Example 45 - Find equation of normal to x2 = 4y which passes - Finding equation of tangent/normal when point and curve is given

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  1. Chapter 6 Class 12 Application of Derivatives
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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 = ﷐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

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