Last updated at April 19, 2021 by Teachoo

Transcript

Example 35 Find the shortest distance of the point (0, c) from the parabola ๐ฆ=๐ฅ2, where 0 โค c โค 5. Let (โ ,๐) be any point on parabola ๐ฆ=๐ฅ2 Let D be required Distance between (โ , ๐) & (0 , ๐) D = โ((0โโ)^2+(๐ โ๐)^2 ) D = โ((โโ)^2+(๐ โ๐)^2 ) D = โ(๐^๐+(๐ โ๐)^๐ ) Distance between two (๐ฅ1,๐ฆ1) & (๐ฅ2 , ๐ฆ2) point is ๐= โ((๐ฅ2โ๐ฆ1)^2+(๐ฅ2 โ๐ฆ1)^2 ) Also, Since point (โ , ๐) is on the parabola ๐ฆ=๐ฅ2 (๐ , ๐) will satisfy the equation of parabola Putting ๐ฅ=โ , ๐ฆ=๐ in equation ๐=๐^๐ Putting value of ๐=โ^2 D = โ(โ^2+(๐ โ๐)^2 ) D = โ(๐+(๐โ๐)^๐ ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = ๐+(๐โ๐)^2 Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =๐+(๐โ๐)^2 Differentiating w.r.t. k Zโ = ๐(๐ + (๐ โ ๐)^2 )/๐๐ Zโ = 1 + 2 (c โ k) ร (c โ k)โ Zโ = 1 + 2 (c โ k) ร (0 โ 1) Zโ = 1 โ 2 (c โ k) Zโ = 1 โ 2c โ 2k Putting Zโ = 0 1 โ 2c โ 2k = 0 2k = 2c โ 1 k = (๐๐ โ ๐)/๐ Now, checking sign of ๐^โฒโฒ " " ๐๐/๐๐=4๐โ2๐ Differentiating again w.r.t k (๐^2 ๐)/(๐โ^2 ) = 4 โ0 (๐ ^๐ ๐)/(๐ ๐^๐ ) = ๐ Since ๐^โฒโฒ > 0 for k = (2๐ โ 1)/2 โด Z is minimum when k = (2๐ โ 1)/2 Thus, D is Minimum at ๐=(๐๐ โ ๐)/๐ Finding Minimum value of D D = โ(๐+(๐โ๐)^2 ) Putting ๐=(2๐ โ 1)/2 D = โ(((2๐ โ 1)/2)+(๐โ((2๐ โ 1)/2))^2 ) D = โ(((2๐ โ 1)/2)+((2๐ โ 2๐ โ 1)/2)^2 ) D = โ(((2๐ โ 1)/2)+((โ1)/2)^2 ) D = โ(((2๐ โ 1)/2)+1/4) D = โ(๐โ1/2+1/4) D = โ(๐โ1/4) D = โ(4๐ โ 1)/2 Hence, shortest distance is โ(๐๐ โ ๐)/๐

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Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.