Finding equation of tangent/normal when point and curve is given

Chapter 6 Class 12 Application of Derivatives
Concept wise

Transcript

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

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.