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Chapter 6 Class 12 Application of Derivatives

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Last updated at April 19, 2021 by Teachoo

Example 35 Find the shortest distance of the point (0, c) from the parabola π¦=π₯2, where 0 β€ c β€ 5. Let (β ,π) be any point on parabola π¦=π₯2 Let D be required Distance between (β , π) & (0 , π) D = β((0ββ)^2+(π βπ)^2 ) D = β((ββ)^2+(π βπ)^2 ) D = β(π^π+(π βπ)^π ) Distance between two (π₯1,π¦1) & (π₯2 , π¦2) point is π= β((π₯2βπ¦1)^2+(π₯2 βπ¦1)^2 ) Also, Since point (β , π) is on the parabola π¦=π₯2 (π , π) will satisfy the equation of parabola Putting π₯=β , π¦=π in equation π=π^π Putting value of π=β^2 D = β(β^2+(π βπ)^2 ) D = β(π+(πβπ)^π ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = π+(πβπ)^2 Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =π+(πβπ)^2 Differentiating w.r.t. k Zβ = π(π + (π β π)^2 )/ππ Zβ = 1 + 2 (c β k) Γ (c β k)β Zβ = 1 + 2 (c β k) Γ (0 β 1) Zβ = 1 β 2 (c β k) Zβ = 1 β 2c β 2k Putting Zβ = 0 1 β 2c β 2k = 0 2k = 2c β 1 k = (ππ β π)/π Now, checking sign of π^β²β² " " ππ/ππ=4πβ2π Differentiating again w.r.t k (π^2 π)/(πβ^2 ) = 4 β0 (π ^π π)/(π π^π ) = π Since π^β²β² > 0 for k = (2π β 1)/2 β΄ Z is minimum when k = (2π β 1)/2 Thus, D is Minimum at π=(ππ β π)/π Finding Minimum value of D D = β(π+(πβπ)^2 ) Putting π=(2π β 1)/2 D = β(((2π β 1)/2)+(πβ((2π β 1)/2))^2 ) D = β(((2π β 1)/2)+((2π β 2π β 1)/2)^2 ) D = β(((2π β 1)/2)+((β1)/2)^2 ) D = β(((2π β 1)/2)+1/4) D = β(πβ1/2+1/4) D = β(πβ1/4) D = β(4π β 1)/2 Hence, shortest distance is β(ππ β π)/π