# Example 45 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

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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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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.