Ex 6.3,27 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Ex 6.3, 27 - Line y = x + 1 is a tangent to y2 = 4x at - Ex 6.3

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Transcript

Ex 6.3,27 The line 𝑦=𝑥+1 is a tangent to the curve 𝑦2=4𝑥 at the point (A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2) Given Curve is 𝑦﷮2﷯=4𝑥 Differentiating w.r.t. 𝑥 𝑑 𝑦﷮2﷯﷯﷮𝑑𝑥﷯= 𝑑 4𝑥﷯﷮𝑑𝑥﷯ 𝑑 𝑦﷮2﷯﷯﷮𝑑𝑦﷯ × 𝑑𝑦﷮𝑑𝑥﷯=4 2𝑦 × 𝑑𝑦﷮𝑑𝑥﷯=4 𝑑𝑦﷮𝑑𝑥﷯= 4﷮2𝑦﷯ 𝑑𝑦﷮𝑑𝑥﷯= 2﷮𝑦﷯ Given line is 𝑦=𝑥+1 The Above line is of the form 𝑦=𝑚𝑥+𝑐 when m is slope of line Slope of line 𝑦=𝑥+1 is 1 Now Slope of tangent = Slope of line 𝑑𝑦﷮𝑑𝑥﷯=1 2﷮𝑦﷯=1 2=𝑦 𝑦=2 Finding x when 𝑦=2 𝑦﷮2﷯=4𝑥 2﷯﷮2﷯=4𝑥 4=4𝑥 4﷮4﷯=𝑥 𝑥=1 Hence the Required point is (x, y) = 1 , 2﷯ Correct Answer is (A)

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.