Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

Slide14.JPG

Slide15.JPG
Slide16.JPG Slide17.JPG

  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‘ฆ/๐‘‘๐‘ฅ = (1 โˆ’ ๐‘ฆ(2๐‘ฅ โˆ’ 5))/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) ) We need to find equation of tangent at the point where the curve cuts the x axis, y = 0 Hence the equation of the curve is 0 = (๐‘ฅ โˆ’ 7)/(๐‘ฅ โˆ’ 2)(๐‘ฅ โˆ’ 3) โˆด x = 7 Thus, curve cuts the x โ€“ axis at point (7, 0) At point (7, 0) the slope is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (1 โˆ’ ๐‘ฆ(2๐‘ฅ โˆ’ 5))/((๐‘ฅ โˆ’ 2) (๐‘ฅ โˆ’ 3) ) [๐‘‘๐‘ฆ/๐‘‘๐‘ฅ]_((7, 0) ) = (1 โˆ’ 0[2(7)โˆ’5])/((7 โˆ’ 2) (7 โˆ’ 3) ) = 1/(5 ร— 4) = 1/20 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 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, Computer Engineer, Marketer
Davneet Singh
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.