Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12

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

Transcript

Misc 4 Find the equation of the normal to the curve x2 = 4y which passes through the point (1, 2). Curve is x2 = 4y Differentiate w.r.t. x 2x = 4๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 2๐‘ฅ/4 = ๐‘ฅ/2 Slope of normal = (โˆ’1)/(๐‘‘๐‘ฆ/๐‘‘๐‘ฅ) = (โˆ’1)/((๐‘ฅ/2) ) = (โˆ’2)/๐‘ฅ Let (h, k) be the point where normal & curve intersect โˆด Slope of normal at (h, k) = (โˆ’2)/โ„Ž Equation of normal passing through (h, k) with slope (โˆ’2)/โ„Ž is y โ€“ y1 = m(x โ€“ x1) y โˆ’ k = (โˆ’2)/โ„Ž (x โˆ’ h) Since normal passes through (1, 2), it will satisfy its equation 2 โˆ’ k = (โˆ’2)/โ„Ž (1 โˆ’ h) k = 2 + 2/โ„Ž (1 โˆ’ h) Since (h, k) lies on curve x2 = 4y h2 = 4k k = โ„Ž^2/4 Comparing (1) and (2) 2 + 2/โ„Ž (1 โˆ’ h) = โ„Ž^2/4 2 + 2/โ„Ž โˆ’ 2 = โ„Ž^2/4 2/โ„Ž = โ„Ž^2/4 โ„Ž^3 = 8 h = ("8" )^(1/3) h = 2 โ€ฆ(2) Putting h = 2 in (2) k = โ„Ž^2/4 = ใ€–(2)ใ€—^2/4 = 4/4 = 1 Hence, h = 2 & k = 1 Putting h = 2 & k = 1 in equation of normal ๐‘ฆโˆ’๐‘˜=(โˆ’2(๐‘ฅ โˆ’ โ„Ž))/โ„Ž ๐‘ฆโˆ’1=(โˆ’2(๐‘ฅ โˆ’ 2))/2 ๐‘ฆโˆ’1=โˆ’1(๐‘ฅโˆ’2) ๐‘ฆโˆ’1=โˆ’๐‘ฅ+2 ๐‘ฅ+๐‘ฆ=2+1 ๐’™+๐’š=๐Ÿ‘

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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.