1. Chapter 6 Class 12 Application of Derivatives
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  3. Tangents and Normals (using Differentiation)
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Question 16 - 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 16 Show that the tangents to the curve ๐‘ฆ=7๐‘ฅ3+11 at the points where ๐‘ฅ=2 and ๐‘ฅ =โˆ’2 are parallel.We know that 2 lines are parallel y Slope of 1st line = Slope of 2nd line ๐‘š1=๐‘š2 We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Given Curve is ๐‘ฆ=7๐‘ฅ^3+11 Differentiating w.r.t.๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(7๐‘ฅ3 + 11)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=21๐‘ฅ^2 We need to show that tangent at ๐‘ฅ=2 & tangent at ๐‘ฅ=โˆ’2 are parallel i.e. we need to show (Slope of tangent at ๐‘ฅ=2) = (Slope of tangent at ๐‘ฅ=โˆ’2) Now, Slope of tangent = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=21๐‘ฅ^2 Slope of tangent at ๐‘ฅ=2 ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_(๐‘ฅ = 2)=21(2)^2=21 ร—4=84 & Slope of tangent at ๐‘ฅ=โˆ’2 ใ€–๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”‚ใ€—_(๐‘ฅ =โˆ’ 2)=21(โˆ’2)^2=21 ร—4=84 Since, (Slope of tangent at ๐‘ฅ=2) = (Slope of tangent at ๐‘ฅ=โˆ’2) Thus, tangent at ๐‘ฅ=2 & tangent at ๐‘ฅ are parallel Hence Proved
  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

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