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

Ex 6.3,27 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.3,27 - Chapter 6 Class 12 Application of Derivatives - Part 3


Transcript

Ex 6.3, 27 The line 𝑦=π‘₯+1 is a tangent to the curve 𝑦2=4π‘₯ at the point (A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (– 1, 2)Given Curve is 𝑦^2=4π‘₯ Differentiating w.r.t. π‘₯ 𝑑(𝑦^2 )/𝑑π‘₯=𝑑(4π‘₯)/𝑑π‘₯ 𝑑(𝑦^2 )/𝑑𝑦 Γ— 𝑑𝑦/𝑑π‘₯=4 2𝑦 Γ— 𝑑𝑦/𝑑π‘₯=4 𝑑𝑦/𝑑π‘₯=4/2𝑦 𝑑𝑦/𝑑π‘₯=2/𝑦 Given line is 𝑦=π‘₯+1 The Above line is of the form 𝑦=π‘šπ‘₯+𝑐 when m is slope of line Slope of line 𝑦=π‘₯+1 is 1 Now Slope of tangent = Slope of line 𝑑𝑦/𝑑π‘₯=1 2/𝑦=1 2=𝑦 𝑦=2 Finding x when 𝑦=2 𝑦^2=4π‘₯ (2)^2=4π‘₯ 4=4π‘₯ 4/4=π‘₯ π‘₯=1 Hence the Required point is (x, y) = (1 , 2) Correct Answer is (A)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.