Question 5 - Examples - Chapter 6 Class 12 Application of Derivatives

Last updated at December 16, 2024 by

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

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Question 5 Find the equation of the tangent to the curve y = (š‘„ āˆ’ 7)/((š‘„ āˆ’ 2)(š‘„ āˆ’ 3)) at the point where it cuts the x-axis.Slope of the tangent to the curve is š‘‘š‘¦/š‘‘š‘„ = ((š‘„ āˆ’ 7)^ā€² [(š‘„ āˆ’ 2) (š‘„ āˆ’ 3)]āˆ’ (š‘„ āˆ’ 7) [(š‘„ āˆ’ 3) (š‘„ āˆ’ 2)]^ā€²)/((š‘„ āˆ’ 2)^2 (š‘„ āˆ’ 3)^2 ) š‘‘š‘¦/š‘‘š‘„ = (1 Ɨ (š‘„ āˆ’ 2) (š‘„ āˆ’ 3) āˆ’ (š‘„ āˆ’ 7)[(š‘„ āˆ’ 3)^ā€² (š‘„ āˆ’ 2) + (š‘„ āˆ’ 3) (š‘„ āˆ’ 2)^ā€² ])/((š‘„ āˆ’ 2)^2 (š‘„ āˆ’ 3)^2 ) š‘‘š‘¦/š‘‘š‘„ = ((1) (š‘„ āˆ’ 2) (š‘„ āˆ’ 3) āˆ’ (š‘„ āˆ’ 7)[1 Ɨ (š‘„ āˆ’ 2) + (š‘„ āˆ’ 3) Ɨ 1])/((š‘„ āˆ’ 2)^2 (š‘„ āˆ’ 3)^2 ) š‘‘š‘¦/š‘‘š‘„ = ((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) āˆ’ (š‘„ āˆ’ 7)(2š‘„ āˆ’ 5))/((š‘„ āˆ’ 2)^2 (š‘„ āˆ’ 3)^2 ) š‘‘š‘¦/š‘‘š‘„ = ((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) )/((š‘„ āˆ’ 2)^2 (š‘„ āˆ’ 3)^2 )āˆ’(š‘„ āˆ’ 7)(2š‘„ āˆ’ 5)/((š‘„ āˆ’ 2)^2 (š‘„ āˆ’ 3)^2 ) š‘‘š‘¦/š‘‘š‘„ = (1 )/((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) )āˆ’((š‘„ āˆ’ 7))/((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) ) Ɨ ((2š‘„ āˆ’ 5))/((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) ) š‘‘š‘¦/š‘‘š‘„ = (1 )/((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) )āˆ’š‘¦ Ɨ ((2š‘„ āˆ’ 5))/((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) ) š’…š’š/š’…š’™ = (šŸ āˆ’ š’š(šŸš’™ āˆ’ šŸ“))/((š’™ āˆ’ šŸ) (š’™ āˆ’ š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 = (š‘„ āˆ’ 7)/(š‘„ āˆ’ 2)(š‘„ āˆ’ 3) āˆ“ x = 7 Thus, curve cuts the x-axis at point (7, 0) Finding Slope at point (7, 0) š‘‘š‘¦/š‘‘š‘„ = (1 āˆ’ š‘¦(2š‘„ āˆ’ 5))/((š‘„ āˆ’ 2) (š‘„ āˆ’ 3) ) Putting x = 7, y = 0 = (1 āˆ’ 0[2(7)āˆ’5])/((7 āˆ’ 2) (7 āˆ’ 3) ) = 1/(5 Ɨ 4) = šŸ/šŸšŸŽ Now, Equation of the tangent at point (7, 0) with slope 1/20 is š‘¦ āˆ’š‘¦_1= š‘š (š‘„ āˆ’ š‘„1) š‘¦āˆ’0= 1/20 (š‘„āˆ’7) 20š‘¦=š‘„āˆ’7 šŸšŸŽš’šāˆ’š’™+šŸ•=šŸŽ

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo