Example 18 - Find equation of tangent at point where it cuts

Example 18 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 18 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 18 - Chapter 6 Class 12 Application of Derivatives - Part 4

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

Transcript

Example 18 Find the equation of the tangent to the curve y = (๐‘ฅ โˆ’ 7)/((๐‘ฅ โˆ’ 2)(๐‘ฅ โˆ’ 3)) at the point where it cuts the x-axis.Slope of the tangent to the curve is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((๐‘ฅ โˆ’ 7)^โ€ฒ [(๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3)]โˆ’ (๐‘ฅ โˆ’ 7) [(๐‘ฅ โˆ’ 3) (๐‘ฅ โˆ’ 2)]^โ€ฒ)/((๐‘ฅ โˆ’ 2)^2 (๐‘ฅ โˆ’ 3)^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (1 ร— (๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) โˆ’ (๐‘ฅ โˆ’ 7)[(๐‘ฅ โˆ’ 3)^โ€ฒ (๐‘ฅ โˆ’ 2) + (๐‘ฅ โˆ’ 3) (๐‘ฅ โˆ’ 2)^โ€ฒ ])/((๐‘ฅ โˆ’ 2)^2 (๐‘ฅ โˆ’ 3)^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((1) (๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) โˆ’ (๐‘ฅ โˆ’ 7)[1 ร— (๐‘ฅ โˆ’ 2) + (๐‘ฅ โˆ’ 3) ร— 1])/((๐‘ฅ โˆ’ 2)^2 (๐‘ฅ โˆ’ 3)^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) โˆ’ (๐‘ฅ โˆ’ 7)(2๐‘ฅ โˆ’ 5))/((๐‘ฅ โˆ’ 2)^2 (๐‘ฅ โˆ’ 3)^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) )/((๐‘ฅ โˆ’ 2)^2 (๐‘ฅ โˆ’ 3)^2 )โˆ’(๐‘ฅ โˆ’ 7)(2๐‘ฅ โˆ’ 5)/((๐‘ฅ โˆ’ 2)^2 (๐‘ฅ โˆ’ 3)^2 ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (1 )/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) )โˆ’((๐‘ฅ โˆ’ 7))/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) ) ร— ((2๐‘ฅ โˆ’ 5))/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (1 )/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) )โˆ’๐‘ฆ ร— ((2๐‘ฅ โˆ’ 5))/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) ) ๐’…๐’š/๐’…๐’™ = (๐Ÿ โˆ’ ๐’š(๐Ÿ๐’™ โˆ’ ๐Ÿ“))/((๐’™ โˆ’ ๐Ÿ) (๐’™ โˆ’ ๐Ÿ‘) ) We need to find Equation of tangent at the point where the curve cuts the x axis, Thus, y = 0 We need to find value of x Putting y = 0 in equation of curve 0 = (๐‘ฅ โˆ’ 7)/(๐‘ฅ โˆ’ 2)(๐‘ฅ โˆ’ 3) โˆด x = 7 Thus, curve cuts the x-axis at point (7, 0) Finding Slope at point (7, 0) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (1 โˆ’ ๐‘ฆ(2๐‘ฅ โˆ’ 5))/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) ) Putting x = 7, y = 0 = (1 โˆ’ 0[2(7)โˆ’5])/((7 โˆ’ 2) (7 โˆ’ 3) ) = 1/(5 ร— 4) = ๐Ÿ/๐Ÿ๐ŸŽ Now, Equation of the tangent at point (7, 0) with slope 1/20 is ๐‘ฆ โˆ’๐‘ฆ_1= ๐‘š (๐‘ฅ โˆ’ ๐‘ฅ1) ๐‘ฆโˆ’0= 1/20 (๐‘ฅโˆ’7) 20๐‘ฆ=๐‘ฅโˆ’7 ๐Ÿ๐ŸŽ๐’šโˆ’๐’™+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.