Example 45 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
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Chapter 6 Class 12 Application of Derivatives
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Transcript
Example 45 Find the equation of the normal to the curve x2 = 4y which passes through the point (1, 2).Given Curve x2 = 4y Differentiating w.r.t. x 2x = 4ππ¦/ππ₯ ππ¦/ππ₯ = π₯/2 β΄ Slope of normal = (β1)/(ππ¦/ππ₯) = (β1)/((π₯/2) ) = (βπ)/π Let (h, k) be the point where normal & curve intersect We need to find equation of the normal to the curve x2 = 4y which passes through the point (1, 2). But to find equationβ¦ we need to find point on curve Let (h, k) be the point where normal & curve intersect β΄ Slope of normal at (h, k) = (βπ)/π Equation of normal passing through (h, k) with slope (β2)/β is y β y1 = m(x β x1) y β k = (βπ)/π (x β h) Since normal passes through (1, 2), it will satisfy its equation 2 β k = (β2)/β (1 β h) k = 2 + π/π (1 β h) Also, (h, k) lies on curve x2 = 4y h2 = 4k k = π^π/π From (1) and (2) 2 + 2/β (1 β h) = β^2/4 2 + 2/β β 2 = β^2/4 2/β = β^2/4 β^3 = 8 h = ("8" )^(1/3) h = 2 Putting h = 2 in (2) k = β^2/4 = γ(2)γ^2/4 = 4/4 = 1 Hence, h = 2 & k = 1 Putting h = 2 & k = 1 in equation of normal π¦βπ=(β2(π₯ β β))/β π¦β1=(β2(π₯ β 2))/2 π¦β1=β1(π₯β2) π¦β1=βπ₯+2 π₯+π¦=2+1 π+π=π
