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Example 23 Find the shortest distance of the point (0, c) from the parabola š‘¦=š‘„2, where 0 ā‰¤ c ā‰¤ 5. Let (ā„Ž ,š‘˜) be any point on parabola š‘¦=š‘„2 Let D be required Distance between (ā„Ž , š‘˜) & (0 , š‘) D = āˆš((0āˆ’ā„Ž)^2+(š‘ āˆ’š‘˜)^2 ) D = āˆš((āˆ’ā„Ž)^2+(š‘ āˆ’š‘˜)^2 ) D = āˆš(š’‰^šŸ+(š’„ āˆ’š’Œ)^šŸ ) Also, Since point (ā„Ž , š‘˜) is on the parabola š‘¦=š‘„2 (š’‰ , š’Œ) will satisfy the equation of parabola Putting š‘„=ā„Ž , š‘¦=š‘˜ in equation š’Œ=š’‰^šŸ Putting value of š‘˜=ā„Ž^2 D = āˆš(ā„Ž^2+(š‘ āˆ’š‘˜)^2 ) D = āˆš(š’Œ+(š’„āˆ’š’Œ)^šŸ ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = š‘˜+(š‘āˆ’š‘˜)^2 Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =š‘˜+(š‘āˆ’š‘˜)^2 Differentiating w.r.t. k Zā€™ = š‘‘(š‘˜ + (š‘ āˆ’ š‘˜)^2 )/š‘‘š‘˜ Zā€™ = 1 + 2 (c āˆ’ k) Ɨ (c āˆ’ k)ā€™ Zā€™ = 1 + 2 (c āˆ’ k) Ɨ (0 āˆ’ 1) Zā€™ = 1 āˆ’ 2 (c āˆ’ k) Zā€™ = 1 āˆ’ 2c āˆ’ 2k Putting Zā€™ = 0 1 āˆ’ 2c āˆ’ 2k = 0 2k = 2c āˆ’ 1 k = (šŸš’„ āˆ’ šŸ)/šŸ Now, checking sign of š’^ā€²ā€² " " š‘‘š‘/š‘‘š‘˜=4š‘˜āˆ’2š‘ Differentiating again w.r.t k (š‘‘^2 š‘)/(š‘‘ā„Ž^2 ) = 4 āˆ’0 (š’…^šŸ š’)/(š’…š’‰^šŸ ) = šŸ’ Since š™^ā€²ā€² > 0 for k = (2š‘ āˆ’ 1)/2 āˆ“ Z is minimum when k = (2š‘ āˆ’ 1)/2 Thus, D is Minimum at š’Œ=(šŸš’„ āˆ’ šŸ)/šŸ Finding Minimum value of D D = āˆš(š‘˜+(š‘āˆ’š‘˜)^2 ) Putting š‘˜=(2š‘ āˆ’ 1)/2 D = āˆš(((2š‘ āˆ’ 1)/2)+(š‘āˆ’((2š‘ āˆ’ 1)/2))^2 ) D = āˆš(((2š‘ āˆ’ 1)/2)+((2š‘ āˆ’ 2š‘ āˆ’ 1)/2)^2 ) D = āˆš(((2š‘ āˆ’ 1)/2)+((āˆ’1)/2)^2 ) D = āˆš(((2š‘ āˆ’ 1)/2)+1/4) D = āˆš(š‘āˆ’1/2+1/4) D = āˆš(š‘āˆ’1/4) D = āˆš(4š‘ āˆ’ 1)/2 Hence, shortest distance is āˆš(šŸ’š’„ āˆ’ šŸ)/šŸ

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo