1. Chapter 6 Class 12 Application of Derivatives
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Question 27 (MCQ) - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

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

    3. Tangents and Normals (using Differentiation)

    Approximations (using Differentiation) →
Approximations (using Differentiation) →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo