Class 11th Chapter -11 Conic Sections | NCERT Maths Solution | NCERT Solution

In This Post we are providing Chapter -11 Conic Sections |NCERT Solutions for Class 11 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Class 11 can be really helpful in the preparation of Conic Section Board exams and will provide you with in depth detail of the chapter.

We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 11 Maths Conic Section NCERT Written Solutions will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.

Table of Contents

Class 11th Chapter -11 Conic Section | NCERT MATHS SOLUTION |

Chapter -11 Conic Section EX 11.1 NCERT Solutions

In each of the following Exercises 1 to 5, find the equation of the circle with

Ex 11.1 Class 11 Maths Question 1.
centre (0, 2) and radius 2
Solution:
Here h = 0,k = 2 and r = 2
The equation of circle is,
(x-h)² + (y- k)² = r²
∴ (x – 0)² + (y – 2)² = (2)²
⇒ x² + y² + 4 – 4y = 4
⇒ x² + y² – 4y = 0

Ex 11.1 Class 11 Maths Question 2.
centre (-2,3) and radius 4
Solution:
Here h=-2,k = 3 and r = 4
The equation of circle is,
(x – h)² + (y – k)² = r²
∴(x + 2)² + (y – 3)² = (4)²
⇒ x² + 4 + 4x + y² + 9 – 6y = 16
⇒ x² + y² + 4x – 6y – 3 = 0

Ex 11.1 Class 11 Maths Question 3.
centre $\left( \frac { 1 }{ 2 } ,\quad \frac { 1 }{ 4 } \right)$

and radius $\frac { 1 }{ 12 }$
Solution:
here h = $\frac { 1 }{ 2 }$ , k = $\frac { 1 }{ 4 }$ and r = $\frac { 1 }{ 12 }$
The equation of circle is,

NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 1

Ex 11.1 Class 11 Maths Question 4.
centre (1, 1) and radius $\sqrt { 2 }$
Solution:
Here h = l, k=l and r = $\sqrt { 2 }$
The equation of circle is,
(x – h)² + (y – k)² = r²
∴ (x – 1)² + (y – 1)² = $\left( \sqrt { 2 } \right)$ ²
⇒ x² + 1 – 2x + y² +1 – 2y = 2
⇒ x² + y² – 2x – 2y = 0

Ex 11.1 Class 11 Maths Question 5.
centre (-a, -b) and radius $\sqrt { { a }^{ 2 }-{ b }^{ 2 } }$ .
Solution:
Here h=-a, k = -b and r = $\sqrt { { a }^{ 2 }-{ b }^{ 2 } }$
The equation of circle is, (x – h)² + (y – k)² = r²
∴ (x + a)² + (y + b)² = $\left( \sqrt { { a }^{ 2 }-{ b }^{ 2 } } \right)$
⇒ x² + a² + 2ax + y² + b² + 2by = a² -b²
⇒ x² + y² + 2ax + 2 by + 2b² = 0

In each of the following exercises 6 to 9, find the centre and radius of the circles.

Ex 11.1 Class 11 Maths Question 6.
(x + 5)² + (y – 3)² = 36
Solution:
The given equation of circle is,
(x + 5)² + (y – 3)² = 36
⇒ (x + 5)² + (y – 3)² = (6)²
Comparing it with (x – h)2² + (y – k)² = r², we get
h = -5, k = 3 and r = 6.
Thus the co-ordinates of the centre are (-5, 3) and radius is 6.

Ex 11.1 Class 11 Maths Question 7.
x² + y² – 4x – 8y – 45 = 0
Solution:
The given equation of circle is
x² + y² – 4x – 8y – 45 = 0
∴ (x² – 4x) + (y² – 8y) = 45
⇒ [x² – 4x + (2)²] + [y² – 8y + (4)²] = 45 + (2)² + (4)²
⇒ (x – 2)² + (y – 4)² = 45 + 4 + 16
⇒ (x – 2)² + (y – 4)² = 65
⇒ (x – 2)² + (y – 4)²= $\left( \sqrt { 65 } \right) ^{ 2 }$
Comparing it with (x – h)² + (y – k)² = r², we
have h = 2,k = 4 and r = $\sqrt { 65 }$ .
Thus co-ordinates of the centre are (2, 4) and radius is $\sqrt { 65 }$ .

Ex 11.1 Class 11 Maths Question 8.
x² + y² – 8x + 10y – 12 = 0
Solution:
The given equation of circle is,
x² + y² – 8x + 10y -12 = 0
∴ (x² – 8x) + (y² + 10y) = 12
⇒ [x² – 8x + (4)²] + [y² + 10y + (5)²] = 12 + (4)² + (5)²
⇒ (x – 4)² + (y + 5)² = 12 + 16 + 25
⇒ (x – 4)² + (y + 5)² = 53
⇒ (x – 4)² + (y + 5)² = $\left( \sqrt { 53 } \right) ^{ 2 }$
Comparing it with (x – h)² + (y – k)² = r², we have h = 4, k = -5 and r = $\sqrt { 53 }$
Thus co-ordinates of the centre are (4, -5) and radius is $\sqrt { 53 }$ .

Ex 11.1 Class 11 Maths Question 9.
2x² + 2y² – x = 0
Solution:
The given equation of circle is,
2x² + 2y² – x = 0
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 2

Ex 11.1 Class 11 Maths 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.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 1)
∴ (4 – h)² + (1 – k)² = r²
⇒ 16 + h² – 8h + 1 + k² – 2k = r²
⇒ h²+ k² – 8h – 2k + 17 = r² …. (ii)
Also, the circle passes through point (6, 5)
∴ (6 – h² + (5 – k)² = r²
⇒ 36 + h² -12h + 25 + k² – 10k = r²
⇒ h² + k² – 12h – 10kk + 61 = r² …. (iii)
From (ii) and (iii), we have h² + k² – 8h – 2k +17
= h² + k²– 12h – 10k + 61
⇒ 4h + 8k = 44 => h + 2k = ll ….(iv)
Since the centre (h, k) of the circle lies on the line 4x + y = 16
∴ 4h + k = 16 …(v)
Solving (iv) and (v), we get h = 3 and k = 4.
Putting value of h and k in (ii), we get
(3)² + (4)² – 8 x 3 – 2 x 4 + 17 = r²
∴ r² = 10
Thus required equation of circle is
(x – 3)² + (y – 4)² = 10
⇒ x² + 9 – 6x + y² +16 – 8y = 10
⇒ x² + y² – 6x – 8y +15 = 0.

Ex 11.1 Class 11 Maths 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.
Solution:
The equation of the circle is,
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (2, 3)
∴ (2 – h)² + (3 – k)² = r²
⇒ 4 + h² – 4h + 9 + k² – 6k = r²
⇒ h²+ k² – 4h – 6k + 13 = r² ….(ii)
Also, the circle passes through point (-1, 1)
∴ (-1 – h)² + (1 – k)² = r²
⇒ 1 + h² + 2h + 1 + k² – 2k = r²
⇒ h² + k² + 2h – 2k + 2 = r² ….(iii)
From (ii) and (iii), we have
h² + k² – 4h – 6k + 13 = h² + k² + 2h – 2k + 2
⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0.
∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)
Solving (iv) and (v), we get
h = $\frac { 7 }{ 2 }$ and k = $\frac { -5 }{ 2 }$
Putting these values of h and k in (ii), we get
$\left( \frac { 7 }{ 2 } \right) ^{ 2 }+\left( \frac { -5 }{ 2 } \right) ^{ 2 }-\frac { 4\times 7 }{ 2 } -6\times \frac { -5 }{ 2 } +13={ r }^{ 2 }$
⇒ $\frac { 49 }{ 4 } +\frac { 25 }{ 4 } -14+15+13$ ⇒ ${ r }^{ 2 }=\frac { 65 }{ 2 }$
Thus required equation of circle is
⇒ $\left( x-\frac { 7 }{ 2 } \right) ^{ 2 }+\left( y+\frac { 5 }{ 2 } \right) ^{ 2 }=\frac { 65 }{ 2 }$
⇒ ${ x }^{ 2 }+\frac { 49 }{ 4 } -7x+{ y }^{ 2 }+\frac { 25 }{ 4 } +5y=\frac { 65 }{ 2 }$
⇒ 4x² + 49 – 28x + 4y² + 25 + 20y = 130
⇒ 4x² + 4y² – 28x + 20y – 56 = 0
⇒ 4(x² + y² – 7x + 5y -14) = 0
⇒ x² + y² – 7x + 5y -14 = 0.

Ex 11.1 Class 11 Maths Question 12.
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
Solution:
Since the centre of the circle lies on x-axis, the co-ordinates of centre are (h, 0). Now the circle passes through the point (2, 3).
∴ Radius of circle
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 3

Ex 11.1 Class 11 Maths Question 13.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the co-ordinate axes.
Solution:
Let the circle makes intercepts a with x-axis and b with y-axis.
∴ OA = a and OB = b
So the co-ordinates of A are (a, 0) and B are (0,b)
Now, the circle passes through three points 0(0, 0), A(a, 0) and B(0, b).
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 4

Ex 11.1 Class 11 Maths Question 14.
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
Solution:
The equation of circle is
(x – h)² + (y – k)² = r² ….(i)
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2)
∴ radius of circle
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 5

Ex 11.1 Class 11 Maths Question 15.
Does the point (-2.5, 3.5) lie inside, outside or on the circle x² + y² = 25?
Solution:
The equation of given circle is x² + y² = 25
⇒ (x – 0)² + (y – 0)² = (5)²
Comparing it with (x – h)² + (y – k)² = r², we
get
h = 0,k = 0, and r = 5
Now, distance of the point (-2.5, 3.5) from the centre (0, 0)
$\sqrt { \left( 0+2.5 \right) ^{ 2 }+\left( 0-3.5 \right) ^{ 2 } } =\sqrt { 6.25+12.25 }$
$\sqrt { 18.5 }$ = 4.3 < 5.
Thus the point (-2.5, 3.5) lies inside the circle.

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.1, drop a comment below and we will get back to you at the earliest.

Chapter -11 Conic Section EX 11.2 NCERT Solutions

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.
y²= 12x
Solution:
The given equation of parabola is y² = 12x which is of the form y² = 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.
x² = 6y
Solution:
The given equation of parabola is x² = 6y which is of the form x² = 4ay.
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 1

Ex 11.2 Class 11 Maths Question 3.
y² = – 8x
Solution:
The given equation of parabola is
y² = -8x, which is of the form y² = – 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.
x² = -16y
Solution:
The given equation of parabola is
x² = -16y, which is of the form x² = -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.
y²= 10x
Solution:
The given equation of parabola is y² = 10x, which is of the form y² = 4ax.
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 2

Ex 11.2 Class 11 Maths Question 6.
x² = -9y
Solution:
The given equation of parabola is
x² = -9y, which is of the form x² = -4ay.
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 3

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 y² = 4ax.
The required equation of parabola is
y² = 4 x 6x ⇒ y² = 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 x² = -4ay.
The required equation of parabola is
x² = – 4 x 3y ⇒ x² = -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 y² = 4ax
The required equation of the parabola is
y² = 4 x 3x ⇒ y² = 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 y² = – 4ax
The required equation of the parabola is
y² = – 4 x 2x ⇒ y² = -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 y² = 4ax
Since the parabola passes through point (2, 3)
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 4

Ex 11.2 Class 11 Maths Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-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 x² = 4ay
Since the parabola passes through point (5, 2)
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 5

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.

Chapter -11 Conic Section EX 11.3 NCERT Solutions

In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.

Ex 11.3 Class 11 Maths Question 1.
$\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 16 } =1$
Solution:
Given equation of ellipse of $\frac { { x }^{ 2 } }{ 36 } +\frac { { y }^{ 2 } }{ 16 } =1$
Clearly, 36 > 16
The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 1

Ex 11.3 Class 11 Maths Question 2.
$\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 25 } =1$
Solution:
Given equation of ellipse is $\frac { { x }^{ 2 } }{ 4 } +\frac { { y }^{ 2 } }{ 25 } =1$
Clearly, 25 > 4
The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 2

Ex 11.3 Class 11 Maths Question 3.
$\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$
Solution:
Given equation of ellipse is $\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$
Clearly, 16 > 9
The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 3

Ex 11.3 Class 11 Maths Question 4.
$\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 100 } =1$
Solution:
Given equation of ellipse is $\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 100 } =1$
Clearly, 100 > 25
The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 4

Ex 11.3 Class 11 Maths Question 5.
$\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 36 } =1$
Solution:
Given equation of ellipse is $\frac { { x }^{ 2 } }{ 49 } +\frac { { y }^{ 2 } }{ 36 } =1$
Clearly, 49 > 36
The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 5

Ex 11.3 Class 11 Maths Question 6.
$\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$
Solution:
Given equation of ellipse is $\frac { { x }^{ 2 } }{ 16 } +\frac { { y }^{ 2 } }{ 9 } =1$
Clearly, 400 > 100
The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 6

Ex 11.3 Class 11 Maths Question 7.
36x² + 4y² = 144
Solution:
Given equation of ellipse is 36x² + 4y² = 144
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 7

Ex 11.3 Class 11 Maths Question 8.
16x² + y² = 16
Solution:
Given equation of ellipse is16x² + y² = 16
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 9

Ex 11.3 Class 11 Maths Question 9.
4x² + 9y² = 36
Solution:
Given equation of ellipse is4x² + 9y² = 36
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 10

In each 0f the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Ex 11.3 Class 11 Maths Question 10.
Vertices (±5, 0), foci (±4,0)
Solution:
Clearly, The foci (±4, o) lie on x-axis.
∴ The equation of ellipse is standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 12

Ex 11.3 Class 11 Maths Question 11.
Vertices (0, ±13), foci (0, ±5)
Solution:
Clearly, The foci (0, ±5) lie on y-axis.
∴ The equation of ellipse is standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 13

Ex 11.3 Class 11 Maths Question 12.
Vertices (±6, 0), foci (±4,0)
Solution:
Clearly, The foci (±4, 0) lie on x-axis.
∴ The equation of ellipse is standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 14

Ex 11.3 Class 11 Maths Question 13.
Ends of major axis (±3, 0), ends of minor axis (0, ±2)
Solution:
Since, ends of major axis (±3, 0) lie on x-axis.
∴ The equation of ellipse in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 15

Ex 11.3 Class 11 Maths Question 14.
Ends of major axis (0, $\pm \sqrt { 5 }$ ), ends of minor axis (±1, 0)
Solution:
Since, ends of major axis (0, $\pm \sqrt { 5 }$ ) lie on i-axis.
∴ The equation of ellipse in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 16

Ex 11.3 Class 11 Maths Question 15.
Length of major axis 26, foci (±5, 0)
Solution:
Since the foci (±5, 0) lie on x-axis.
∴ The equation of ellipse in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 17

Ex 11.3 Class 11 Maths Question 16.
Length of major axis 16, foci (0, ±6)
Solution:
Since the foci (0, ±6) lie on y-axis.
∴ The equation of ellipse in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 18

Ex 11.3 Class 11 Maths Question 17.
Foci (±3, 0) a = 4
Solution:
since the foci (±3, 0) on x-axis.
∴ The equation of ellipse in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 19

Ex 11.3 Class 11 Maths Question 18.
b = 3, c = 4, centre at the origin; foci on the x axis.
Solution:
Since the foci lie on x-axis.
∴ The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 20

Ex 11.3 Class 11 Maths Question 19.
Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6)
Solution:
Since the major axis is along y-axis.
∴ The equation of ellipse in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 21

Ex 11.3 Class 11 Maths Question 20.
Major axis on the x-axis and passes through the points (4, 3) and (6, 2).
Solution:
Since the major axis is along the x-axis.
∴ The equation of ellipse in standard form is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 22

We hope the NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.3 help you. If you have any query regarding NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections EX 11.3, drop a comment below and we will get back to you at the earliest.

Chapter -11 Conic Section EX 11.4 NCERT Solutions

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

Ex 11.4 Class 11 Maths Question 1.
$\frac { { x }^{ 2 } }{ 16 } -\frac { { y }^{ 2 } }{ 9 } =1$
Solution:
Given equation of hyperbola is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 1

Ex 11.4 Class 11 Maths Question 2.
$\frac { { y }^{ 2 } }{ 9 } -\frac { x^{ 2 } }{ 27 } =1$
Solution:
Given equation of hyperbola is
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 2

Ex 11.4 Class 11 Maths Question 3.
9y² – 4x² = 36
Solution:
Given equation of hyperbola is9y² – 4x² = 36
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 4

Ex 11.4 Class 11 Maths Question 4.
16x² – 9y² = 576
Solution:
Given equation of hyperbola is16x² – 9y² = 576
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 5

Ex 11.4 Class 11 Maths Question 5.
5y² – 9x² = 36
Solution:
Given equation of hyperbola is
5y² – 9x² = 36
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 6

Ex 11.4 Class 11 Maths Question 6.
49y² – 16x² = 784
Solution:
Given equation of hyperbola is49y² – 16x² = 784
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 8

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions.

Ex 11.4 Class 11 Maths Question 7.
Vertices (±2,0), foci (±3,0)
Solution:
Vertices are (±2, 0) which lie on x-axis. So the equation of hyperbola in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 9

Ex 11.4 Class 11 Maths Question 8.
Vertices (0, ±5), foci (0, ±8)
Solution:
Vertices are (0, ±5) which lie on x-axis. So the equation of hyperbola in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 10

Ex 11.4 Class 11 Maths Question 9.
Vertices (0, ±3), foci (0, ±5)
Solution:
Vertices are (0, ±3) which lie on x-axis. So the equation of hyperbola in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 11

Ex 11.4 Class 11 Maths Question 10.
Foci (±5, 0), the transverse axis is of length 8.
Solution:
Here foci are (±5, 0) which lie on x-axis. So the equation of the hyperbola in standard
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 13

Ex 11.4 Class 11 Maths Question 11.
Foci (0, ±13), the conjugate axis is of length 24.
Solution:
Here foci are (0, ±13) which lie on y-axis.
So the equation of hyperbola in standard
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 14

Ex 11.4 Class 11 Maths Question 12.
Foci ( $\pm 3\sqrt { 5 }$ ,0) , the latus rectum is of length 8.
Solution:
Here foci are ( $\pm 3\sqrt { 5 }$ , 0) which lie on x-axis.
So the equation of the hyperbola in standard
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 15

Ex 11.4 Class 11 Maths Question 13.
Foci (±4, 0), the latus rectum is of length 12.
Solution:
Here foci are (±4, 0) which lie on x-axis.
So the equation of the hyperbola in standard
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 17

Ex 11.4 Class 11 Maths Question 14.
Vertices (+7, 0), e = $\frac { 4 }{ 3 }$
Solution:
Here vertices are (±7, 0) which lie on x-axis.
So, the equation of hyperbola in standard
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 18

Ex 11.4 Class 11 Maths Question 15.
Foci (0, $\pm \sqrt { 10 }$ ), passing through (2, 3).
Solution:
Here foci are (0, $\pm \sqrt { 10 }$ ) which lie on y-axis.
So the equation of hyperbola in standard form
NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.4 19

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Class 11th Chapter -11 Conic Section | NCERT MATHS SOLUTION |

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Class 11th Chapter -11 Conic Sections | NCERT Maths Solution | NCERT Solution | Edugrown

Class 11th Chapter -11 Conic Section | NCERT MATHS SOLUTION |

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