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NCERT Solutions for Class 11 Maths Chapter 11 : Conic Sections

Q: Q1. How many exercises are there in the Conic Sections Class 11 NCERT Solutions?

Answer: Chapter 11 Conic Sections Class 11 Maths has a total of 70 questions, 55 of which are easy, 10 of which are moderate, and the remaining are extended answer type problems. These are required to solve because they provide additional information on the ideas learnt. The solutions to the exercise-specific problems are also accessible in Conic Sections Class 11 PDF format, with the goal of assisting students in performing well in the yearly exam.

Q: Q2. What are the key subtopics in Conic Sections Class 11 Maths Chapter 11 that could be tested?

Answer: NCERT Solutions Class 11 Maths Chapter 11 begins with a brief overview of the Conical Section concepts. The NCERT Solutions Class 11 Maths Chapter 11 focuses on the fundamentals of several forms of conical sections such as circles, ellipses, parabolas, and hyperbolas. It also discusses the conventional equation for each of these curves. It also defines words relating to these curves, such as the latus rectum and eccentricity. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

Q: Q3. Is it necessary to practice all of the questions in Conic Sections Class 11 NCERT Solutions?

Answer: Conic Sections Class 11 NCERT Solutions are well-designed to help students study different types of conical sections along with the equations of different types of curves, be it parabola, circle, ellipse or hyperbola. It thoroughly covers all essential principles, allowing students to grasp the concepts. These solutions use examples to clarify the subjects presented so that students may easily relate to the concepts being addressed.

Conic Sections Class 11 Maths Chapter 11 deals with the study of essential topics related to different types of conical sections. Class 11 Maths Chapter 11 explains the key concepts related to the Introduction to Conic Sections, Circle Sections, as well as Circles, Ellipses, Parabolas, and Hyperbolas. The concepts taught in Class 11 Chapter 11 Conic Sections will also help students to gain a thorough understanding of this concept and its practical applications. The expert faculty of Toppr produced these solutions to assist students with their first-term exam preparations.

Conic Sections Class 11 Chapter 11 also covers topics such as circle, ellipse, parabola, hyperbola, degenerated conic sections, standard equations of such curves, the relationship between semi-major axis, semi-minor axis, and the distance of the focus from the center of the ellipse, as well as special cases of the ellipse, eccentricity and latus rectum of each conical section, etc. Students will quickly learn about these concepts if they practice Chapter 11 Maths Class 11 NCERT Solutions of Conic Sections. All of these solutions have been created with the new CBSE pattern in mind, ensuring that students have a thorough understanding of their exams.

Chapter 11 Maths Class 11 Questions and Answers can help you get good grades in tests and properly prepare for all of the important concepts. Several examples provided in the solutions will assist students with a better understanding of conic sections. Conic Sections Class 11 NCERT Solutions Chapter 11 are available below in pdf format, and a few solutions are also included in the exercises. These solutions explain the topics covered with examples so that students can easily relate to the notion being discussed.

Table of Content

Introduction

Download PDF of NCERT Solutions for Class 11 Maths Chapter 11 : Conic Sections

Access NCERT Solutions for Class 11 Maths Chapter 11 : Conic Sections

EXERCISE 11.1

Question 1

In each of the following Exercises

1

5,

find the equation of the circle with

(0, 2) and radius 2

centre (- 2, 3) and r a d i u s 4

c e n t r e (\frac{1}{2}, \frac{1}{4}) and r a d i u s \frac{1}{1 2}

c e n t r e (1, 1) and r a d i u s 2

Medium

Solution

Verified by Toppr

The equation of circle with centre

(a, b)

and radius

r

(x - a)^{2} + (y - b)^{2} = r^{2}

A)

The equation of circle is

(x - 0)^{2} + (y - 2)^{2} = 2^{2} ⟹ x^{2} + y^{2} - 4 y = 0

B)

The equation of circle is

(x + 2)^{2} + (y - 3)^{2} = 4^{2} ⟹ x^{2} + y^{2} + 4 x - 6 y + 9 = 0

C)

The equation of circle is

(x - 1 / 2)^{2} + (y - 1 / 4)^{2} = (1 / 12)^{2} ⟹ x^{2} + y^{2} - x - \frac{y}{2} + \frac{1 1}{3 6} = 0

D)

The equation of circle is

(x - 1)^{2} + (y - 1)^{2} = (2)^{2} ⟹ x^{2} + y^{2} - 2 x - 2 y = 0

Question 2

Find the equation of the circle with centre

(- 2, 3)

and radius

4

Easy

Solution

Verified by Toppr

Here 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}

\Rightarrow x^{2} + 4 x + 4 + y^{2} - 6 y + 9 = 16

\Rightarrow x^{2} + y^{2} + 4 x - 6 y - 3 = 0

Question 3

Find the equation of the circle with centre

(\frac{1}{2}, \frac{1}{4})

and radius

\frac{1}{1 2}

Easy

Solution

Verified by Toppr

Here it is given that centre

(h, k) = (\frac{1}{2}, \frac{1}{4})

and radius

r = \frac{1}{2}

Therefore the equation of the circle is

(x - \frac{1}{2})^{2} + (y - \frac{1}{4})^{2} = (\frac{1}{1 2})^{2}

\Rightarrow x^{2} - x + \frac{1}{4} + y^{2} - \frac{y}{2} + \frac{1}{1 6} = \frac{1}{1 4 4}

\Rightarrow x^{2} - x + \frac{1}{4} + y^{2} - \frac{y}{2} + \frac{1}{1 6} - \frac{1}{1 4 4} = 0

\Rightarrow 144 x^{2} - 144 x + 36 + 144 y^{2} - 72 y + 9 - 1 = 0

\Rightarrow 36 x^{2} + 36 y^{2} - 36 x - 18 y + 11 = 0

Question 4

Find the equation of centre

(1, 1)

and radius

2

Easy

Solution

Verified by Toppr

centre (1,1) (p,a)

radius

2

(x - p)^{2} + (y - q)^{2} = r^{2}

\Rightarrow (x - 1)^{2} + (y - 1)^{2} = (2)^{2}

\Rightarrow x^{2} + 1 + 2 x + y^{2} + 1 + 2 y = 2

\Rightarrow x^{2} + y^{2} + 2 x + 2 y = 0

Question 5

Equation of circle with center (-a, -b) and radius

a^{2} - b^{2}

is.

x^{2} + y^{2} - 2 a x - 2 b y - 2 b^{2} = 0

x^{2} + y^{2} - 2 a x - 2 b y + 2 a^{2} = 0

x^{2} + y^{2} + 2 a x + 2 b y + 2 b^{2} = 0

x^{2} + y^{2} - 2 a x - 2 b y + 2 b^{2} = 0

Medium

Solution

Verified by Toppr

Fact: equation of circle with centre

(h, k)

and radius

r

is given by

(x - h)^{2} + (y - k)^{2} = r^{2}

Hence equation of circle with centre

(- a, - b)

having radius

a^{2} - b^{2}

is given by,

(x + a)^{2} + (y + b)^{2} = a^{2} - b^{2}

\Rightarrow x^{2} + y^{2} + 2 a x + 2 b y + a^{2} + b^{2} = a^{2} - b^{2}

\Rightarrow x^{2} + y^{2} + 2 a x + 2 b y + 2 b^{2} = 0

Question 6

Find the centre and radius of the circle

(x + 5)^{2} + (y - 3)^{2} = 36

Easy

Solution

Verified by Toppr

The equation of the given circle is,

(x + 5)^{2} + (y - 3)^{2} = 36

\Rightarrow {x - (- 5)}^{2} + (y - 3)^{2} = 6^{2}

which is of the form

(x - h)^{2} + (y - k)^{2} = r^{2}

where

h = - 5, k = 3

and

r = 6

Thus the centre of the given circle is

(- 5, 3)

while its radius is

6

Question 7

Find the centre and radius of the circle

x^{2} + y^{2} - 4 x - 8 y - 45 = 0

(2, 6), 63

(2, 4), 65

(2, - 4), 66

None

Medium

Solution

Verified by Toppr

Given circle equation $x^{2} + y^{2} - 4 x - 8 y - 45 = 0 h e r e 2 g = - 4 ∴ g = - 2 a n d 2 f = - 8 ∴ f = - 4 ∴ c e n t r e (- g, - f) = (2, 4) r a d i u s = g^{2} + f^{2} - c = 2^{2} + 4^{2} - (- 45) = 4 + 16 + 45 = 65 u n i t$

Question 8

Find the centre and radius of the circle

x^{2} + y^{2} - 8 x + 10 y - 12 = 0

(4, - 5), 53

(- 4, - 5), 53

(- 4, 5), 53

(4, 5), 53

Medium

Solution

Verified by Toppr

Given circle equation

$x^{2} + y^{2} - 8 x + 10 y - 12 = 0 h e r e 2 g = - 8 ∴ g = - 4 a n d 2 f = 10 ∴ f = 5 ∴ c e n t r e (- g, - f) = (4, - 5) r a d i u s = g^{2} + f^{2} - c = 4^{2} + 5^{2} - (- 12) = 16 + 25 + 12 = 53 u n i t$

Question 9

Find the centre and radius of the circles.

2 x^{2} + 2 y^{2} - x = 0

Medium

Solution

Verified by Toppr

Given the equation of the circle is

2 x^{2} + 2 y^{2} - x = 0

or,

x^{2} + y^{2} - \frac{x}{2} = 0

or,

x^{2} - 2 . x . \frac{1}{4} + \frac{1}{4 ^{2}} + y^{2} = \frac{1}{4 ^{2}}

or,

(x - \frac{1}{4})^{2} + y^{2} = \frac{1}{4 ^{2}}

From this equation we've the center is

(\frac{1}{4}, 0)

and radius is

\frac{1}{4}

Question 10

The equation of the circle passing through the points

(4, 1), (6, 5)

and having the centre on the line

4 x + y - 16 = 0

x^{2} + y^{2} - 6 x - 8 y + 15 = 0

15 (x^{2} + y^{2}) - 94 x + 18 y + 55 = 0

x^{2} + y^{2} - 4 x - 3 y = 0

x^{2} + y^{2} + 6 x - 4 y = 0

Medium

Solution

Verified by Toppr

Given points,

(4, 1), (6, 5)

equation of circle

(x - h)^{2} + (y - k)^{2} = r^{2}

\Rightarrow (4 - h)^{2} + (1 - k)^{2} = r^{2}

....(1)

\Rightarrow (6 - h)^{2} + (5 - k)^{2} = r^{2}

....(2)

solving the above 2 equations, we get,

h + 2 k = 11

....(3)

given,

4 h + k = 16

.....(4)

solving the above 2 equations, we get,

h = 3, k = 4

substituting the above values in (1), we get,

(4 - 3)^{2} + (1 - 4)^{2} = r^{2}

∴ r = 10

Hence, the equation is,

(x - 3)^{2} + (y - 4)^{2} = (10)^{2}

x^{2} + y^{2} - 6 x - 8 y + 15 = 0

Practice more questions

Fast Revision

NCERT Solutions for Class 11 Maths Chapter 11 : Conic Sections

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections – Brief Overview

Circle

A circle is defined as the collection of all points in a plane that are equidistant from a fixed point in the plane.

Standard Equation of a Circle –

If C(h, k) is the center and r is the radius of a circle, while P(x, y) is any point on the circle

(x – h)² + (y – k)² = r²

Parabola

A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point (not on the line) in the plane.

Standard Equation of a Parabola –

Considering focus at (a, 0) a > 0; and directrix x = – a

y² = 4ax

Ellipse

An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant.

Standard Equation of an Ellipse –

The center of the ellipse is the origin and ‘a’ and ‘b’ are the X and Y intercepts respectively.

x²/a² + y²/b² = 1

Hyperbola

A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

Standard Equation of a Hyperbola –

x²/a² - y²/b² = 1

Latus Rectum

The latus rectum is a line that runs parallel to the conic's directrix and passes through its foci. The latus rectum is also the focal chord that runs perpendicular to the conic axis.

Length of Latus Rectum of a Parabola – 4a
Length of Latus Rectum of an Ellipse and a Hyperbola – 2b²/a

Related Chapters

Chapter 1 : Sets

Chapter 2 : Relations and Functions

Chapter 3 : Trigonometric Functions

Chapter 4 : Principle of Mathematical Induction

Chapter 5 : Complex Numbers and Quadratic Equations

Chapter 6 : Linear Inequalities

Chapter 7 : Permutations and Combinations

Chapter 8 : Binomial Theorem

Chapter 9 : Sequences and Series

Chapter 10 : Straight Lines

Chapter 12 : Introduction to Three Dimensional Geometry

Chapter 13 : Limits and Derivatives

Chapter 14 : Mathematical Reasoning

Chapter 15 : Statistics

Chapter 16 : Probability

Frequently Asked Questions on NCERT Class 11 Maths Chapter 11 : Conic Sections

Q1. How many exercises are there in the Conic Sections Class 11 NCERT Solutions?

Answer: Chapter 11 Conic Sections Class 11 Maths has a total of 70 questions, 55 of which are easy, 10 of which are moderate, and the remaining are extended answer type problems. These are required to solve because they provide additional information on the ideas learnt. The solutions to the exercise-specific problems are also accessible in Conic Sections Class 11 PDF format, with the goal of assisting students in performing well in the yearly exam.

Q2. What are the key subtopics in Conic Sections Class 11 Maths Chapter 11 that could be tested?

Answer: NCERT Solutions Class 11 Maths Chapter 11 begins with a brief overview of the Conical Section concepts. The NCERT Solutions Class 11 Maths Chapter 11 focuses on the fundamentals of several forms of conical sections such as circles, ellipses, parabolas, and hyperbolas. It also discusses the conventional equation for each of these curves. It also defines words relating to these curves, such as the latus rectum and eccentricity. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

Q3. Is it necessary to practice all of the questions in Conic Sections Class 11 NCERT Solutions?

Answer: Conic Sections Class 11 NCERT Solutions are well-designed to help students study different types of conical sections along with the equations of different types of curves, be it parabola, circle, ellipse or hyperbola. It thoroughly covers all essential principles, allowing students to grasp the concepts. These solutions use examples to clarify the subjects presented so that students may easily relate to the concepts being addressed.

Related Chapters

Chapter 1 : Sets

Chapter 2 : Relations and Functions

Chapter 3 : Trigonometric Functions

Chapter 4 : Principle of Mathematical Induction

Chapter 5 : Complex Numbers and Quadratic Equations

Chapter 6 : Linear Inequalities

Chapter 7 : Permutations and Combinations

Chapter 8 : Binomial Theorem

Chapter 9 : Sequences and Series

Chapter 10 : Straight Lines

Chapter 12 : Introduction to Three Dimensional Geometry