NCERT Solutions of Class 11 maths exercise 11.4-Conic Section

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section are created for the purpose of clearing the doubts of class 11 students on the chapter 11 Conic Section. All Class 11 NCERT maths questions of the exercise 11.4 Conic Sections are solved as per the CBSE norms by an experienced maths teacher.NCERT solutions of class 11 exercise 11.4 Conic Section will help you in achieving excellent marks in the exams and in doing your assignments of the chapter 11 Conic Section.

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NCERT Solutions of Class 11 maths  of chapter 11-Conic Section

Exercise 11.1- Conic Section

Exercise 11.2 – Conic Section

Exercise 11.3- Conic Section

Exercise 11.4-Conic Section

NCERT solutions of class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

In each of the exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbolas.

Ans.We are given the equation of parabola

Comparing the given equation with the standard equation of the parabola

Rewriting the given equation in the form of a standard equation

Comparing the given equation with the standard equation, we get     a =4 and b = 3

The relationship between minor axis(b),major axis(a) and foci(c) is given as follows

c² = a² + b²

c² =4² +3² = 16 +9 =25

c = ±5

Therefore the coordinates of foci are= (±5,0)

Since the hyperbola is symmetric about the x-axis,so the coordinates of vertices are = (±4,0)

Eccentricity (e) of the hyperbola is given as follows

e = c/a =5/4

The latus rectum of the hyperbola is given as follows

Length of latus rectum is =2b²/a = 2(3²)/4 =9/2

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Ans.We are given the equation of the parabola

Comparing the given equation with the standard equation of the parabola

Rewriting the given equation in the form of a standard equation

Comparing the given equation with the standard equation, we get    a =3 and b = 3√3

The relationship between minor axis(b),major axis(a) and foci(c) is given as follows

c² = a² + b²

c² =3² +(3√3)² = 9 +27 =36

c = ±6

Since the hyperbola is symmetric about the y-axis,

Therefore the coordinates of foci are = (0,±6)

So the coordinates of vertices are = (0,±6)

Eccentricity (e) of the hyperbola is given as follows

e = 6/3 =2

The latus rectum of the hyperbola is given as follows

Length of latus rectum is =2b²/a = 2(3√3)²/3 =54/3=18

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Q3.9y² – 4x² = 36

Ans.The given equation is 9y² – 4x² = 36

Rewriting the given equation in the form of standard equation by dividing 36 both sides of equation

The standard form of the equation is as follows

Comparing the given equation with the standard equation, we get  a =2 and b = 3

The relationship between minor axis(b),major axis(a) and foci(c) is given as follows

c² = a² + b²

c² =2² +3² = 4 +9 =13

c = ±√13

Since the hyperbola lies on  the y-axis,

Therefore the coordinates of foci are= (0,±√13)

The coordinates of vertices are = (0,±2)

Eccentricity (e) of the hyperbola is given as follows

e = √13/2

The latus rectum of the hyperbola is given as follows

Length of latus rectum is =2b²/a = 2(3)²/2 =9

Length of latus rectum is =2b²/a = 2(3)²/2 =9

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Q4.16x²- 9y² = 576

Ans.Rewriting the given equation in the form of a standard equation by dividing 36 both sides of the equation

Standard form of the equation is as follows

Comparing the given equation with the standard equation, we get                         a =6 and b = 8

The relationship between minor axis(b),major axis(a) and foci(c) is given as follows

c² = a² + b²

c² =6² +8² = 36 +64 =100

c = ±10

Since the hyperbola lies on  the x-axis,

Therefore the coordinates of foci are= (±10,0)

The coordinates of vertices are = (±6,0)

Eccentricity (e) of the hyperbola is given as following

e = c/a =10/6 =5/3

The latus rectum of the hyperbola is given as follows

Length of latus rectum is =2b²/a = 2(8)²/6 =64/3

Q5. 5y² – 9x² = 36

Ans. Given: The equation is 5y² – 9x² = 36

Let us divide the whole equation by 36, we get

5y²/36 – 9x²/36 = 36/36

y²/(36/5) – x²/4 = 1

y²/[√(36/5)]² – x²/2² = 1

On comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1

We get a = 6/√5 and b = 2

It is known that a² + b² = c²

So, c² = 36/5 + 4

c² = 56/5

c = √(56/5)

= 2√14/√5

Then the coordinates of the foci are (0, 2√14/√5 ) and (0, – 2√14/√5)

The coordinates of the vertices are (0, 6/√5) and (0, – 6/√5)

Eccentricity, e = c/a = (2√14/√5) / 6/√5 = √14/3

Length of lacrus rectum = 2b²/a = (2 × 2²)/6/√5 = (2 × 4)/6/√5 = 4√5/3

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Q6. 49y² – 16x² = 784

Ans. The given  equation  is 49y² – 16x² = 784

Dividing the whole equation by 784, we get

49y²/784 – 16x²/784 = 1

y²/(784/49) – x²/(784/16) = 1

y²/16 – x²/49 = 1

y²/4² – x²/7² = 1

Comparing this equation with the standard equation of hyperbola y²/a² – x²/b² = 1

We get a = 4 and b = 7

It is known that a² + b² = c²

So, c² =4² + 7²

c² = 16 +49=65

c = ±√65

Since the hyperbola lies on Y-axis

The coordinates of  the foci are (0, ±c)

The coordinates of the foci are (0, √65 ) and (0, – √65)

The coordinates of the vertices are (0, ±a)

The coordinates of the vertices are (0, 4) and (0, -4)

Eccentricity, e = c/a =√65 / 4

Length of lacrus rectum = 2b²/a = (2 × 7²)/4 = 98/4 = 49/2

In each exercises 7 to 15 ,find the equations of the hyperbola satisfying the given condition.

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Q7.Vertices (±2,0), foci (±3,0)

Ans. The given  vertices of the hyperbola are (±2,0) and of foci (±3,0)

The vertices (±a,0) and of foci (±c,0) shows that hyperbola lies on x -axis.

Therefore the equation of hyperbola is

x²/a² – y²/b² = 1

Where  (±a,0) = (±2,0) and (±c,0) =(±3,0) indicates that a = 2 and c = 3

It is known to us

a² + b² = c²

b = √( c² – a²) = √( 3² – 2²) = √5

Therefore the equation of parabola is

x²/2² – y²/(√5²)² = 1

x²/4- y²/5 = 1

Q8.Vertices (0,±5), foci (0,±8)

Ans. The given  vertices of the hyperbola are (0,±5) and foci (0,±8)

The vertices (0,±a) and of foci (0,±c) shows that hyperbola lies on y-axis.

Therefore the equation of hyperbola is

y²/a² – x²/b² = 1

Where  (0,±a) =(0,±5) and (0,±c) =(0,±8) indicates that a = 5 and c = 8

It is known to us

a² + b² = c²

b = √( 8² – 5²) = √( 64 – 25) = √39

Therefore the equation of parabola is

y²/5² – x²/(√39)² = 1

x²/25- y²/39 = 1

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Q9.Vertices (0,±3), foci (0,±5)

Ans. The given  vertices of the hyperbola are (0,±3) and foci (0,±5)

The vertices (0,±a) and of foci (0,±c) shows that hyperbola lies on y-axis.

Therefore the equation of hyperbola is

y²/a² – x²/b² = 1

Where  (0,±a) =(0,±3) and (0,±c) =(0,±5) indicates that a² = 9 and c² = 25

It is known to us

a² + b² = c²

b² = 5² – 3² = 16

Therefore the equation of parabola is

y²/9 – x²/16 = 1

Q10. Foci (±5,0), the transverse axis is of length 8

Ans. The  coordinates of   foci (±5,0), shows that the hyperbola is on axis

Foci (±c,0) =(±5,0) indicates c = 5

The length of transverse axis is = 8

We know the length of transverse axis is 2a

∴ 2a = 8 ⇒ a = 4⇒a² = 16

Since a ² + b² = c²

4 ² + b² = 5²⇒ b ² = 9

Therefore the equation of hyperbola is

x²/16 – y²/9 = 1

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Q11. Foci (0,±13), the conjugate axis is of length 24

Ans. The  coordinates of   foci (0,±13) shows that the hyperbola is on y axis

∴Standard equation of the hyperbola is y²/a² – x²/b²= 1

Foci (0,±c) =(0,±13) indicates c = 13

The length ofconjugate axis is = 24

We know the length of conjugate axis is 2b

∴ 2b = 24 ⇒ b = 12⇒a² = 144

Since a ² + b² = c²

a ² + 144 = 13²⇒ a ² = 169 -144 = 25

Therefore the equation of hyperbola is

y²/25 – x²/144 = 1

NCERT Solutions of Class 11 maths exercise 11.4 of chapter 11-Conic Section

Q12.Foci (± 3√5, 0), the latus rectum is of length 8.

Ans. The coordinates of Foci (± 3√5, 0) indicates that hyperbola is on x -axis

∴Standard equation of the hyperbola is x²/a² – y²/b²= 1

Foci (±c,0) =(± 3√5, 0)⇒c = 3√5

Latus rectum of hyperbola is=2b²/a

latus rectum is given of length 8

∴ 2b²/a = 8

b² = 4a

It is known to us

a ² + b² = c²

Putting b² = 4a and c = 3√5  in the above equation

a ² + 4a = (3√5)²=45

a ² + 4a -45 = 0

a ² + 9a -5a-45 = 0

a(a +9) – 5(a +9) = 0

(a +9)(a – 5) = 0

a = -9, a  = 5

Since a is + ve therefore a = 5

Putting a = 5 in 2b²/a = 8

2b²/5 = 8⇒ b² = 20

Substituting the value of a² = 25 and b² = 20 in the standard equation of hyperbola

x²/25 – y²/20= 1

Q13. Foci (± 4, 0), the latus rectum is of length 12

Ans. The coordinates of Foci (± 4, 0) indicates that hyperbola is on x -axis

∴Standard equation of the hyperbola is x²/a² – y²/b²= 1

Foci (±c,0) = (± 4, 0)⇒c = 4

Latus rectum of hyperbola is=2b²/a

latus rectum is given of length 12

∴ 2b²/a = 12

b² = 6a

It is known to us

a ² + b² = c²

Putting b² = 6a and c = 4  in the above equation

a ² + 6a = 4²=16

a ² + 6a -16 = 0

a ² + 8a -2a-16 = 0

a(a +8) – 2(a +8) = 0

(a +8)(a – 2) = 0

a = -8, a  = 2

Since a is + ve therefore a = 2

Putting a = 2 in 2b²/a = 12

2b²/2 = 12⇒ b² = 12

Substituting the value of a² = 4 and b² =12 in the standard equation of hyperbola

x²/4 – y²/12= 1

Q14.Vertices (±7, 0), e = 4/3

Ans. The coordinates of Vertices (±7, 0) indicates that hyperbola is on x -axis

∴Standard equation of the hyperbola is x²/a² – y²/b²= 1

Vertices (±a, 0) = (± 7, 0)⇒a= 4

Ecenricity of hyperbola is=c/a

Ecenricity is given to us = 4/3

∴c/a = 4/3⇒c = 4a/3 = 4×7/3 = 28/3

c² =784/9

It is known to us

a ² + b² = c²

Putting a² =49 and c = 784 /9  in the above equation

b ² +49 = 784/9

b ² = (784/9 ) -49 =(784 -441)/9 =343/9

Putting a² = 49 and b² =343/9 in the standard equation of the parabola

x²/49 – y²/(343/9)= 1

x²/49 – 9y²/343= 1 is the required equation of the parabola

Q15.Foci (0, ±√10), passing through (2, 3)

Ans. The coordinates of Foci (0, ±√10) indicates that hyperbola is on y -axis

∴Standard equation of the hyperbola is y²/a² – x²/b²= 1

Foci (0, c) = (0, ±√10) ⇒c= √10

We know a ² + b² = c²

⇒ a ² + b² = 10

b² = 10 – a²…….(i)

Since it is given to us that hyperbola is passing through (2.3), therefore putting x =2 and y = 3 in the standard equation y²/a² – x²/b²= 1

3²/a² -2²/b² = 1

9/a² – 4/b² = 1……(ii)

Substituting the value   b² = 10 – a² from equation (i) and putting it into equation (ii)

9/a² – 4/(10-a²) = 1

(90 -9a² -4a²)/a²(10-a²) = 1

90 -13a²  = a²(10-a²)

90 -13a² = 10 a² – a⁴

a⁴ -23a²+ 90 = 0

a⁴ -18a²-5a²+ 90 = 0

a²( a² – 18) – 5(a² – 18) = 0

( a² – 18)(a² – 5) = 0

a² =18, a² = 5

Since in hyperbola c > a⇒c² > a²,here c² = 10, therefore a² = 5

Putting a² = 5 in equation (i)

b² = 10 – a²⇒ 10 – 5 = 5

Putting the value of a² and b² in the standard equation of the parabola

y²/5 – x²/5= 1 is the required equation of the hyperabola

NCERT Solutions of  Science and Maths for Class 9,10,11 and 12

NCERT Solutions of class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions of class 9 science 

Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

NCERT Solutions of class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT solutions of class 10 science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

NCERT solutions of class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals