NCERT Solutions for Class 11-science Maths Chapter 11 - Conic Sections

Chapter 11 - Conic Sections Ex. 11.1

Solution 1
Solution 2
Solution 3

Solution 4
Solution 5
Solution 6

Solution 7
Solution 8
Solution 9
Solution 10

Solution 11

Solution 12
Solution 13

Solution 14
Solution 15

Chapter 11 - Conic Sections Ex. 11.2

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12

Chapter 11 - Conic Sections Ex. 11.3

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7

Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20

Chapter 11 - Conic Sections Ex. 11.4

Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9

Vertices (0, ± 3), foci (0, ± 5)

Here, the vertices are on the y-axis.

begin mathsize 12px style therefore the space equation space of space the space hyperbola space is space of space the space form space straight y squared over straight a squared minus straight x squared over straight b squared equals 1. end style

Since the Vertices are (0, ± 3), a = 3

Since the foci are (0, ± 5), c = 5

We know that a+ b2 = c2

∴ 32 + b2 = 52

∴ b2 = 25 - 9 = 16

begin mathsize 12px style therefore the space equation space of space the space hyperbola space is space of space the space form space straight y squared over 9 minus straight x squared over 16 equals 1. end style

Solution 10

Solution 11
Solution 12
Solution 13
Solution 14
Solution 15

Chapter 11 - Conic Sections Misc. Ex.

Solution 1
Solution 2

Solution 3

Solution 4

Solution 5
Solution 6

Solution 7

Solution 8