Check sibling questions

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

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Example 19 Find the equations of the tangent and normal to the curve π‘₯^(2/3) + 𝑦^(2/3) = 2 at (1, 1).Given curve π‘₯^(2/3) + 𝑦^(2/3) = 2 Differentiating both sides w.r.t x 2/3 π‘₯^(1 βˆ’ 2/3)+2/3 𝑦^(1 βˆ’ 2/3) 𝑑𝑦/𝑑π‘₯ = 0 2/3 π‘₯^((βˆ’1)/3)+2/3 𝑦^((βˆ’1)/3) 𝑑𝑦/𝑑π‘₯ = 0 2/3 𝑦^((βˆ’1)/3) 𝑑𝑦/𝑑π‘₯ = (βˆ’2)/3 π‘₯^((βˆ’1)/3) 1/𝑦^(1/3) 𝑑𝑦/𝑑π‘₯ = (βˆ’1)/π‘₯^(1/3) π’…π’š/𝒅𝒙 = βˆ’ (π’š/𝒙)^(𝟏/πŸ‘) Thus, Slope of tangent to the curve = βˆ’ (𝑦/π‘₯)^(1/3) At point (1, 1) Slope = βˆ’ (𝟏/𝟏)^(𝟏/πŸ‘) = βˆ’1 Hence, Equation of tangent at point (1, 1) and with slope βˆ’1 is π‘¦βˆ’1=βˆ’1 (π‘₯βˆ’1) π‘¦βˆ’1=βˆ’π‘₯+1 π’š+π’™βˆ’πŸ = 𝟎 Also, Slope of Normal = (βˆ’1)/(π‘†π‘™π‘œπ‘π‘’ π‘œπ‘“ π‘‘π‘Žπ‘›π‘”π‘’π‘›π‘‘) = (βˆ’1)/(βˆ’1) = 1 Thus, Equation of normal at point (1, 1) and with slope 1 is 𝑦 βˆ’ 1 = 1 (π‘₯ βˆ’ 1) 𝑦 βˆ’ 1 = π‘₯ βˆ’ 1 𝑦 =π‘₯ π’š βˆ’π’™=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.