# Misc 4 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 4 Find the equation of the normal to curve x2 = 4y which passes through the point (1, 2). Curve is x2 = 4y Differentiate w.r.t. x 2x = 4ðð¦ï·®ðð¥ï·¯ ðð¦ï·®ðð¥ï·¯ = 2ð¥ï·®4ï·¯ = ð¥ï·®2ï·¯ Slope of normal = â1ï·® ðð¦ï·®ðð¥ï·¯ï·¯ = â1ï·® ð¥ï·®2ï·¯ï·¯ï·¯ = â2ï·®ð¥ï·¯ Let (h, k) be the point where normal & curve intersect â´ Slope of normal at (h, k) = â2ï·®âï·¯ Equation of normal passing through (h, k) with slope â2ï·®âï·¯ is y â y1 = m(x â x1) y â k = â2ï·®âï·¯ (x â h) Since normal passes through (1, 2), it will satisfy its equation 2 â k = â2ï·®âï·¯ (1 â h) k = 2 + 2ï·®âï·¯ (1 â h) Since (h, k) lies on curve x2 = 4y h2 = 4k k = âï·®2ï·¯ï·®4ï·¯ Using (1) and (2) 2 + 2ï·®âï·¯ (1 â h) = âï·®2ï·¯ï·®4ï·¯ 2 + 2ï·®âï·¯ â 2 = âï·®2ï·¯ï·®4ï·¯ 2ï·®âï·¯ = âï·®2ï·¯ï·®4ï·¯ âï·®3ï·¯ = 8 h = 8ï·¯ï·® 1ï·®3ï·¯ï·¯ h = 2 Putting h = 2 in (2) k = âï·®2ï·¯ï·®4ï·¯ = (2)ï·®2ï·¯ï·®4ï·¯ = 4ï·®4ï·¯ = 1 Hence, h = 2 & k = 1 Putting h = 2 & k = 1 in equation of normal ð¦âð= â2 ð¥ â âï·¯ï·®âï·¯ ð¦â1= â2 ð¥ â 2ï·¯ï·®2ï·¯ ð¦â1=â1 ð¥â2ï·¯ ð¦â1=âð¥+2 ð¥+ð¦=2+1 ð+ð=ð

Chapter 6 Class 12 Application of Derivatives

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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.