Misc 4 - Chapter 6 Class 12 Application of Derivatives

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Misc 4 Find the equation of the normal to 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﷯ Using (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 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