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 ๐‘ฆ=๐‘ฅ+1 is a tangent to the curve ๐‘ฆ2=4๐‘ฅ at the point (A) (1, 2) (B) (2, 1) (C) (1, โ€“ 2) (D) (โ€“ 1, 2)Given Curve is ๐‘ฆ^2=4๐‘ฅ Differentiating w.r.t. ๐‘ฅ ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฅ=๐‘‘(4๐‘ฅ)/๐‘‘๐‘ฅ ๐‘‘(๐‘ฆ^2 )/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4 2๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4/2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2/๐‘ฆ Given line is ๐‘ฆ=๐‘ฅ+1 The Above line is of the form ๐‘ฆ=๐‘š๐‘ฅ+๐‘ when m is slope of line Slope of line ๐‘ฆ=๐‘ฅ+1 is 1 Now Slope of tangent = Slope of line ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=1 2/๐‘ฆ=1 2=๐‘ฆ ๐‘ฆ=2 Finding x when ๐‘ฆ=2 ๐‘ฆ^2=4๐‘ฅ (2)^2=4๐‘ฅ 4=4๐‘ฅ 4/4=๐‘ฅ ๐‘ฅ=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.