Ex 6.3,24 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 6.3,24 Find the equations of the tangent and normal to the hyperbola 2 2 2 2 = 1 at the point ( 0 , 0) We know that Slope of tangent is 2 2 2 2 =1 2 2 =1 2 2 2 2 = 2 2 1 Differentiating w.r.t. 2 2 = 2 1 1 2 2 = 2 1 1 2 2 = 1 2 2 0 1 2 2 = 1 2 . 2 1 2 2 = 1 2 2 = 2 2 2 2 = 2 2 2 2 = 2 2 Slope of tangent at 0 , 0 is 0 , 0 = 2 0 2 0 We know that Slope of tangent Slope of Normal = 1 2 0 2 0 Slope of Normal = 1 Slope of Normal = 1 2 0 2 0 Slope of Normal = 2 0 2 0 Finding equation of tangent & normal
Ex 6.3
Ex 6.3,1
Ex 6.3,2
Ex 6.3,3
Ex 6.3,4
Ex 6.3,5 Important
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9
Ex 6.3,10
Ex 6.3,11
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18
Ex 6.3,19
Ex 6.3,20
Ex 6.3,21
Ex 6.3,22
Ex 6.3,23
Ex 6.3,24 You are here
Ex 6.3,25
Ex 6.3,26 Important
Ex 6.3,27
Chapter 6 Class 12 Application of Derivatives
Serial order wise
About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.