Ex 6.3, 12 - Find equations of all lines having slope 0, tangent

Ex 6.3,12 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,12 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.3,12 - Chapter 6 Class 12 Application of Derivatives - Part 4

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

Transcript

Ex 6.3, 12 Find the equations of all lines having slope 0 which are tangent to the curve ๐‘ฆ = 1/(๐‘ฅ2 โˆ’2๐‘ฅ + 3) Given Curve is ๐‘ฆ = 1/(๐‘ฅ2 โˆ’2๐‘ฅ + 3) Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(1/(๐‘ฅ2 โˆ’ 2๐‘ฅ + 3))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘‘(๐‘ฅ2 โˆ’ 2๐‘ฅ + 3)^(โˆ’1))/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=โˆ’1(๐‘ฅ2โˆ’2๐‘ฅ+3)^(โˆ’2) . ๐‘‘(๐‘ฅ^2โˆ’ 2๐‘ฅ + 3)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=โˆ’(๐‘ฅ2โˆ’2๐‘ฅ+3)^(โˆ’2) (2๐‘ฅโˆ’2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=โˆ’2(๐‘ฅ2โˆ’2๐‘ฅ+3)^(โˆ’2) (๐‘ฅโˆ’1) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(โˆ’2(๐‘ฅ โˆ’ 1))/(๐‘ฅ2 โˆ’ 2๐‘ฅ + 3)^2 Hence Slope of tangent is (โˆ’2(๐‘ฅ โˆ’ 1))/(๐‘ฅ2 โˆ’ 2๐‘ฅ + 3)^2 Given Slope of tangent is 0 โ‡’ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=0 (โˆ’2(๐‘ฅ โˆ’ 1))/(๐‘ฅ2 โˆ’ 2๐‘ฅ + 3)^2 =0 โˆ’2(๐‘ฅ โˆ’ 1)=0 ร—(๐‘ฅ2 โˆ’ 2๐‘ฅ + 3)^2 โˆ’2(๐‘ฅ โˆ’ 1)=0 (๐‘ฅโˆ’1)=0 ๐‘ฅ=1 Finding y when ๐‘ฅ=1 ๐‘ฆ=1/(๐‘ฅ2 โˆ’ 2๐‘ฅ + 3) ๐‘ฆ=1/((1)^2 โˆ’ 2(1) + 3) ๐‘ฆ=1/(1 โˆ’ 2 + 3) ๐‘ฆ=1/(2 ) Point is (๐Ÿ , ๐Ÿ/๐Ÿ) Thus , tangent passes through (1 , 1/2) Equation of tangent at (1 , 1/2) & having Slope zero is (๐‘ฆ โˆ’1/2)=0(๐‘ฅโˆ’1) ๐‘ฆ โˆ’1/2=0 ๐‘ฆ=1/2 Hence Equation of tangent is ๐’š=๐Ÿ/๐Ÿ We know that Equation of time passing through (๐‘ฅ , ๐‘ฆ) & having Slope m is (๐‘ฆ โˆ’๐‘ฆ1)=๐‘š(๐‘ฅ โˆ’๐‘ฅ2)

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.