Example 19 - Find equations of tangent, normal to x2/3 + y2/3 = 2 - Examples

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  1. Chapter 6 Class 12 Application of Derivatives
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Example 19 Find the equations of the tangent and normal to the curve ﷐𝑥﷮﷐2﷮3﷯﷯ + ﷐𝑦﷮﷐2﷮3﷯﷯ = 2 at (1, 1). The curve is given as ﷐𝑥﷮﷐2﷮3﷯﷯ + ﷐𝑦﷮﷐2﷮3﷯﷯ = 2 ﷐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﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = ﷐−2﷮3﷯ ﷐𝑥﷮﷐−1﷮3﷯﷯ ×﷐𝟑﷮𝟐﷯﷐𝒚﷮﷐𝟏﷮𝟑﷯﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯ = − ﷐﷐﷐𝑦﷮𝑥﷯﷯﷮﷐1﷮3﷯﷯ Thus, Slope of tangent to the curve = − ﷐﷐﷐𝑦﷮𝑥﷯﷯﷮﷐1﷮3﷯﷯ At point (1, 1) ﷐﷐𝑑𝑦﷮𝑑𝑥﷯﷯(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 = ﷐−1﷮𝑠𝑙𝑜𝑝𝑒 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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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.