Example 45 - Chapter 6 Class 12 Application of Derivatives
Last updated at May 29, 2018 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 = 4𝑑𝑦𝑑𝑥 𝑑𝑦𝑑𝑥 = 2𝑥4 = 𝑥2 Slope of normal = −1𝑑𝑦𝑑𝑥 = −1𝑥2 = −2𝑥 Let (h, k) be the point where normal & curve intersect ∴ Slope of normal at (h, k) = −2ℎ Equation of normal passing through (h, k) with slope −2ℎ is y – y1 = m(x – x1) y − k = −2ℎ (x − h) Since normal passes through (1, 2), it will satisfy its equation 2 − k = −2ℎ (1 − h) k = 2 + 2ℎ (1 − h) Since (h, k) lies on curve x2 = 4y h2 = 4k k = ℎ24 Using (1) and (2) 2 + 2ℎ (1 − h) = ℎ24 2 + 2ℎ − 2 = ℎ24 2ℎ = ℎ24 ℎ3 = 8 h = 813 h = 2 Putting h = 2 in (2) k = ℎ24 = (2)24 = 44 = 1 Hence, h = 2 & k = 1 Putting h = 2 & k = 1 in equation of normal 𝑦−𝑘=−2𝑥 − ℎℎ 𝑦−1=−2𝑥 − 22 𝑦−1=−1𝑥−2 𝑦−1=−𝑥+2 𝑥+𝑦=2+1 𝒙+𝒚=𝟑
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Example 35 Important
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Example 38 Important
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Example 44
Example 45 You are here
Example 46 Important
Example 47 Important
Example 48
Example 49
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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.