Example 19 - Find equations of tangent, normal to x2/3 + y2/3 = 2

Example 19 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 19 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 19 - Chapter 6 Class 12 Application of Derivatives - Part 4

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

Transcript

Example 19 Find the equations of the tangent and normal to the curve ๐‘ฅ^(2/3) + ๐‘ฆ^(2/3) = 2 at (1, 1).Given curve ๐‘ฅ^(2/3) + ๐‘ฆ^(2/3) = 2 Differentiating both sides w.r.t x 2/3 ๐‘ฅ^(1 โˆ’ 2/3)+2/3 ๐‘ฆ^(1 โˆ’ 2/3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 2/3 ๐‘ฅ^((โˆ’1)/3)+2/3 ๐‘ฆ^((โˆ’1)/3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 0 2/3 ๐‘ฆ^((โˆ’1)/3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’2)/3 ๐‘ฅ^((โˆ’1)/3) 1/๐‘ฆ^(1/3) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = (โˆ’1)/๐‘ฅ^(1/3) ๐’…๐’š/๐’…๐’™ = โˆ’ (๐’š/๐’™)^(๐Ÿ/๐Ÿ‘) Thus, Slope of tangent to the curve = โˆ’ (๐‘ฆ/๐‘ฅ)^(1/3) At point (1, 1) Slope = โˆ’ (๐Ÿ/๐Ÿ)^(๐Ÿ/๐Ÿ‘) = โˆ’1 Hence, Equation of tangent at point (1, 1) and with slope โˆ’1 is ๐‘ฆโˆ’1=โˆ’1 (๐‘ฅโˆ’1) ๐‘ฆโˆ’1=โˆ’๐‘ฅ+1 ๐’š+๐’™โˆ’๐Ÿ = ๐ŸŽ Also, Slope of Normal = (โˆ’1)/(๐‘†๐‘™๐‘œ๐‘๐‘’ ๐‘œ๐‘“ ๐‘ก๐‘Ž๐‘›๐‘”๐‘’๐‘›๐‘ก) = (โˆ’1)/(โˆ’1) = 1 Thus, Equation of normal at point (1, 1) and with slope 1 is ๐‘ฆ โˆ’ 1 = 1 (๐‘ฅ โˆ’ 1) ๐‘ฆ โˆ’ 1 = ๐‘ฅ โˆ’ 1 ๐‘ฆ =๐‘ฅ ๐’š โˆ’๐’™=๐ŸŽ

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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.