# Example 19 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 19 Find the equations of the tangent and normal to the curve 𝑥23 + 𝑦23 = 2 at (1, 1). The curve is given as 𝑥23 + 𝑦23 = 2 23 𝑥−13+23𝑦−13 𝑑𝑦𝑑𝑥 = 0 23𝑦−13 𝑑𝑦𝑑𝑥 = −23 𝑥−13 𝑑𝑦𝑑𝑥 = −23 𝑥−13 ×𝟑𝟐𝒚𝟏𝟑 𝑑𝑦𝑑𝑥 = − 𝑦𝑥13 Thus, Slope of tangent to the curve = − 𝑦𝑥13 At point (1, 1) 𝑑𝑦𝑑𝑥(1, 1) = −1 Hence, equation of the 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 the normal = −1𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = −1−1 = 1 Hence, equation of the 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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Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.