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Ex 6.3, 27 (Method 1) The point on the curve š‘„2=2š‘¦ which is nearest to the point (0, 5) is (A) (2 āˆš2 ,4) (B) (2 āˆš2,0) (C) (0, 0) (D) (2, 2) Let (ā„Ž , š‘˜) be the point on the curve š‘„2 = 2š‘¦ Where is nearest to the point (0, 5) Since (ā„Ž, š‘˜) lie on the curve š‘„2= 2š‘¦ ā‡’ (ā„Ž š‘˜) will satisfy the equation of curve š‘„2=2š‘¦ ā‡’ Putting š‘„=ā„Ž & y=š‘˜ in equation ā„Ž^2=2š‘˜ We need to minimize the distance of a point (ā„Ž ,š‘˜) from(0, 5) Let D be the distant between (ā„Ž,š‘˜) & (0,5) D = āˆš((0āˆ’ā„Ž)^2+(5āˆ’š‘˜)^2 ) D = āˆš(ā„Ž^2+(5āˆ’š‘˜^2 ) ) From (1) ā„Ž^2=2š‘˜ D = āˆš(2š‘˜+(5āˆ’š‘˜)^2 ) Diff w.r.t š‘˜ š‘‘š·/š‘‘š‘˜=š‘‘(āˆš(2š‘˜ + (5 āˆ’ š‘˜)^2 ))/š‘‘š‘˜ =1/(2āˆš(2š‘˜ + (5 āˆ’ š‘˜)^2 )) Ɨš‘‘(2š‘˜ + (5 āˆ’ š‘˜)^2 )/š‘‘š¾ =1/(2āˆš(2š‘˜ + (5 āˆ’ š‘˜)^2 )) Ɨ [2+2(5āˆ’š‘˜).š‘‘(5 āˆ’ š‘˜)/š‘‘š¾] Let D be the distant between (ā„Ž,š‘˜) & (0,5) D = āˆš((0āˆ’ā„Ž)^2+(5āˆ’š‘˜)^2 ) D = āˆš(ā„Ž^2+(5āˆ’š‘˜^2 ) ) From (1) ā„Ž^2=2š‘˜ D = āˆš(2š‘˜+(5āˆ’š‘˜)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2š‘˜+(5āˆ’š‘˜)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2š‘˜+(5āˆ’š‘˜)^2 Diff w.r.t. k Zā€™ = š‘‘(2š‘˜ + (5 āˆ’ š‘˜)^2 )/š‘‘š‘˜ Zā€™ = 2 + 2 (5 āˆ’ k) Ɨ (5 āˆ’ k)ā€™ Zā€™ = 2 + 2 (5 āˆ’ k) Ɨ (0 āˆ’ 1) Zā€™ = 2 āˆ’ 2 (5 āˆ’ k) Zā€™ = 2 āˆ’ 10 + 2k Zā€™ = āˆ’8 + 2k Putting Zā€™ = 0 āˆ’8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (š‘‘^2 š‘)/(š‘‘ā„Ž^2 ) " " š‘‘š‘/š‘‘š‘˜=āˆ’8+2š‘˜ Differentiating again w.r.t k (š‘‘^2 š‘)/(š‘‘ā„Ž^2 ) = 0+2 (š‘‘^2 š‘)/(š‘‘ā„Ž^2 ) = 2 āˆ“ (š‘‘^2 š‘)/(š‘‘ā„Ž^2 ) > 0 for k = 4 āˆ“ Z is minimum when k = 4 Thus, D is Minimum at š‘˜= 4 Finding h From (1) h^2=2š‘˜ h^2=2(4) h=āˆš8 h=2āˆš2 Hence, Required Point is (ā„Ž,š‘˜)=(2āˆš(2 ,) 4) Correct answer is A Ex 6.3, 27 (Method 2) The point on the curve š‘„2= 2š‘¦ which is nearest to the point (0, 5) is (A) (2 āˆš2,4) (B) (2 āˆš2,0) (C) (0, 0) (D) (2, 2)Since points given lie on the curve, it will satisfy equation of curve Option 1 Point is (2āˆš(2 ,) 4) Putting š‘„=2āˆš2 , & š‘¦=4 in š‘„2=2š‘¦ ā‡’ (2āˆš2)^2=2(4) ā‡’ 4 Ɨ 2 = 8 Which is true Thus, (2āˆš2,4) lie on the curve Now, finding distance between (2āˆš(2 ,) 4" " ) & (0 ,5) D = āˆš((0āˆ’2āˆš(2 ))^2+(5āˆ’4)^2 ) = āˆš(8+1) = āˆš9 = 3 Option 2 Point (2āˆš(2 ,) 0) Putting š‘„=2āˆš2 & š‘¦=0 in š‘„2=2š‘¦ (2āˆš(2 ))^2=2(0) (4 Ɨ2)=0 8 = 0 Since 8 ā‰ 0 ā‡’ (2āˆš(2 ,) 0) is not the required point Option 3 Point (0, 0) Putting š‘„=0 & š‘¦=0 in š‘„2=2š‘¦ (0)^2=2(0) 0=0 āˆ“ (0 , 0) lie on the curve Now, Finding distance between (0, 0) š‘Žš‘›š‘‘ (0 , 5) D = āˆš((0āˆ’0)^2+(5āˆ’0)^2 ) = āˆš(0+5^2 ) = 5 Option 4 Point (2 ,2) Putting š‘„=2 & š‘¦=2 in š‘„2=2š‘¦ (2)^2=2(2) 4=4 āˆ“ (2, 2) lie on the curve Now, Finding distance between (2, 2) š‘Žš‘›š‘‘ (0 , 5) D = āˆš((0āˆ’2)^2+(5āˆ’2)^2 ) = āˆš((āˆ’2)^2+(3)^2 ) = āˆš(4+9) = āˆš13 Thus, Point (2āˆš2,4) is on the curve š‘„2=2š‘¦ & nearest to the point (0, 5) Hence correct answer is A

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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