Misc 4 - Chapter 6 Class 12 Application of Derivatives

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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 = 4𝑑𝑦/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 2𝑥/4 = 𝑥/2 Slope of normal = (−1)/(𝑑𝑦/𝑑𝑥) = (−1)/((𝑥/2) ) = (−2)/𝑥 Let (h, k) be the point where normal & curve intersect ∴ Slope of normal at (h, k) = (−2)/ℎ Equation of normal passing through (h, k) with slope (−2)/ℎ is y – y1 = m(x – x1) y − k = (−2)/ℎ (x − h) Since normal passes through (1, 2), it will satisfy its equation 2 − k = (−2)/ℎ (1 − h) k = 2 + 2/ℎ (1 − h) Since (h, k) lies on curve x2 = 4y h2 = 4k k = ℎ^2/4 Comparing (1) and (2) 2 + 2/ℎ (1 − h) = ℎ^2/4 2 + 2/ℎ − 2 = ℎ^2/4 2/ℎ = ℎ^2/4 ℎ^3 = 8 h = ("8" )^(1/3) h = 2 …(2) Putting h = 2 in (2) k = ℎ^2/4 = 〖(2)〗^2/4 = 4/4 = 1 Hence, h = 2 & k = 1 Putting h = 2 & k = 1 in equation of normal 𝑦−𝑘=(−2(𝑥 − ℎ))/ℎ 𝑦−1=(−2(𝑥 − 2))/2 𝑦−1=−1(𝑥−2) 𝑦−1=−𝑥+2 𝑥+𝑦=2+1 𝒙+𝒚=𝟑