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


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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.