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

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.