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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

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 𝑦2=4𝑥 Differentiating w.r.t. 𝑥 𝑑 𝑦2𝑑𝑥= 𝑑 4𝑥𝑑𝑥 𝑑 𝑦2𝑑𝑦 × 𝑑𝑦𝑑𝑥=4 2𝑦 × 𝑑𝑦𝑑𝑥=4 𝑑𝑦𝑑𝑥= 42𝑦 𝑑𝑦𝑑𝑥= 2𝑦 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 𝑑𝑦𝑑𝑥=1 2𝑦=1 2=𝑦 𝑦=2 Finding x when 𝑦=2 𝑦2=4𝑥 22=4𝑥 4=4𝑥 44=𝑥 𝑥=1 Hence the Required point is (x, y) = 1 , 2 Correct Answer is (A)

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Ex 6.3,27 You are here

Chapter 6 Class 12 Application of Derivatives

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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.