Question 6 - Examples - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 6 Find the equations of the tangent and normal to the curve 𝑥^(2/3) + 𝑦^(2/3) = 2 at (1, 1).Given curve 𝑥^(2/3) + 𝑦^(2/3) = 2 Differentiating both sides w.r.t x 2/3 𝑥^(1 − 2/3)+2/3 𝑦^(1 − 2/3) 𝑑𝑦/𝑑𝑥 = 0 2/3 𝑥^((−1)/3)+2/3 𝑦^((−1)/3) 𝑑𝑦/𝑑𝑥 = 0 2/3 𝑦^((−1)/3) 𝑑𝑦/𝑑𝑥 = (−2)/3 𝑥^((−1)/3) 1/𝑦^(1/3) 𝑑𝑦/𝑑𝑥 = (−1)/𝑥^(1/3) 𝒅𝒚/𝒅𝒙 = − (𝒚/𝒙)^(𝟏/𝟑) Thus, Slope of tangent to the curve = − (𝑦/𝑥)^(1/3) At point (1, 1) Slope = − (𝟏/𝟏)^(𝟏/𝟑) = −1 Hence, Equation of tangent at point (1, 1) and with slope −1 is 𝑦−1=−1 (𝑥−1) 𝑦−1=−𝑥+1 𝒚+𝒙−𝟐 = 𝟎 Also, Slope of Normal = (−1)/(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡) = (−1)/(−1) = 1 Thus, Equation of normal at point (1, 1) and with slope 1 is 𝑦 − 1 = 1 (𝑥 − 1) 𝑦 − 1 = 𝑥 − 1 𝑦 =𝑥 𝒚 −𝒙=𝟎
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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