1. Chapter 6 Class 12 Application of Derivatives
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  3. Tangents and Normals (using Differentiation)
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Question 9 - Tangents and Normals (using Differentiation) - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

Transcript

Question 9 Find the point on the curve ๐‘ฆ=๐‘ฅ^3โˆ’11๐‘ฅ+5 at which the tangent is ๐‘ฆ=๐‘ฅ โˆ’11.Equation of Curve is ๐‘ฆ=๐‘ฅ^3โˆ’11๐‘ฅ+5 We know that Slope of tangent is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^3 โˆ’ 11๐‘ฅ + 5)/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=ใ€–3๐‘ฅใ€—^2โˆ’11 Also, Given tangent is ๐‘ฆ=๐‘ฅโˆ’12 Comparing with ๐‘ฆ=๐‘š๐‘ฅ+๐‘ , when m is the Slope Slope of tangent =1 From (1) and (2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=1 3๐‘ฅ^2โˆ’11=1 3๐‘ฅ^2=1+11 3๐‘ฅ^2=12 ๐‘ฅ^2=12/3 ๐‘ฅ^2=4 ๐‘ฅ=ยฑ2 When ๐’™=๐Ÿ ๐‘ฆ=(2)^3โˆ’11(2)+5 ๐‘ฆ=8โˆ’22+5 ๐‘ฆ=โˆ’ 9 So, Point is (2 , โˆ’9) When ๐’™=โˆ’๐Ÿ ๐‘ฆ=(โˆ’2)^3โˆ’11(โˆ’2)+5 ๐‘ฆ=โˆ’ 8+22+5 ๐‘ฆ=19 So, Point is (โˆ’2 , 19) Hence , Points (2 , โˆ’9) & (โˆ’2 , 19) But (โ€“2, 19) does not satisfy line y = x โ€“ 11 As 19 โ‰  โ€“2 โ€“ 11 โˆด Only point is (2, โ€“9)
  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