Example 16 - Find equation of all lines having slope 2, tangent

Example 16 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 16 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 16 - Chapter 6 Class 12 Application of Derivatives - Part 4

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

Transcript

Example 16 Find the equation of all lines having slope 2 and being tangent Equation of curve is y + 2/(๐‘ฅ โˆ’ 3) = 0 Differentiating both sides w.r.t x (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ + (๐‘‘ )/๐‘‘๐‘ฅ (2/(๐‘ฅ โˆ’ 3))=0 (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ =โˆ’(๐‘‘ )/๐‘‘๐‘ฅ (2/(๐‘ฅ โˆ’ 3)) (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ =โˆ’2 (๐‘‘ )/๐‘‘๐‘ฅ (๐‘ฅโˆ’3)^(โˆ’1) (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ =โˆ’2ใ€– ร— โˆ’(๐‘ฅโˆ’3)ใ€—^(โˆ’1โˆ’1) (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ =2(๐‘ฅโˆ’3)^(โˆ’2) (๐’…๐’š )/๐’…๐’™ =๐Ÿ/(๐’™ โˆ’ ๐Ÿ‘)^๐Ÿ Given that slope = 2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2 2/(๐‘ฅ โˆ’ 3)^2 = 2 1/(๐‘ฅ โˆ’ 3)^2 = 1 (๐‘ฅโˆ’3)^2 = 1 ๐‘ฅโˆ’3 = ยฑ1 x โˆ’ 3 = 1 x = 4 x โˆ’ 3 = โˆ’ 1 x = 2 x โˆ’ 3 = โˆ’ 1 x = 2 So, x = 4 & x = 2 Finding value of y If x = 2 y = (โˆ’2)/(๐‘ฅ โˆ’ 3) y = (โˆ’2)/(2 โˆ’ 3) ๐‘ฆ=2 Thus, point is (2, 2) If x = 4 y = (โˆ’2)/(๐‘ฅ โˆ’ 3) y = (โˆ’2)/(4 โˆ’ 3) ๐‘ฆ=โˆ’2 Thus, point is (4, โ€“2) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 2) and (4, โˆ’ 2) We know that Equation of line at (๐‘ฅ1 , ๐‘ฆ1)& having Slope m is ๐‘ฆโˆ’๐‘ฆ1=๐‘š(๐‘ฅโˆ’๐‘ฅ1) Equation of tangent through (2, 2) is ๐‘ฆ โˆ’ 2 = 2 (๐‘ฅ โˆ’2) ๐‘ฆ โˆ’ 2 = 2๐‘ฅ โˆ’ 4 ๐’šโˆ’๐Ÿ๐’™+๐Ÿ = ๐ŸŽ Equation of tangent through (4, โˆ’2) is ๐‘ฆ โˆ’(โˆ’2) = 2 (๐‘ฅ โˆ’4) ๐‘ฆ + 2 = 2๐‘ฅ โˆ’ 8 ๐’š โˆ’ ๐Ÿ๐’™ + ๐Ÿ๐ŸŽ = ๐ŸŽ to the curve y + 2/(๐‘ฅ โˆ’ 3) = 0

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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.