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
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
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
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
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 |
