Example 35 - Find shortest distance of (0, c) from parabola

Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 3 Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 4 Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 5 Example 35 - Chapter 6 Class 12 Application of Derivatives - Part 6

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Example 35 Find the shortest distance of the point (0, c) from the parabola ๐‘ฆ=๐‘ฅ2, where 0 โ‰ค c โ‰ค 5. Let (โ„Ž ,๐‘˜) be any point on parabola ๐‘ฆ=๐‘ฅ2 Let D be required Distance between (โ„Ž , ๐‘˜) & (0 , ๐‘) D = โˆš((0โˆ’โ„Ž)^2+(๐‘ โˆ’๐‘˜)^2 ) D = โˆš((โˆ’โ„Ž)^2+(๐‘ โˆ’๐‘˜)^2 ) D = โˆš(๐’‰^๐Ÿ+(๐’„ โˆ’๐’Œ)^๐Ÿ ) Distance between two (๐‘ฅ1,๐‘ฆ1) & (๐‘ฅ2 , ๐‘ฆ2) point is ๐‘‘= โˆš((๐‘ฅ2โˆ’๐‘ฆ1)^2+(๐‘ฅ2 โˆ’๐‘ฆ1)^2 ) Also, Since point (โ„Ž , ๐‘˜) is on the parabola ๐‘ฆ=๐‘ฅ2 (๐’‰ , ๐’Œ) will satisfy the equation of parabola Putting ๐‘ฅ=โ„Ž , ๐‘ฆ=๐‘˜ in equation ๐’Œ=๐’‰^๐Ÿ Putting value of ๐‘˜=โ„Ž^2 D = โˆš(โ„Ž^2+(๐‘ โˆ’๐‘˜)^2 ) D = โˆš(๐’Œ+(๐’„โˆ’๐’Œ)^๐Ÿ ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = ๐‘˜+(๐‘โˆ’๐‘˜)^2 Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =๐‘˜+(๐‘โˆ’๐‘˜)^2 Differentiating w.r.t. k Zโ€™ = ๐‘‘(๐‘˜ + (๐‘ โˆ’ ๐‘˜)^2 )/๐‘‘๐‘˜ Zโ€™ = 1 + 2 (c โˆ’ k) ร— (c โˆ’ k)โ€™ Zโ€™ = 1 + 2 (c โˆ’ k) ร— (0 โˆ’ 1) Zโ€™ = 1 โˆ’ 2 (c โˆ’ k) Zโ€™ = 1 โˆ’ 2c โˆ’ 2k Putting Zโ€™ = 0 1 โˆ’ 2c โˆ’ 2k = 0 2k = 2c โˆ’ 1 k = (๐Ÿ๐’„ โˆ’ ๐Ÿ)/๐Ÿ Now, checking sign of ๐’^โ€ฒโ€ฒ " " ๐‘‘๐‘/๐‘‘๐‘˜=4๐‘˜โˆ’2๐‘ Differentiating again w.r.t k (๐‘‘^2 ๐‘)/(๐‘‘โ„Ž^2 ) = 4 โˆ’0 (๐’…^๐Ÿ ๐’)/(๐’…๐’‰^๐Ÿ ) = ๐Ÿ’ Since ๐™^โ€ฒโ€ฒ > 0 for k = (2๐‘ โˆ’ 1)/2 โˆด Z is minimum when k = (2๐‘ โˆ’ 1)/2 Thus, D is Minimum at ๐’Œ=(๐Ÿ๐’„ โˆ’ ๐Ÿ)/๐Ÿ Finding Minimum value of D D = โˆš(๐‘˜+(๐‘โˆ’๐‘˜)^2 ) Putting ๐‘˜=(2๐‘ โˆ’ 1)/2 D = โˆš(((2๐‘ โˆ’ 1)/2)+(๐‘โˆ’((2๐‘ โˆ’ 1)/2))^2 ) D = โˆš(((2๐‘ โˆ’ 1)/2)+((2๐‘ โˆ’ 2๐‘ โˆ’ 1)/2)^2 ) D = โˆš(((2๐‘ โˆ’ 1)/2)+((โˆ’1)/2)^2 ) D = โˆš(((2๐‘ โˆ’ 1)/2)+1/4) D = โˆš(๐‘โˆ’1/2+1/4) D = โˆš(๐‘โˆ’1/4) D = โˆš(4๐‘ โˆ’ 1)/2 Hence, shortest distance is โˆš(๐Ÿ’๐’„ โˆ’ ๐Ÿ)/๐Ÿ

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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.