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

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.