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Example 45 - Chapter 6 Class 12 Application of Derivatives
Last updated at Jan. 7, 2020 by Teachoo
Transcript
Example 45 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 Same as Misc 4 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
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Chapter 6 Class 12 Application of Derivatives
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About the Author
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.