misc. Exercise 10 in Hindi

**Related Links**

Class 11 Maths NCERT Solutions

Class 11 all Subject NCERT Solutions

## NCERT Solutions for Class 11 Maths Chapter 10

Students can do directly 12th class after doing 10th though NIOS Admission 2023-24. For the other subjects, there are also available NCERT Solutions 2023-24 to free download without any login password. Download Offline Apps for all CBSE and UP Board (intermediate) students for their March exams. Download UP Board Solutions for Class 11 Maths Chapter 10 in PDF file format.

### 11th Maths Chapter 10 Solutions

NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections all exercise and miscellaneous are given below to free download in PDF format updated for new academic session. NCERT Solutions and Offline Apps 2023-24 are based on latest 11th textbooks available on NCERT (https://ncert.nic.in/) website 2023-24 following new CBSE Syllabus.

Class: 11 | Mathematics |

Chapter 10: | Conic Sections |

Number of Exercises: | Five (4 + Miscellaneous) |

Content: | NCERT Exercise Solution |

Mode: | Videos and Text format |

Academic Year: | CBSE 2023-24 |

Medium: | Hindi and English Medium |

#### Important Terms on Conic Sections

1. Circle, ellipse, parabola and hyperbola are curves which are obtained by intersection of a plane and cone in different positions

2. Circle: It is the set of all points in a plane that are equidistant from a fixed point in that plane.

3. Parabola: It is the set of all points in a plane which are equidistant from a fixed point (focus) and a fixed line in the plane. Fixed point does not lie on the line.

4. Latus Rectum: A chord through focus perpendicular to axis of parabola is called its latus rectum.

5. Ellipse: It is the set of points in a plane the sum of whose distances from two fixed points in the plane is a constant and is always greater than the distances between the fixed points.

6. Hyperbola: It is the set of all points in a plane, the differences of whose distance from two fixed points in the plane is a constant.

##### Questions from Exam Papers

1. Find the centre and radius of the circle x² + y² – 6x + 4y – 12 = 0. [Answer: (3, -2), 5]

2. Find the equation of hyperbola satisfying given conditions foci (5, 0) and transverse axis is of length 8. [Answer: x²/16 – y²/9 = 1.]

3. Find the coordinates of points on parabola y² = 8x whose focal distance is 4. [Answer: (2, 4), (2, -4)]

4. If one end of a diameter of the circle x² + y² – 4x – 6y + 11 = 0 is (3, 4), then find the coordinates of the other end of diameter. [Answer: (1, 2)]

5. Find equation of an ellipse having vertices (0, 5) and foci (0, 4). [Answer: x²/9 + y²/25 = 1]

6. Find the equation of a circle whose centre is at (4, –2) and 3x – 4y+ 5 = 0 is tangent to circle. [Answer: x² + y² – 8x + 4y – 5 = 0]

###### Questions for Practice

1. If the distance between the foci of a hyperbola is 16 and its eccentricity is 2, then obtain the equation of a hyperbola. [Answer: x² – y² = 32]

2. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity. [Answer: e = √3/2]

3. Find the length of major and minor axis of the following ellipse, 16x² + 25y² = 400. [Answer: 10, 8]

### Important Questions on 11th Maths Chapter 10

### Find the equation of the circle with centre (0, 2) and radius 2.

The equation of a circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r²

It is given that centre (h, k) = (0, 2) and radius (r) = 2.

Therefore, the equation of the circle is

(x – 0)² + (y – 2)² = 2²

⟹ x² + y² + 4 – 4 y = 4

⟹ x² + y² – 4y = 0

### Find the equation of the circle with centre (–2, 3) and radius 4.

The equation of a circle with centre (h, k) and radius r is given as (x – h)² + (y – k)² = r²

It is given that centre (h, k) = (–2, 3) and radius (r) = 4.

Therefore, the equation of the circle is (x + 2)² + (y – 3)² = (4)²

⟹ x² + 4x + 4 + y² – 6y + 9 = 16

⟹ x² + y² + 4x – 6y – 3 = 0

### Find the centre and radius of the circle (x + 5)² + (y – 3)² = 36

The equation of the given circle is

(x + 5)² + (y – 3)² = 36. (x + 5)² + (y – 3)² = 36

⇒ {x – (–5)}² + (y – 3)² = 6², which is of the form (x – h)² + (y – k)² = r², where h = – 5, k = 3, and r = 6.

Thus, the centre of the given circle is (–5, 3), while its radius is 6.

### Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Let the equation of the required circle be (x – h)² + (y – k)² = r².

Since the circle passes through points (4, 1) and (6, 5), (4 – h)² + (1 – k)² = r² …… (1)

(6 – h)² + (5 – k)² = r² …… (2)

Since the centre (h, k) of the circle lies on line 4x + y = 16, 4h + k = 16 …… (3)

From equations (1) and (2), we obtain (4 – h)² + (1 – k)² = (6 – h)² + (5 – k)²

⇒ 16 – 8h + h² + 1 – 2k + k² = 36 – 12h + h² + 25 – 10k + k²

⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k

⇒ 4h + 8k = 44 ⇒ h + 2k = 11 …… (4)

On solving equations (3) and (4), we obtain h = 3 and k = 4.

On substituting the values of h and k in equation (1), we obtain

(4 – 3)² + (1 – 4)² = r² ⇒ (1)² + (– 3)² = r²

⇒ 1 + 9 = r² ⇒ r² = 10

⇒ r =√10

Thus, the equation of the required circle is (x – 3)² + (y – 4)² = (√10)²

x² – 6x + 9 + y² – 8y + 16 = 10 x² + y² – 6x – 8y + 15 = 0

### Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y² = 12x.

The given equation is y² = 12x.

Here, the coefficient of x is positive.

Hence, the parabola opens towards the right.

On comparing this equation with y² = 4ax, we obtain 4a = 12 ⇒ a = 3

∴ Coordinates of the focus = (a, 0) = (3, 0)

Since the given equation involves y², the axis of the parabola is the x-axis. Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0

Length of latus rectum = 4a = 4 × 3 = 12.

### Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6.

Focus (6, 0); directrix, x = –6

Since the focus lies on the x-axis, the x-axis is the axis of the parabola. Therefore, the equation of the parabola is either of the form y² = 4ax or y² = – 4ax.

It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.

Hence, the parabola is of the form y² = 4ax. Here, a = 6

Thus, the equation of the parabola is y² = 24x.

4. Find equation of circle concentric with circle 4x² + 4y² – 12x – 16y – 21 = 0 and of half its area. [Answer: 2x² + 2y² – 6x + 8y + 1 = 0]

5. Find the equation for the ellipse that satisfies the given condition Major axis on the x-axis and passes through the points (4, 3) and (6, 2). [Answer: x²/52 + y²/13 = 1]

### What are the main concepts to learn in chapter 10 Class 11 Maths?

The teacher teaches the following topics in chapter 10 of grade 11th Maths:

- Sections of a cone.
- Circle, ellipse, parabola, and hyperbola
- Degenerated conic sections
- All about circles
- Parabola
- Standard equations of parabola
- Length of the latus rectum of the parabola
- Ellipse
- Relationship between semi-major axis, semi-minor axis, and the distance of the focus from the centre of the ellipse.
- Special cases of an ellipse
- The eccentricity of an ellipse
- Standard equations of an ellipse
- Length of the latus rectum of the ellipse
- Hyperbola
- The eccentricity of the hyperbola
- Standard equation of hyperbola
- Length of the latus rectum of the hyperbola.

### How much time is required to complete chapter 10 of class 11th Maths?

Students need a maximum of 10-12 days to do chapter 10 of class 11th Maths if they honestly and seriously give 1-2 hours per day to this chapter. This time also depends on student’s ability, efficiency, and working speed. This chapter is not very tough and not very easy. This chapter is moderate that is, this chapter lies in the mid of easy and tough.

### Can students skip chapter 10 of 11th standard Maths?

No, students can’t skip chapter 10 of class 11th Maths. Chapter 10 (Conic sections) of grade 11th Math is important. Students will study conic sections in graduation also if they do B.Sc. (H) Maths in future. Some important problems of chapter 10 of grade 11th Maths are questions 3, 5, 9, 12, 13, and 15 of exercise 10.1, questions 2, 3, 5, 6, 8, 10, and 12 of exercise 10.2, questions 5, 11, 12, 14, 16, 18, and 19 of exercise 10.3, questions 5, 7, 9, 11, 12, 14, and 15 of exercise 10.4, questions 1, 2, 3, 4, 6, 7, and 8 of last (miscellaneous) exercise and examples 3, 4, 6, 7, 8, 10, 12, 13, 14, 16, 18, and 19.

### Which sums of chapter 10 of grade 11th Maths are of the same kind?

There are five exercises in chapter 10 of grade 11th Maths.

- In the first exercise (Ex 10.1), examples 1, 2 and questions 1, 2, 3, 4, 5 are of the same kind, example 3 and questions 6, 7, 8, 9 are of the same type, example 4 and questions 10, 11 are similar.
- In the second exercise (Ex 10.2), example 5 and questions 1, 2, 3, 4, 5, 6 are the same, example 6 and questions 7, 8 are similar, example 7 and questions 9, 10 are of the same kind, example 8 and questions 11, 12 are of the same type.
- In the third exercise (Ex 10.3), examples 9, 10 and questions 1, 2, 3, 4, 5, 6, 7, 8, 9 are similar, example 11 and questions 10, 11, 12 are the same, example 12 and questions 15, 16 are of the same kind, example 13 and questions 19, 20 are of the same type, questions 13, 14 are same, questions 17, 18 are based on the same pattern.
- In the fourth exercise (Ex 10.4), example 14 and questions 1, 2, 3, 4, 5, 6 are the same, example 15 and questions 7, 8, 9 are of the same type, example 16 and questions 12, 13 are similar, questions 10, 11 are based on the same pattern.
- In the last (Miscellaneous) exercise, all the examples and questions are word problems, and all the examples and questions are of a different type.