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


Misc 21 The line ๐‘ฆ=๐‘š๐‘ฅ+1 is a tangent to the curve ๐‘ฆ^2=4๐‘ฅ if the value of ๐‘š is (A) 1 (B) 2 (C) 3 (D) 1/2 Let (โ„Ž , ๐‘˜) be the point at which tangent is to be taken & Given Equation of tangent ๐‘ฆ=๐‘š๐‘ฅ+1 & Curve is ๐‘ฆ^2=4๐‘ฅ We know that Slope of tangent to the Curve is ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ 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‘ฅ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ=4 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=2/๐‘ฆ Since tangent to be taken from (โ„Ž , ๐‘˜) So, Slope of tangent at (โ„Ž , ๐‘˜) is โ”œ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅโ”ค|_((โ„Ž , ๐‘˜) )=2/๐‘˜ But Given Equation of tangent ๐‘ฆ=๐‘š๐‘ฅ+1 Hence Slope of tangent = m โ‡’ ๐‘š =2/๐‘˜ Now, Point (โ„Ž , ๐‘˜) lie on the Curve ๐‘ฆ^2=4๐‘ฅ โ‡’(โ„Ž , ๐‘˜) will Satisfies the Equation of Curve Putting ๐‘ฅ=โ„Ž , ๐‘ฆ=๐‘˜ in equation โ‡’ ๐‘˜^2=4โ„Ž โ€ฆ(1) โ€ฆ(2) Also, Point (โ„Ž , ๐‘˜) lie on the tangent โ‡’ (โ„Ž , ๐‘˜) will satisfies the Equation of tangent ๐‘ฆ=๐‘š๐‘ฅ+1 Putting ๐‘ฅ=โ„Ž , ๐‘ฆ=๐‘˜ in equation ๐‘˜=๐‘šโ„Ž+1 From (1) Putting Value of ๐‘š=2/๐‘˜ ๐‘˜=2/๐‘˜ ร—โ„Ž+1 ๐‘˜=(2โ„Ž + ๐‘˜)/๐‘˜ ๐‘˜^2=2โ„Ž+๐‘˜ ๐‘˜^2โˆ’๐‘˜=2โ„Ž Now Our Equation are ๐‘˜^2=4โ„Ž โ€ฆ(3) โ€ฆ(2) ๐‘˜^2โˆ’๐‘˜=2โ„Ž Solving (2) ๐‘˜^2=4โ„Ž ๐‘˜^2=2(2โ„Ž) Putting 2โ„Ž=๐‘˜^2โˆ’๐‘˜ from (3) ๐‘˜^2=2(๐‘˜^2โˆ’๐‘˜) ๐‘˜^2=2๐‘˜^2โˆ’2๐‘˜ 0=2๐‘˜^2โˆ’2๐‘˜โˆ’๐‘˜^2 2๐‘˜^2โˆ’๐‘˜^2โˆ’2๐‘˜=0 ๐‘˜^2โˆ’2๐‘˜=0 ๐‘˜(๐‘˜โˆ’2)=0 Hence ๐‘˜=0, 2 When ๐’Œ=๐ŸŽ From (1) ๐‘š=2/๐‘˜ ๐‘š=2/0 ๐‘š=โˆž When ๐’Œ=๐Ÿ From (1) ๐‘š=2/๐‘˜ ๐‘š=2/2 ๐‘š=1 Since ๐‘š=โˆž not in the Option So, m = 1 Hence 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 9 years. He provides courses for Maths and Science at Teachoo.