# Misc 4 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

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 = 4๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ = 2๐ฅ/4 = ๐ฅ/2 Slope of normal = (โ1)/(๐๐ฆ/๐๐ฅ) = (โ1)/((๐ฅ/2) ) = (โ2)/๐ฅ Let (h, k) be the point where normal & curve intersect โด Slope of normal at (h, k) = (โ2)/โ Equation of normal passing through (h, k) with slope (โ2)/โ is y โ y1 = m(x โ x1) y โ k = (โ2)/โ (x โ h) Since normal passes through (1, 2), it will satisfy its equation 2 โ k = (โ2)/โ (1 โ h) k = 2 + 2/โ (1 โ h) Since (h, k) lies on curve x2 = 4y h2 = 4k k = โ^2/4 Comparing (1) and (2) 2 + 2/โ (1 โ h) = โ^2/4 2 + 2/โ โ 2 = โ^2/4 2/โ = โ^2/4 โ^3 = 8 h = ("8" )^(1/3) h = 2 โฆ(2) Putting h = 2 in (2) k = โ^2/4 = ใ(2)ใ^2/4 = 4/4 = 1 Hence, h = 2 & k = 1 Putting h = 2 & k = 1 in equation of normal ๐ฆโ๐=(โ2(๐ฅ โ โ))/โ ๐ฆโ1=(โ2(๐ฅ โ 2))/2 ๐ฆโ1=โ1(๐ฅโ2) ๐ฆโ1=โ๐ฅ+2 ๐ฅ+๐ฆ=2+1 ๐+๐=๐

Miscellaneous

Misc 1
Important

Misc 2 Important

Misc 3 Important

Misc 4 You are here

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc. 19

Misc 20 Important

Misc 21 Important

Misc 22

Misc. 23 Important

Misc 24 Important

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.