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

Transcript

Ex 6.5, 27 (Method 1) The point on the curve ๐‘ฅ2=2๐‘ฆ which is nearest to the point (0, 5) is (A) (2 โˆš2 ,4) (B) (2 โˆš2,0) (C) (0, 0) (D) (2, 2) Let (โ„Ž , ๐‘˜) be the point on the curve ๐‘ฅ2 = 2๐‘ฆ Where is nearest to the point (0, 5) Since (โ„Ž, ๐‘˜) lie on the curve ๐‘ฅ2= 2๐‘ฆ โ‡’ (โ„Ž ๐‘˜) will satisfy the equation of curve ๐‘ฅ2=2๐‘ฆ โ‡’ Putting ๐‘ฅ=โ„Ž & y=๐‘˜ in equation โ„Ž^2=2๐‘˜ We need to minimize the distance of a point (โ„Ž ,๐‘˜) from(0, 5) Let D be the distant between (โ„Ž,๐‘˜) & (0,5) D = โˆš((0โˆ’โ„Ž)^2+(5โˆ’๐‘˜)^2 ) D = โˆš(โ„Ž^2+(5โˆ’๐‘˜^2 ) ) From (1) โ„Ž^2=2๐‘˜ D = โˆš(2๐‘˜+(5โˆ’๐‘˜)^2 ) Diff w.r.t ๐‘˜ ๐‘‘๐ท/๐‘‘๐‘˜=๐‘‘(โˆš(2๐‘˜ + (5 โˆ’ ๐‘˜)^2 ))/๐‘‘๐‘˜ =1/(2โˆš(2๐‘˜ + (5 โˆ’ ๐‘˜)^2 )) ร—๐‘‘(2๐‘˜ + (5 โˆ’ ๐‘˜)^2 )/๐‘‘๐พ =1/(2โˆš(2๐‘˜ + (5 โˆ’ ๐‘˜)^2 )) ร— [2+2(5โˆ’๐‘˜).๐‘‘(5 โˆ’ ๐‘˜)/๐‘‘๐พ] Let D be the distant between (โ„Ž,๐‘˜) & (0,5) D = โˆš((0โˆ’โ„Ž)^2+(5โˆ’๐‘˜)^2 ) D = โˆš(โ„Ž^2+(5โˆ’๐‘˜^2 ) ) From (1) โ„Ž^2=2๐‘˜ D = โˆš(2๐‘˜+(5โˆ’๐‘˜)^2 ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 2๐‘˜+(5โˆ’๐‘˜)^2 Since D is positive, D is minimum if d2 is minimum So, we minimize Z = D2 Differentiating Z Z = 2๐‘˜+(5โˆ’๐‘˜)^2 Diff w.r.t. k Zโ€™ = ๐‘‘(2๐‘˜ + (5 โˆ’ ๐‘˜)^2 )/๐‘‘๐‘˜ Zโ€™ = 2 + 2 (5 โˆ’ k) ร— (5 โˆ’ k)โ€™ Zโ€™ = 2 + 2 (5 โˆ’ k) ร— (0 โˆ’ 1) Zโ€™ = 2 โˆ’ 2 (5 โˆ’ k) Zโ€™ = 2 โˆ’ 10 + 2k Zโ€™ = โˆ’8 + 2k Putting Zโ€™ = 0 โˆ’8 + 2k = 0 2k = 8 k = 8/2 = 4 Now, checking sign of (๐‘‘^2 ๐‘)/(๐‘‘โ„Ž^2 ) " " ๐‘‘๐‘/๐‘‘๐‘˜=โˆ’8+2๐‘˜ Differentiating again w.r.t k (๐‘‘^2 ๐‘)/(๐‘‘โ„Ž^2 ) = 0+2 (๐‘‘^2 ๐‘)/(๐‘‘โ„Ž^2 ) = 2 โˆด (๐‘‘^2 ๐‘)/(๐‘‘โ„Ž^2 ) > 0 for k = 4 โˆด Z is minimum when k = 4 Thus, D is Minimum at ๐‘˜= 4 Finding h From (1) h^2=2๐‘˜ h^2=2(4) h=โˆš8 h=2โˆš2 Hence, Required Point is (โ„Ž,๐‘˜)=(2โˆš(2 ,) 4) Correct answer is A Ex 6.5, 27 (Method 2) The point on the curve ๐‘ฅ2= 2๐‘ฆ which is nearest to the point (0, 5) is (A) (2 โˆš2,4) (B) (2 โˆš2,0) (C) (0, 0) (D) (2, 2) Since points given lie on the curve, it will satisfy equation of curve Option 1 Point is (2โˆš(2 ,) 4) Putting ๐‘ฅ=2โˆš2 , & ๐‘ฆ=4 in ๐‘ฅ2=2๐‘ฆ โ‡’ (2โˆš2)^2=2(4) โ‡’ 4 ร— 2 = 8 Which is true Thus, (2โˆš2,4) lie on the curve Now, finding distance between (2โˆš(2 ,) 4" " ) & (0 ,5) D = โˆš((0โˆ’2โˆš(2 ))^2+(5โˆ’4)^2 ) = โˆš(8+1) = โˆš9 = 3 Option 2 Point (2โˆš(2 ,) 0) Putting ๐‘ฅ=2โˆš2 & ๐‘ฆ=0 in ๐‘ฅ2=2๐‘ฆ (2โˆš(2 ))^2=2(0) (4 ร—2)=0 8 = 0 Since 8 โ‰ 0 โ‡’ (2โˆš(2 ,) 0) is not the required point Option 3 Point (0 ,0) Putting ๐‘ฅ=0 & ๐‘ฆ=0 in ๐‘ฅ2=2๐‘ฆ (0)^2=2(0) 0=0 โˆด (0 , 0) lie on the curve Now, Finding distance between (0 , 0) ๐‘Ž๐‘›๐‘‘ (0 , 5) D = โˆš((0โˆ’0)^2+(5โˆ’0)^2 ) = โˆš(0+5^2 ) = 5 Option 4 Point (2 ,2) Putting ๐‘ฅ=2 & ๐‘ฆ=2 in ๐‘ฅ2=2๐‘ฆ (2)^2=2(2) 4=4 โˆด (2, 2) lie on the curve Now, Finding distance between (2, 2) ๐‘Ž๐‘›๐‘‘ (0 , 5) D = โˆš((0โˆ’2)^2+(5โˆ’2)^2 ) = โˆš((โˆ’2)^2+(3)^2 ) = โˆš(4+9) = โˆš13 Thus, Point (2โˆš2,4) is on the curve ๐‘ฅ2=2๐‘ฆ & nearest to the point (0, 5) Hence correct answer is A

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.