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Ex 6.3,20 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2018 by Teachoo
Transcript
Ex 6.3,20 Find the equation of the normal at the point ( 2 , 3 ) for the curve 2 = 3 We know that Slope of tangent is Given 2 = 3 Differentiating w.r.t. 2 = 3 2 = 3 . 2 =3 2 .2 =3 2 = 3 2 2 Slope of tangent at 2 , 3 is 2 , 3 = 3 2 2 2 3 = 3 2 4 2 2 3 = 3 2 We know that Slope of tangent Slope of Normal = 1 3 2 Slope of Normal = 1 Slope of Normal = 1 3 2 Slope of Normal = 2 3 Finding equation of normal Equation of Normal at 2 , 3 & having Slope 2 3 is 3 = 2 3 2 3 3 = 2 2 3 3 4 = 2 +2 2 2 +3 3 4 2 2 =0 2 +3 2 3 2 +2 =0 Required Equation of Normal is : + + =
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Chapter 6 Class 12 Application of Derivatives
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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.