Check sibling questions

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

This video is only available for Teachoo black users

Solve all your doubts with Teachoo Black (new monthly pack available now!)


Transcript

Example 18 Find the equation of the tangent to the curve y = (π‘₯ βˆ’ 7)/((π‘₯ βˆ’ 2)(π‘₯ βˆ’ 3)) at the point where it cuts the x-axis.Slope of the tangent to the curve is 𝑑𝑦/𝑑π‘₯ = ((π‘₯ βˆ’ 7)^β€² [(π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3)]βˆ’ (π‘₯ βˆ’ 7) [(π‘₯ βˆ’ 3) (π‘₯ βˆ’ 2)]^β€²)/((π‘₯ βˆ’ 2)^2 (π‘₯ βˆ’ 3)^2 ) 𝑑𝑦/𝑑π‘₯ = (1 Γ— (π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) βˆ’ (π‘₯ βˆ’ 7)[(π‘₯ βˆ’ 3)^β€² (π‘₯ βˆ’ 2) + (π‘₯ βˆ’ 3) (π‘₯ βˆ’ 2)^β€² ])/((π‘₯ βˆ’ 2)^2 (π‘₯ βˆ’ 3)^2 ) 𝑑𝑦/𝑑π‘₯ = ((1) (π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) βˆ’ (π‘₯ βˆ’ 7)[1 Γ— (π‘₯ βˆ’ 2) + (π‘₯ βˆ’ 3) Γ— 1])/((π‘₯ βˆ’ 2)^2 (π‘₯ βˆ’ 3)^2 ) 𝑑𝑦/𝑑π‘₯ = ((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) βˆ’ (π‘₯ βˆ’ 7)(2π‘₯ βˆ’ 5))/((π‘₯ βˆ’ 2)^2 (π‘₯ βˆ’ 3)^2 ) 𝑑𝑦/𝑑π‘₯ = ((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) )/((π‘₯ βˆ’ 2)^2 (π‘₯ βˆ’ 3)^2 )βˆ’(π‘₯ βˆ’ 7)(2π‘₯ βˆ’ 5)/((π‘₯ βˆ’ 2)^2 (π‘₯ βˆ’ 3)^2 ) 𝑑𝑦/𝑑π‘₯ = (1 )/((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) )βˆ’((π‘₯ βˆ’ 7))/((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) ) Γ— ((2π‘₯ βˆ’ 5))/((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) ) 𝑑𝑦/𝑑π‘₯ = (1 )/((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) )βˆ’π‘¦ Γ— ((2π‘₯ βˆ’ 5))/((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) ) π’…π’š/𝒅𝒙 = (𝟏 βˆ’ π’š(πŸπ’™ βˆ’ πŸ“))/((𝒙 βˆ’ 𝟐) (𝒙 βˆ’ π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 = (π‘₯ βˆ’ 7)/(π‘₯ βˆ’ 2)(π‘₯ βˆ’ 3) ∴ x = 7 Thus, curve cuts the x-axis at point (7, 0) Finding Slope at point (7, 0) 𝑑𝑦/𝑑π‘₯ = (1 βˆ’ 𝑦(2π‘₯ βˆ’ 5))/((π‘₯ βˆ’ 2) (π‘₯ βˆ’ 3) ) Putting x = 7, y = 0 = (1 βˆ’ 0[2(7)βˆ’5])/((7 βˆ’ 2) (7 βˆ’ 3) ) = 1/(5 Γ— 4) = 𝟏/𝟐𝟎 Now, Equation of the tangent at point (7, 0) with slope 1/20 is 𝑦 βˆ’π‘¦_1= π‘š (π‘₯ βˆ’ π‘₯1) π‘¦βˆ’0= 1/20 (π‘₯βˆ’7) 20𝑦=π‘₯βˆ’7 πŸπŸŽπ’šβˆ’π’™+πŸ•=𝟎

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.