Ex 6.3, 20 - Find equation of normal at (am2, am3) for ay2 = x3

Advertisement

Ex 6.3,20 - Chapter 6 Class 12 Application of Derivatives - Part 2

Advertisement

Ex 6.3,20 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,20 - Chapter 6 Class 12 Application of Derivatives - Part 4

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Ex 6.3, 20 Find the equation of the normal at the point (๐‘Ž๐‘š^2,๐‘Ž๐‘š^3) for the curve ๐‘Ž๐‘ฆ^2=๐‘ฅ^3We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Given ๐‘Ž๐‘ฆ^2=๐‘ฅ^3 Differentiating w.r.t.๐‘ฅ ๐‘‘(๐‘Ž๐‘ฆ^2 )/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^3 )/๐‘‘๐‘ฅ ๐‘Ž ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^3 )/๐‘‘๐‘ฅ ๐‘Ž . ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ=3๐‘ฅ^2 ๐‘Ž .2๐‘ฆ ร—๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=3๐‘ฅ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(3๐‘ฅ^2)/2๐‘Ž๐‘ฆ Slope of tangent at (๐‘Ž๐‘š^2,๐‘Ž๐‘š^3 ) is ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_((๐‘Ž๐‘š^2,๐‘Ž๐‘š^3 ) )=(3(๐‘Ž๐‘š^2 )^2)/2๐‘Ž(๐‘Ž๐‘š^3 ) =(3๐‘Ž^2 ๐‘š^4)/(2๐‘Ž^2 ๐‘š^3 )=3/2 ๐‘š We know that Slope of tangent ร— Slope of Normal =โˆ’1 3๐‘š/2 ร— Slope of Normal =โˆ’1 Slope of Normal =(โˆ’1)/(3๐‘š/2) Slope of Normal =(โˆ’2)/3๐‘š Finding equation of normal Equation of Normal at (๐‘Ž๐‘š^2, ๐‘Ž๐‘š^3 ) & having Slope (โˆ’2)/3๐‘š is (๐‘ฆโˆ’๐‘Ž๐‘š^3 )=(โˆ’2)/3๐‘š (๐‘ฅโˆ’๐‘Ž๐‘š^2 ) 3๐‘š(๐‘ฆโˆ’๐‘Ž๐‘š^3 )=โˆ’2(๐‘ฅโˆ’๐‘Ž๐‘š^2 ) 3๐‘š๐‘ฆโˆ’3 ๐‘Ž๐‘š^4=โˆ’2๐‘ฅ+2๐‘Ž๐‘š^2 2๐‘ฅ+3๐‘š๐‘ฆโˆ’3๐‘Ž๐‘š^4โˆ’2๐‘Ž๐‘š^2=0 2๐‘ฅ+3๐‘š๐‘ฆโˆ’๐‘Ž๐‘š^2 (3๐‘š^2+2)=0 We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1)& having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Required Equation of Normal is : ๐Ÿ๐’™+๐Ÿ‘๐’Ž๐’šโˆ’๐’‚๐’Ž^๐Ÿ (๐Ÿ‘๐’Ž^๐Ÿ+๐Ÿ)=๐ŸŽ

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.