# 11.1: Prelude to Conics - Mathematics

Five, Four. Then you will use what you learn to investigate systems of nonlinear equations.

## Exercise 11.1 Chapter 11 Conic Sections Class 11 Maths

In this lecture I have discussed ncert solutions for Exercise 11.1 of Chapter 11 Conic Sections Class 11 Maths.

00:00:05 How to study Conic Sections in Class 11

00:05:31 Introduction to Circle, Parabola, Ellipse and Hyperbola

00:07:31 Animation to understand why they are called conic sections

00:09:51 How to derive equation of circle
In each of the following Exercises 1 to 5, find the equation of the circle with

00:13:51 NCERT Solutions Exercise 11.1 Question 1 centre (0,2) and radius 2

00:16:41 NCERT Solutions Exercise 11.1 Question 2 centre (–2,3) and radius 4

00:18:31 NCERT Solutions Exercise 11.1 Question 3 centre left ( frac<1><2>, frac<1> <4> ight ) and radius frac <1>

00:21:41 NCERT Solutions Exercise 11.1 Question 4 centre (1,1) and radius sqrt

00:22:52 NCERT Solutions Exercise 11.1 Question 5 centre (–a, –b) and radius sqrt
In each of the following Exercises 6 to 9, find the centre and radius of the circles.

00:25:32 NCERT Solutions Exercise 11.1 Question 6 (x+5)^2+(y-3)^2=36

00:26:52 NCERT Solutions Exercise 11.1 Question 7 x^2+y^2-4x-8y-45=0

00:31:02 NCERT Solutions Exercise 11.1 Question 8 x^2+y^2-8x+10y-12=0

00:33:22 NCERT Solutions Exercise 11.1 Question 9 2x^2+2y^2-x=0

00:36:02 NCERT Solutions Exercise 11.1 Question 15 Does the point (–2.5, 3.5) lie inside, outside or on the circle x^2+ y^2 = 25 ?

00:42:42 NCERT Solutions Exercise 11.1 Question 14 Find the equation of a circle with centre (2,2) and passes through the point (4,5).

00:45:42 NCERT Solutions Exercise 11.1 Question 12 Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2,3).

00:51:42 NCERT Solutions Exercise 11.1 Question 13 Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

00:59:42 NCERT Solutions Exercise 11.1 Question 10 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.

01:08:32 NCERT Solutions Exercise 11.1 Question 11 Find the equation of the circle passing through the points (2,3) and (–1,1) and whose centre is on the line x – 3y – 11 = 0.

## Chapter 11 Class 11 Conic Sections

Learn Chapter 11 Conic Sections of Class 11 free with solutions of all NCERT Questions, Examples and Miscelleanous exercises. All solutions are provided with step-by-step explanation for your reference.

Let's see what conic section is.

We learned Straight Lines in the last chapter, but straight lines are not the only type of curves we have.

In this chapter, we talk about Conic Sections,

that is, sections of the cone

Circles, Ellipse, Parabola and Hyperbola

So, the topics of the chapter include

• Circles - How to find equation of circle, center of circle
• Parabola - Equation of parabola, its directrix, eccentricity and focus
• Ellipse - Equation of ellipse, its directrix, eccentricity, focus and vertices
• Hyperbola - Equation of hyperbola, its directrix, eccentricity, focus and vertices
• Other questions like mirror problem , Triangle in parabola problem, Beam problem, Locus, Path traced problems

Click on an exercise to start with answers of the questions, or the topic to learn the concepts with the questions

## Conic Sections Class 11 Notes Maths Chapter 11

Circle
A circle is the set of all points in a plane, which are at a fixed distance from a fixed point in the plane. The fixed point is called the centre of the circle and the distance from centre to any point on the circle is called the radius of the circle.
The equation of a circle with radius r having centre (h, k) is given by (x – h) 2 + (y – k) 2 = r 2 .

The general equation of the circle is given by x 2 + y 2 + 2gx + 2fy + c = 0 , where, g, f and c are constants.

The general equation of the circle passing through origin is x 2 + y 2 + 2gx + 2fy = 0.

The parametric equation of the circle x 2 + y 2 = r 2 are given by x = r cos θ, y = r sin θ, where θ is the parametre and the parametric equation of the circle (x – h) 2 + (y – k) 2 = r 2 are given by x = h + r cos θ, y = k + r sin θ.

Note: The general equation of the circle involves three constants which implies that at least three conditions are required to determine a circle uniquely.

Parabola
A parabola is the set of points P whose distances from a fixed point F in the plane are equal to their distance from a fixed line l in the plane. The fixed point F is called focus and the fixed line l is the directrix of the parabola.

 Forms of parabola y 2 = 4ax y 2 = -4ax x 2 = 4ay x 2 = -4ay Axis of parabola y = 0 y = 0 x = 0 x = 0 Directrix of parabola x = -a x = a y = -a y = a Vertex (0, 0) (0, 0) (0, 0) (0, 0) Focus (a, 0) (-a, 0) (0, a) (0, -a) Length of latus rectum 4a 4a 4a 4a Focal length |x + a| |x – a| |y + a| |y – a|

Ellipse
An ellipse is the set of all points in a plane such that the sum of whose distances from two fixed points is constant.
or
An ellipse is the set of all points in the plane whose distances from a fixed point in the plane bears a constant ratio, less than to their distance from a fixed point in the plane. The fixed point is called focus, the fixed line a directrix and the constant ratio(e) the eccentricity of the ellipse. We have two standard forms of ellipse i.e.

Hyperbola
A hyperbola is the locus of a point in a plane which moves in such a way that the ratio of its distance from a fixed point in the same plane to its distance from a fixed line is always constant which is always greater than unity. The fixed point is called the focus, the fixed line is called the directrix and the constant ratio, generally denoted bye, is known as the eccentricity of the hyperbola.
We have two standard forms of hyperbola i.e.

Exploring Patterns and Sequences
TI-83 Plus Calculator: Generating the Terms of a
Sequence
Sequences and Recursive Formulas
TI-83 Plus Calculator: Graphing Sequences
Investigating Ways of Cutting "Vegetables"
Arithmetic Sequences
Geometric Sequences
Compound Interest: Amount and Present Value
Rational Exponents
Simplifying Expressions Involving Exponents
Solving Exponential Equations

Chapter 2: Series and Financial Applications

Arithmetic Series
Investigating a Sequence using the TI-83 Plus Calculator    and the TI Calculator-Based Ranger (CBR)
Geometric Series
Using a Spreadsheet to Represent the Value of a Deposit
Using Spreadsheets to Analyze and Represent Financial    Situations: Future Value and Amortization Tables
TI-83 Plus Calculator: Finding the Sum and the Using Series to Analyze Financial Situations: Future Value
Using Series to Analyze Financial Situations: Present TI-83 Plus Calculator: Analyzing Financial Situations Using    the TVM Solver
Equivalent Rates and General Annuities
TI-83 Plus Calculator: Creating Repayment Schedules
Using Technology to Analyze Canadian Mortgages

Chapter 2 Review
Chapter 2 Review Test
Cumulative Review Test

Performance Task for Chapters 1 and 2

Review of Essential Skills and Knowledge - Part 2

Chapter 3: Introducing Functions

Investigating a Special Kind of Relationship
Functions: Concept and Notation
Solving Inequalities
The Inverse Function
TI-83 Plus Calculator: Graphing Functions and Inverse Investigating Properties of Inverse Functions
Transformations and Function Notation

Chapter 4: Quadratic and Rational Functions

Extending Algebra Skills: Completing the Square
Maximum and Minimum Values of Quadratic Functions
Introducing Complex Numbers
Adding, Subtracting, and Multiplying Complex Numbers
Reciprocal Functions
TI-83 Plus Calculator: Exploring the Behaviour of Functions Near the Asymptotes
Simplifying Rational Expressions
Multiplying and Dividing Rational Expressions
Dividing Complex Numbers
Extending Algebra Skills: Working with Polynomials

Chapter 4 Review
Chapter 4 Review Test
Cumulative Review Test

Performance Task for Chapters 3 and 4

PART 3 - TRIGONOMETRIC FUNCTIONS

Review of Essential Skills and Knowledge - Part 3

Chapter 5: Modeling Periodic Functions

Periodic Phenomena
Understanding Angles
Trigonometric Functions
TI-83 Plus Calculator: Graphing Trigonometric Functions
Investigating Transformations
Modeling Periodic Phenomena
Solving Linear Trigonometric Equations
TI-83 Plus Calculator: Using Sinusodial Regression
to Find the Curve of Best Fit
Using Digital Probes to Collect Periodic Data

Chapter 5 Review
Chapter 5 Review Test

Chapter 6: Extending Skills With Trigonometry

Extending Trigonometry Skills With Oblique Triangles
Solving Trigonometry Problems in Two and Three Dimensions
Using Special Triangles to Determine Exact Values
Investigating Identical Expressions
Trigonometric Identities

PART 4 - LOCI AND CONICS

Review of Essential Skills and Knowledge - Part 4

Chapter 7: Investigating Loci and Conics

Introducing Locus Definitions
Revisiting the Circle
TI-83 Plus Calculator: Graphing Circles
The Ellipse
Waxed-Paper Models
The Parabola
Reflective Properties of Conics
Graphing Conics Using Technology
The Hyperbola
The General Form of Conics
When Lines Meet Conics

## 11.1: Prelude to Conics - Mathematics

Given: Centre (0, 2) and radius (r) = 2

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

As, Centre (h, k) = (0, 2) and radius (r) = 2

Thus, the equation of the circle is

(x – 0) 2 + (y – 2) 2 = 2 2 [by using formula (a – b) 2 = a 2 – 2ab + b 2 ]

x 2 + y 2 + 4 – 4y = 4

x 2 + y 2 – 4y = 0

Therefore, the equation of the circle is x 2 + y 2 – 4y = 0

### Question 2: Centre (–2, 3) and radius 4

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

As, centre (h, k) = (-2, 3) and radius (r) = 4

Thus, the equation of the circle is

(x + 2) 2 + (y – 3) 2 = (4) 2 [by using formula (a – b) 2 = a 2 – 2ab + b 2 ]

x 2 + 4x + 4 + y 2 – 6y + 9 = 16

x 2 + y 2 + 4x – 6y – 3 = 0

Therefore, the equation of the circle is x 2 + y 2 + 4x – 6y – 3 = 0

### Question 3: Centre (1/2, 1/4) and radius (1/12)

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

So, centre (h, k) = (1/2, 1/4) and radius (r) = 1/12

Thus, the equation of the circle is

(x – 1/2) 2 + (y – 1/4) 2 = (1/12) 2 [by using formula (a – b) 2 = a 2 – 2ab + b 2 ]

x 2 – x + 1/4 + y 2 – y/2 + 1/16 = 1/144

x 2 – x + 1/4 + y 2 – y/2 + 1/16 = 1/144

144x 2 – 144x + 36 + 144y 2 – 72y + 9 – 1 = 0

144x 2 – 144x + 144y 2 – 72y + 44 = 0

36x 2 + 36x + 36y2 – 18y + 11 = 0

36x 2 + 36y2 – 36x – 18y + 11= 0

Therefore, the equation of the circle is 36x 2 + 36y 2 – 36x – 18y + 11= 0

### Question 4: Centre (1, 1) and radius √2

Given: Centre (1, 1) and radius √2

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

So, centre (h, k) = (1, 1) and radius (r) = √2

Thus, the equation of the circle is

(x-1) 2 + (y-1) 2 = (√2) 2 [by using formula (a – b) 2 = a 2 – 2ab + b 2 ]

x 2 – 2x + 1 + y 2 -2y + 1 = 2

x 2 + y 2 – 2x -2y = 0

Therefore, the equation of the circle is x 2 + y 2 – 2x -2y = 0

### Question 5: Centre (–a, –b) and radius √(a 2 – b 2 )

Given: Centre (-a, -b) and radius √(a 2 – b 2 )

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

So, centre (h, k) = (-a, -b) and radius (r) = √(a 2 – b 2 )

Thus, the equation of the circle is

(x + a) 2 + (y + b) 2 = (√(a 2 – b 2 ) 2 ) [by using formula (a + b) 2 = a 2 + 2ab + b 2 ]

x 2 + 2ax + a 2 + y 2 + 2by + b 2 = a 2 – b 2

x 2 + y 2 +2ax + 2by + 2b 2 = 0

Therefore, the equation of the circle is x 2 + y 2 +2ax + 2by + 2b 2 = 0

### Question 6: (x + 5) 2 + (y – 3) 2 = 36

Given equation: (x + 5) 2 + (y – 3) 2 = 36

(x – (-5)) 2 + (y – 3) 2 = 6 2

The equation is of the form (x – h) 2 + (y – k) 2 = r 2 where, h = -5, k = 3 and r = 6

Therefore, the centre is (-5, 3) and its radius is 6.

### Question 7: x 2 + y 2 – 4x – 8y – 45 = 0

Given equation: x 2 + y 2 – 4x – 8y – 45 = 0.

x 2 + y 2 – 4x – 8y – 45 = 0

(x 2 – 4x) + (y 2 -8y) = 45

(x 2 – 2(x) (2) + 2 2 ) + (y 2 – 2(y) (4) + 4 2 ) – 4 – 16 = 45

(x – 2) 2 + (y – 4) 2 = 65

(x – 2) 2 + (y – 4) 2 = (√65) 2

The equation is of the form (x-h) 2 +(y-k) 2 = r 2 ,where h = 2, k = 4 and r = √65

Therefore, the centre is (2, 4) and its radius is √65.

### Question 8: x 2 + y 2 – 8x + 10y – 12 = 0

Given equation: x 2 + y 2 -8x + 10y -12 = 0.

x 2 + y 2 – 8x + 10y – 12 = 0

(x 2 – 8x) + (y 2 + 10y) = 12

(x 2 – 2(x) (4) + 4 2 ) + (y 2 – 2(y) (5) + 5 2 ) – 16 – 25 = 12

(x – 4) 2 + (y + 5) 2 = 53

(x – 4) 2 + (y – (-5)) 2 = (√53) 2

The equation is of the form (x-h) 2 +(y-k) 2 = r 2 ,where h = 4, k= -5 and r = √53

Therefore, the centre is (4, -5) and its radius is √53.

### Question 9: 2x 2 + 2y 2 – x = 0

Given equation: 2x 2 + 2y 2 – x = 0.

2x 2 + 2y 2 – x = 0

(2x 2 – x) + 2y 2 = 0

(x 2 – 2 (x) (1/4) + (1/4) 2 ) + y 2 – (1/4) 2 = 0

(x – 1/4) 2 + (y – 0) 2 = (1/4) 2

The equation is of the form (x-h) 2 +(y-k) 2 = r 2 , where, h = 1/4, k = 0, and r = 1/4

Therefore, the centre is (1/4, 0) and its radius is 1/4.

### Question 10: 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.

The equation of the circle is (x – h) 2 + (y – k) 2 = r 2

As the circle passes through points (4,1) and (6,5)

So, When the circle passes through (4,1)

(4 – h) 2 + (1 – k) 2 = r 2 ……………..(1)

When the circle passes through (6,5)

(6 – h) 2 + (5 – k) 2 = r 2 ………………(2)

Given that, the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k =16 ………………… (3)

From equation (1) and (2), we get

(4 – h) 2 + (1 – k) 2 =(6 – h) 2 + (5 – k) 2

16 – 8h + h 2 +1 -2k +k 2 = 36 -12h +h 2 +15 – 10k + k 2

16 – 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k = 11 ……………. (4)

Now let us multiply equation (3) by 2, and subtracting it with equation (4), we get

(h + 2k) – 2(4h + k) = 11 – 32

h + 2k – 8h – 2k = -21

-7h = -21

h = 3

Substitute this value of h in equation (4), we get

3 + 2k = 11

2k = 11 – 3

2k = 8

k = 4

We get h = 3 and k = 4

When we substitute the values of h and k in equation (1), we get

(4 – 3) 2 + (1 – 4) 2 = r 2

(1) 2 + (-3) 2 = r 2

1+9 = r 2

r = √10

Now, the equation of the circle is,

(x – 3) 2 + (y – 4) 2 = (√10) 2

x 2 – 6x + 9 + y 2 – 8y + 16 =10

x 2 + y 2 – 6x – 8y + 15 = 0

Therefore, the equation of the circle is x 2 + y 2 – 6x – 8y + 15 = 0

### Question 11: Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

The equation of the circle is (x – h) 2 + (y – k) 2 = r 2

As the circle passes through points (2,3) and (-1,1)

So, When the circle passes through (2,3)

(2 – h) 2 + (3 – k) 2 =r 2 ……………..(1)

When the circle passes through (-1,1)

(-1 – h) 2 + (1– k) 2 =r 2 ………………(2)

Given that, the centre (h, k) of the circle lies on line x – 3y – 11= 0,

h – 3k =11 ………………… (3)

From the equation (1) and (2), we get

(2 – h) 2 + (3 – k) 2 =(-1 – h) 2 + (1 – k) 2

4 – 4h + h 2 +9 -6k +k 2 = 1 + 2h +h 2 +1 – 2k + k 2

4 – 4h +9 -6k = 1 + 2h + 1 -2k

6h + 4k =11 ……………. (4)

Now let us multiply equation (3) by 6, and subtracting it with equation 4, we get

6h+ 4k – 6(h-3k) = 11 – 66

6h + 4k – 6h + 18k = 11 – 66

22 k = – 55

k = -5/2

Substitute this value of k in equation (4), we get

6h + 4(-5/2) = 11

6h – 10 = 11

6h = 21

h = 21/6

h = 7/2

We get h = 7/2 and k = -5/2

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

(2 – 7/2) 2 + (3 + 5/2) 2 = r 2

[(4-7)/2] 2 + [(6+5)/2] 2 = r 2

(-3/2) 2 + (11/2) 2 = r 2

9/4 + 121/4 = r 2

130/4 = r 2

Now, the equation of the circle is,

(x – 7/2) 2 + (y + 5/2) 2 = 130/4

[(2x-7)/2] 2 + [(2y+5)/2] 2 = 130/4

4ࡨ -28x + 49 +4y 2 + 20y + 25 =130

4x 2 +4y 2 -28x + 20y – 56 = 0

4(x 2 +y 2 -7x + 5y – 14) = 0

x 2 + y 2 – 7x + 5y – 14 = 0

Therefore, the equation of the circle is x 2 + y 2 – 7x + 5y – 14 = 0

### Question 12: Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

The equation of the circle is (x – h) 2 + (y – k) 2 = r 2

Given the radius of the circle is 5 and it’s centre lies on the x-axis, k = 0 and r = 5.

So now, the equation of the circle is (x – h) 2 + y 2 = 25.

Also given that the circle passes through the point (2, 3).

Therefore,

(2 – h) 2 + 3 2 = 25

(2 – h) 2 = 25-9

(2 – h) 2 = 16

2 – h = ± √16 = ± 4

If 2-h = 4, then h = -2

If 2-h = -4, then h = 6

Now, when h = -2, the equation of the circle is

(x + 2) 2 + y 2 = 25

x 2 + 4x + 4 + y 2 = 25

x 2 + y 2 + 4x – 21 = 0

Now, when h = 6, the equation of the circle is

(x – 6) 2 + y 2 = 25

x 2 -12x + 36 + y 2 = 25

x 2 + y 2 -12x + 11 = 0

Therefore, the equation of the circle is x 2 + y 2 – 4x + 21 = 0 and x 2 + y 2 -12x + 11 = 0

### Question 13: Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

The equation of the circle is (x – h) 2 + (y – k) 2 = r 2

When the circle passes through (0, 0),we get,

(0 – h) 2 + (0 – k) 2 = r 2

h 2 + k 2 = r 2

The equation of the circle is (x – h) 2 + (y – k) 2 = h 2 + k 2 .

Given that the circle intercepts points a and b on the coordinate axes.

Since, the circle passes through points (a, 0) and (0, b).

So the equations are,

(a – h) 2 + (0 – k) 2 = h 2 +k 2 ……………..(1)

(0 – h) 2 + (b– k) 2 = h 2 +k 2 ………………(2)

From equation (1), we get

a 2 – 2ah + h 2 + k 2 = h 2 + k 2

a 2 – 2ah = 0

a(a – 2h) =0

a = 0 or (a -2h) = 0

As, a ≠ 0 hence, (a -2h) = 0

h = a/2

From equation (2), we get

h 2 – 2bk + k 2 + b 2 = h 2 + k 2

b 2 – 2bk = 0

b(b– 2k) = 0

b= 0 or (b-2k) =0

As, a ≠ 0 hence, (b -2k) = 0

k = b/2

Now, substituting the value of h and k, we get

(x – a/2) 2 + (y – b/4) 2 = (a/2) 2 + (b/2) 2

[(2x-a)/2] 2 + [(2y+b)/2] 2 = (a2 + b2)/4

4x 2 – 4ax + a 2 +4y 2 – 4by + b 2 = a 2 + b 2

4x 2 + 4y 2 -4ax – 4by = 0

4(x 2 +y 2 -7x + 5y – 14) = 0

x 2 + y 2 – ax – by = 0

Therefore, the equation of the circle is x 2 + y 2 – ax – by = 0

### Question 14: Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Given the centre of the circle as (h, k) = (2,2)

Also given that the circle passes through the point (4,5),

the radius (r) of the circle is the distance between the points (2,2) and (4,5).

r = √[(2-4) 2 + (2-5) 2 ]

= √[(-2) 2 + (-3) 2 ]

= √[4+9]

= √13

Now, the equation of the circle is

(x– h) 2 + (y – k) 2 = r 2

(x –h) 2 + (y – k) 2 = (√13) 2

(x –2) 2 + (y – 2) 2 = (√13) 2

x 2 – 4x + 4 + y 2 – 4y + 4 = 13

x 2 + y 2 – 4x – 4y = 5

Therefore, the equation of the circle is x 2 + y 2 – 4x – 4y = 5

### Question 15: Does the point (–2.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25?

Given equation of the circle is x 2 +y 2 = 25.

x 2 + y 2 = 25

(x – 0) 2 + (y – 0) 2 = 5 2

The equation is of the form (x – h) 2 + (y – k) 2 = r 2 ,where, h = 0, k = 0 and r = 5.

Now the distance between the point (-2.5, 3.5) and the centre (0,0) is

= √[(-2.5 – 0) 2 + (-3.5 – 0) 2 ]

= √(6.25 + 12.25)

= √18.5

= 4.3 [which is < 5]

Since, the distance between the point (-2.5, -3.5) and the centre (0, 0) of the circle is less than the radius of the circle.

Therefore, the point (-2.5, -3.5) lies inside the circle.

In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.

Ex 11.2 Class 11 Maths Question 1.
y 2 = 12x
Solution:
The given equation of parabola is y 2 = 12x which is of the form y 2 = 4ax.
∴ 4a = 12 ⇒ a = 3
∴ Coordinates of focus are (3, 0)
Axis of parabola is y = 0
Equation of the directrix is x = -3 ⇒ x + 3 = 0
Length of latus rectum = 4 x 3 = 12.

Ex 11.2 Class 11 Maths Question 2.
x 2 = 6y
Solution:
The given equation of parabola is x 2 = 6y which is of the form x 2 = 4ay.

Ex 11.2 Class 11 Maths Question 3.
y 2 = – 8x
Solution:
The given equation of parabola is
y 2 = -8x, which is of the form y 2 = – 4ax.
∴ 4a = 8 ⇒ a = 2
∴ Coordinates of focus are (-2, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 2 ⇒ x – 2 = 0
Length of latus rectum = 4 x 2 = 8.

Ex 11.2 Class 11 Maths Question 4.
x 2 = -16y
Solution:
The given equation of parabola is
x 2 = -16y, which is of the form x 2 = -4ay.
∴ 4a = 16 ⇒ a = 4
∴ Coordinates of focus are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is y = 4 ⇒ y – 4 = 0
Length of latus rectum = 4 x 4 = 16.

Check the Parabola Calculator to solve Parabola Equation.

Ex 11.2 Class 11 Maths Question 5.
y 2 = 10x
Solution:
The given equation of parabola is y 2 = 10x, which is of the form y 2 = 4ax.

Ex 11.2 Class 11 Maths Question 6.
x 2 = -9y
Solution:
The given equation of parabola is
x 2 = -9y, which is of the form x 2 = -4ay.

In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:

Ex 11.2 Class 11 Maths Question 7.
Focus (6, 0) directrix x = -6
Solution:
We are given that the focus (6, 0) lies on the x-axis, therefore x-axis is the axis of parabola. Also, the directrix is x = -6 i.e. x = -a and focus (6, 0) i.e. (a, 0). The equation of parabola is of the form y 2 = 4ax.
The required equation of parabola is
y 2 = 4 x 6x ⇒ y 2 = 24x.

Ex 11.2 Class 11 Maths Question 8.
Focus (0, -3) directri xy=3
Solution:
We are given that the focus (0, -3) lies on the y-axis, therefore y-axis is the axis of parabola. Also the directrix is y = 3 i.e. y = a and focus (0, -3) i.e. (0, -a). The equation of parabola is of the form x 2 = -4ay.
The required equation of parabola is
x 2 = – 4 x 3y ⇒ x 2 = -12y.

Ex 11.2 Class 11 Maths Question 9.
Vertex (0, 0) focus (3, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (3, 0)
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y 2 = 4ax
The required equation of the parabola is
y 2 = 4 x 3x ⇒ y 2 = 12x.

Ex 11.2 Class 11 Maths Question 10.
Vertex (0, 0) focus (-2, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (-2, 0).
∴ y = 0 ⇒ The axis of parabola is along x-axis
∴ The equation of the parabola is of the form y 2 = – 4ax
The required equation of the parabola is
y 2 = – 4 x 2x ⇒ y 2 = -8x.

Ex 11.2 Class 11 Maths Question 11.
Vertex (0, 0), passing through (2, 3) and axis is along x-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.
∴ The equation of the parabola is of the form y 2 = 4ax
Since the parabola passes through point (2, 3)

Ex 11.2 Class 11 Maths Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect toy-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ The equation of the parabola is of the form x 2 = 4ay
Since the parabola passes through point (5, 2)

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.2, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.3

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3

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## Conic Sections – Exercise 11.1 – Class XI

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

Therefore, the equation of the circle is

The equation of a circle with centre (h, k) and radius r is given as

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

Therefore, the equation of the circle is

⟹ x 2 + 4x + 4 + y 2 – 6y + 9 = 16

The equation of a circle with centre (h, k) and radius r is given as

It is given that centre (h, k) =( 1 /2, 1 /4) and radius (r) = ( 1 /12)

Therefore, the equation of the circle is

144x 2 – 144x + 36 + 144y 2 – 72y + 44 = 0

36x 2 – 36x + 36y 2 – 18y + 11 = 0

36x 2 + 36y 2 – 36x – 18y + 11=0

The equation of a circle with centre (h, k) and radius r is given as

It is given that centre (h, k) = (1, 1) and radius (r) = √2.

Therefore, the equation of the circle is

The equation of a circle with centre (h, k) and radius r is given as

It is given that centre (h, k) = (-a, -b) and radius (r) = √(a 2 – b 2 ).

Therefore, the equation of the circle is

(x + a) 2 + (y + b) 2 = (√(a 2 – b 2 )) 2

x 2 + 2ax + a 2 +y 2 + 2by + b 2 = a 2 – b 2

x 2 + y 2 + 2ax + 2by + 2b 2 = 0

The equation of the given circle is (x + 5) 2 + (y – 3) 2 = 36.

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.

The equation of the given circle is x 2 + y 2 – 4x – 8y – 45 = 0.

Which is of the form (x – h) 2 + (y – k) 2 = r 2 , where h = 2, k = 4 and since r 2 = 65, we have r = √65 ,

Thus, the centre of the given circle is (2, 4), while its radius is √65 .

The equation of the given circle is x 2 + y 2 – 8x + 10y – 12 = 0.

⟹ (x − 4) 2 + 2 = (√53) 2 which is of the form (x – h) 2 + (y – k) 2 = r 2 , where h = 4, k = – 5 and r = √53

Thus, the centre of the given circle is (4, –5), while its radius is √53.

The equation of the given circle is 2x 2 + 2y 2 – x = 0.

which is of the form (x – h) 2 + (y – k) 2 = r 2 , where h = 1/4, k = 0 and r = 1/4.

Thus, the centre of the given circle is (1/4, 0), while its radius is 1/4.

1. 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 = r 2 .

Since the circle passes through points (4, 1) and (6, 5),

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

From equations (1) and (2), we obtain

(4 – h) 2 + (1 – k) 2 = (6 – h) 2 + (5 – k) 2

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

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

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 = r 2

Thus, the equation of the required circle is

x 2 – 6x + 9 + y 2 – 8y + 16 = 10

1. Find the equation of the circle passing through the points (2, 3) and (–1, 1) and whose centre is on the line x – 3y – 11 = 0.

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

Since the circle passes through points (2, 3) and (–1, 1),

Since the centre (h, k) of the circle lies on line x – 3y – 11 = 0,

From equations (1) and (2), we obtain

(2 – h) 2 + (3 – k) 2 = (–1 – h) 2 + (1 – k) 2

⇒ 4 – 4h + h 2 + 9 – 6k + k 2 = 1 + 2h + h 2 + 1 – 2k + k 2

⇒ 4 – 4h + 9 – 6k = 1 + 2h + 1 – 2k

On solving equations (3) and (4), we obtain h = 7 /2 and k =− 5 /2

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

Thus the equation of the required circle is

4x 2 – 28x + 49 + 4y 2 + 20y + 25 = 130

4x 2 + 4y 2 – 28x + 20y – 56 = 0

1. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

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

Since the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

Now, the equation of the circle becomes (x – h) 2 + y 2 = 25.

It is given that the circle passes through point (2, 3).

when h = -2 the equation of the circle becomes

When h = 6 the equation of the circle becomes

1. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

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

Since the centre of the circle passes through (0, 0),

The equation of the circle now becomes (x – h) 2 + ( y- k) 2 =h 2 + k 2

It is given that the circle makes intercepts a and b on the coordinate axes. This means that the circle passes through points (a, 0) and (0, b). Therefore,

From equation (1), we obtain a 2 – 2ah + h 2 + k 2 = h 2 + k 2

However, a ≠ 0 hence, (a – 2h) = 0 ⇒ h =a/2.

From equation (2), we obtain h 2 + b 2 – 2bk + k 2 = h 2 + k 2

However, b ≠ 0 hence, (b – 2k) = 0 ⇒ k =b/2.

Thus, the equation of the required circle is

4x 2 – 4ax +a 2 + 4y 2 – 4by + b 2 = a 2 + b 2

1. Find the equation of a circle with centre (2, 2) and passes through the point (4, 5) .

The centre of the circle is given as (h, k) = (2, 2).

Since the circle passes through point (4, 5), the radius (r) of the circle is the distance between the points (2, 2) and (4, 5).

r = √[(2-4) 2 + (2-5) 2 ] = √[(2) 2 + (3) 2 ] = √(4+9) = √13

Thhus the equation of the circle is

The equation of the given circle is x 2 + y 2 = 25.

⇒ (x – 0) 2 + (y – 0) 2 = 5 2 ,which is of the form (x – h) 2 + (y – k) 2 = r 2 , where h = 0, k = 0, and r = 5.

∴ Centre = (0, 0) and radius = 5

Distance between point (–2.5, 3.5) and centre (0, 0)

Since the distance between point (–2.5, 3.5) and centre (0, 0) of the circle is less than the radius of the circle, point (–2.5, 3.5) lies inside the circle.

## NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Exercise 11.1

NCERT Solutions of Chapter 11 Conic Sections Exercise 11.1 is provided here which will help you solving difficult questions easily and completing your homework in no time. NCERT Solutions for Class 11 Maths are prepared by Studyrankers experts that are detailed and correct so you can improve your score in the examinations.

Here h = 𔃀, k = 3 and r = 4. Therefore, the required equation of the circle is
[x – (𔃀)] 2 + (y – 3) 2 = (4) 2
or (x + 2) 2 + (y – 3) 2 = 16
or x 2 + 4x + 4 + y 2 – 6y + 9 = 16.

3. Find the equation of the circle with centre (1/2 , 1/4) and radius 1/12.

Equation of the circle is
(x – 1/2) 2 + (y – 1/4) 2 = 1/12 2 = 1/144
=> x 2 + y 2 – x – y/2 + 1/4 + 1/16 = 1/144
=> 36x 2 + 36y 2 – 36x – 18y + 11 = 0.

4. Find the equation of the circle with centre (1, 1) and radius 𕔆.

Centre of circle is (1, 1), radius = 𕔆
Equation of circle is
(x – 1) 2 + (y – 1) 2 = (𕔆) 2 =2
or x 2 + y 2 – 2x – 2y + 2 = 2
or x 2 + y 2 – 2x – 2y = 0.

5. Find the equation of the circle with centre (–a, –b) and radius √a 2 + b 2

Centre of circle is (–a, –b), radius =√a2 + b2
∴ Equation of the circle is
(x + a) 2 + (y + b) 2 = a 2 – b 2
Or x 2 + y 2 + 2xa + 2yb + a 2 + b 2 = a 2 – b 2
Or x 2 + y 2 + 2ax + 2by + 2b 2 = 0.

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

Comparing the equation of the circle
(x + 5) 2 + (y – 3) 2 = 36
with (x – h) 2 + (y – k) 2 = 2
∴ –h = 5 or h = 𔃃, k = 3, r 2 = 36, r = 6
∴ Centre of the circle is (𔃃, 3) and radius = 6

7. Find the centre and radius of the circle.
x 2 + y 2 – 4x – 8y – 45 = 0

The given equation is
x 2 + y 2 – 4x – 8y – 45 = 0
or (x 2 – 4x) + (y 2 – 5y) = 45
Now completing the squares with in the parenthesis, we get
(x 2 – 4x + 4) + (y 2 – 8y + 16) = 4 + 16 + 45
or (x – 2) 2 + (y – 4) 2 = 65
Therefore, the given circle has centre at (2, 4) and radius 󕉩.

8. Find the centre and radius of the circle.
x 2 + y 2 – 8x + 10y – 12 = 0.

The given equation is
x 2 + y 2 – 8x + 10y – 12 = 0
or (x 2 – 8x) + (y 2 + 10) = 12
or (x 2 – 8x + 16) + (y 2 + 10y + 25) = 12 + 16 + 25
or (x – 4) 2 + (y + 5) 2 = 53
Therefore, the given circle has centre at (4, 𔃃) and radius 󕉝.

9. Find the centre and radius of the given circle
2x 2 + 2y 2 – x = 0.

Equation of circle is 2x 2 + 2y 2 – x = 0
=> x 2 + y 2 – x/2 = 0 => (x 2 – x/2) + y 2 = 0
=> (x 2 – x/ 2 + 1/16) + y 2 = 1/16
=> (x – 1/4) 2 + y 2 = 1/16
Centre is (1/4 , 0)and radius is 1.

10. 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 circle be
(x – h) 2 + (y – k) 2 = r 2 … (i)
The points (4, 1) and (6, 5) lies on it
∴ (4 – h) 2 + (1 – k) 2 = r 2
=> h 2 + k 2 – 8h – 2k + 17 = r 2 …..(ii)
and (6 – h) 2 + (5 – k) 2 = r 2 …..(iii)
The centre (h, k) lies on
4x + y = 16
4h + k = 16 … (iv)
Subtracting (iii) from (ii),
∴ 4h + 8k – 44 = 0 Þh + 2k = 11 … (v)
Multiplying (v) by 4, 4h + 8k = 44
Subtracting eqn (iv) from it
7k = 44 – 16 = 28 ∴ k = 4
From (v) h + 8 = 11 ∴ h = 3
Putting h = 3, k = 4 in (ii)
9 + 16 – 24 – 8 + 17 = r 2
=> 42 – 32 = r 2
∴ r 2 = 10
∴ Equation of the circle is
(x – 3) 2 + (y – 4) 2 = 10
=> x 2 + y 2 – 6x – 8y + 15 = 0

11. Find the equation of the circle passing through the points (2, 3) and (𔂿, 1) and whose centre is on the line x – 3y – 11 = 0.

Let the equation of the circle be
(x – h) 2 + (y – k) 2 = r 2 … (i)
Since, the points (2, 3) and (𔂿, 1) lies on it.
∴ (2 – h) 2 + (3 – k) 2 = r 2
h 2 + k 2 – 4h – 6k + 13 = r 2 … (ii)
Centre (h, k) lies on x – 3y – 11 = 0
h – 3k – 11 = 0 … (iii)
Subtracting (ii) from (i) 6h + 4 k – 11 = 0 … (iv)
Multiply eqn (iii) by 6 6h – 18 k – 66 = 0 … (v)
Subtracting (v) from (iv) 22k + 55 = 0
∴ k = -(55/22) = -(5/2)
from (iii) h = 3k + 11 = -(15/2) + 11 = 7/2
Put the value of h and k in (2 – h) 2 + (3 – k) 2 = r 2
(2 – 7/2) 2 + (3 + 5/2) 2 = r 2
=> r 2 = 9/4 + 121/4 130/4 = 65/2
∴ Equation of the circle
(x – 7/2) 2 + (y + 5/2) 2 = 65/2
=> x 2 + y 2 – 7x + 5y + 49/4 + 25/4 – 65/2 = 0
=> x 2 + y 2 – 7x + 5y – 14 = 0

12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).

Let the equation of the circle be
(x – h) 2 + (y – k) 2 = r 2 … (i)
r = 5 ∴ r 2 = 25
Centre lies on x -axis is k = 0
Equation (i) becomes (x – h) 2 + y 2 = 25 (2, 3) lies on it
∴ (2 – h) 2 + 9 = 25 => (2 – h) 2 = 16,
∴ 2 – h = ۮ => h = 𔃀, 6
When h = 𔃀, equation of circle
(x + 2) 2 + y 2 = 25
=> x 2 + y 2 + 4x – 21 = 0
When h = 6, (x – 6) 2 + y 2 = 25,
x 2 + y 2 – 12x + 11 = 0
Thus, required circles are x 2 + y 2 + 4x – 21 = 0
and x 2 + y 2 – 12x + 11 = 0

13. Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

a, b are the intercepts made by the circle on the co-ordinate axes at A and B, C the mid point of AB is the centre of the circle
∴ centre (a/2 , b/2)
=
∴ Equation of the circle is
(x – a/2) 2 + (y – b/2) 2 =

x 2 + y 2 – ax – by + a 2 /4 + b 2 /4 = (a 2 + b 2 )4
=> x 2 + y 2 – ax – by = 0

14. Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).

Let the centre of the circle C(2, 2), and P(4, 5) is a point on the circle
∴ radius CP = √(4 - 2)2 + (5 - 2)2
= √4 + 9 = √13
∴ Equation of the circle is
(x – 2) 2 + (y – 2) 2 = 13
=> x 2 + y 2 – 4x – 4y = 5

15. Does the point (𔃀.5, 3.5) lie inside, outside or on the circle x 2 + y 2 = 25?