Slide1.JPG

Slide2.JPG
Slide3.JPG Slide4.JPG Slide5.JPG Slide6.JPG Slide7.JPG Slide8.JPG Slide9.JPG Slide10.JPG Slide11.JPG Slide12.JPG Slide13.JPG Slide14.JPG

Subscribe to our Youtube Channel - https://you.tube/teachoo

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

Transcript

Misc 8 Find the maximum area of an isosceles triangle inscribed in the ellipse ๐‘ฅ^2/๐‘Ž^2 + ๐‘ฆ^2/๐‘^2 = 1 with its vertex at one end of the major axis. Given equation of ellipse is ๐‘ฅ^2/๐‘Ž^2 +๐‘ฆ^2/๐‘^2 =1 Where Major axis of ellipse is AAโ€™ (along x-axis) Length of major axis = 2a โ‡’ AAโ€™ = 2a And of OAโ€™ = OA = a Hence coordinate of A = (๐‘Ž , 0) An isosceles triangle inscribed in the ellipse its vertex at one end of major axis Let โˆ†๐‘ƒ๐ด๐‘ƒ^โ€ฒ be isosceles triangle We need to maximize area of triangle Let A be the area of isosceles โˆ†๐‘ƒ๐ด๐‘ƒโ€ฒ A = 1/2 ร— ๐ต๐‘Ž๐‘ ๐‘’ ร— ๐ป๐‘’๐‘–๐‘”โ„Ž๐‘ก A = 1/2 ร— ใ€–๐‘ƒ๐‘ƒใ€—^โ€ฒร— ๐ด๐‘€ Let (โ„Ž, ๐‘˜) ๐‘๐‘’ ๐‘กโ„Ž๐‘’ ๐‘๐‘œ๐‘Ÿ๐‘‘๐‘–๐‘›๐‘Ž๐‘ก๐‘’ ๐‘œ๐‘“ ๐‘๐‘œ๐‘–๐‘›๐‘ก ๐‘ƒ Since, ellipse is symmetric So Coordinate of Pโ€™ is (โ„Ž , โˆ’๐‘˜) Since point M lie on the ๐‘ฅ โ€“ axis So, coordinate of point M = (โ„Ž , 0) Since point (โ„Ž, ๐‘˜) ๐‘™๐‘–๐‘’ ๐‘œ๐‘› ๐‘กโ„Ž๐‘’ ๐‘’๐‘™๐‘™๐‘–๐‘๐‘ ๐‘’ โˆด (โ„Ž , ๐‘˜) ๐‘ค๐‘–๐‘™๐‘™ satisfy the equation of ellipse Putting ๐‘ฅ = โ„Ž & y = k in equation ๐‘ฅ^2/๐‘Ž^2 +๐‘ฆ^2/๐‘^2 =1 (โ„Ž)^2/๐‘Ž^2 +(๐‘˜)^2/๐‘^2 =1 (๐‘^2 โ„Ž^2 + ๐‘Ž^2 ๐‘˜^2)/(๐‘Ž^2 ๐‘^2 )=1 ๐‘2โ„Ž2+๐‘Ž2๐‘˜2 = ๐‘Ž2๐‘2 ๐‘Ž2๐‘˜2 = ๐‘Ž2๐‘2 โ€“ ๐‘2โ„Ž2 ๐‘˜2 = (๐‘Ž2๐‘2" โ€“ " ๐‘2โ„Ž2" " )/๐‘Ž2 ๐‘˜ = โˆš((๐‘Ž2๐‘2" โ€“ " ๐‘2โ„Ž2" " )/๐‘Ž2) ๐‘˜ = 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘โ„Ž)^2 ) Now we have P = (โ„Ž , ๐‘˜) , ๐‘ƒ^โ€ฒ=(โ„Ž , โˆ’๐‘˜) A = (๐‘Ž , 0) & M = (โ„Ž,0) Finding PPโ€™ PPโ€™ = โˆš((โ„Žโˆ’โ„Ž)^2+(โˆ’๐‘˜โˆ’๐‘˜)^2 ) PPโ€™ = โˆš((โˆ’2๐‘˜)^2 ) PPโ€™ = โˆš(ใ€–4๐‘˜ใ€—^2 ) PPโ€™ = 2๐‘˜ Finding AM AM = โˆš((๐‘Žโˆ’โ„Ž)^2+(0โˆ’0)^2 ) AM = โˆš((๐‘Žโˆ’โ„Ž)^2 ) AM = a โ€“ h Thus, Area of Triangle A = 1/2 ร—๐ด๐‘€ ร— ๐‘ƒ๐‘ƒ^โ€ฒ A = 1/2 ร— (๐‘Žโˆ’โ„Ž) ร—2๐‘˜ A = (๐‘Žโˆ’โ„Ž) ร—1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’๐‘^2 โ„Ž^2 ) Let Z = A2 Z = ((๐‘Žโˆ’โ„Ž) ร— 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘ โ„Ž)^2 ))^2 Z = (๐‘Žโˆ’โ„Ž)^2 ร— 1/๐‘Ž^2 ((๐‘Ž๐‘)^2โˆ’(๐‘^2 โ„Ž^2 )) Z = (๐‘Žโˆ’โ„Ž)^2 ร— 1/๐‘Ž^2 ร—๐‘^2 (๐‘Ž^2โˆ’โ„Ž^2 ) Z = (๐‘Žโˆ’โ„Ž)^2 ร— (๐‘/๐‘Ž)^2 (๐‘Ž^2โˆ’โ„Ž^2 ) A is maximum if Z is maximum So, maximizing Z Z = (๐‘Žโˆ’โ„Ž)^2 ร—(๐‘/๐‘Ž)^2 (๐‘Ž^2โˆ’โ„Ž^2 ) Z = (๐‘/๐‘Ž)^2 (๐‘Žโˆ’โ„Ž)^2 (๐‘Ž^2โˆ’โ„Ž^2 ) Differentiating w.r.t h ๐‘‘๐‘/๐‘‘โ„Ž= ๐‘‘((๐‘/๐‘Ž)^2 (๐‘Ž โˆ’ โ„Ž)^2 (๐‘Ž^2 โˆ’ โ„Ž^2 ))/๐‘‘โ„Ž ๐‘‘๐‘/๐‘‘โ„Ž= ๐‘‘((๐‘/๐‘Ž)^2 (๐‘Ž โˆ’ โ„Ž)^2 (๐‘Ž^2 โˆ’ โ„Ž^2 ))/๐‘‘โ„Ž ๐‘‘๐‘/๐‘‘โ„Ž=(๐‘/๐‘Ž)^2. [(๐‘‘(๐‘Žโˆ’โ„Ž)^2)/๐‘‘โ„Ž.(๐‘Ž^2โˆ’โ„Ž^2 )+๐‘‘(๐‘Ž^2โˆ’ โ„Ž^2 )/๐‘‘โ„Ž . (๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž=(๐‘/๐‘Ž)^2 [2(๐‘Žโˆ’โ„Ž). ๐‘‘(โˆ’โ„Ž)/๐‘‘โ„Ž (๐‘Ž^2โˆ’โ„Ž^2 )+(0โˆ’2โ„Ž) (๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= (๐‘/๐‘Ž)^2 [โˆ’2(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2 )โˆ’2โ„Ž(๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= (๐‘/๐‘Ž)^2 (โˆ’2)[(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2 )+โ„Ž(๐‘Žโˆ’โ„Ž)^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 [(๐‘Žโˆ’โ„Ž)(๐‘Ž^2โˆ’โ„Ž^2+โ„Ž(๐‘Žโˆ’โ„Ž))] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 (๐‘Žโˆ’โ„Ž)[๐‘Ž^2โˆ’โ„Ž^2+โ„Ž๐‘Žโˆ’โ„Ž^2 ] ๐‘‘๐‘/๐‘‘โ„Ž= ใ€–โˆ’2(๐‘/๐‘Ž)ใ€—^2 (๐‘Žโˆ’โ„Ž)[๐‘Ž^2+โ„Ž๐‘Žโˆ’2โ„Ž^2 ] Putting ๐‘‘๐‘/๐‘‘โ„Ž= 0 โ€“2(๐‘/๐‘Ž)^2 (๐‘Žโˆ’โ„Ž)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )=0 (๐‘Žโˆ’โ„Ž)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )=0 Thus, h = a & h = (โˆ’๐‘Ž)/2 If h = a then k = 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘โ„Ž)^2 )= 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘๐‘Ž)^2 )= 0 Which is not possible a โ€“ h = 0 h = a a2 + ah โ€“ 2h2 = 0 a2 + 2ah โ€“ ah โ€“ 2h2 = 0 a(๐‘Ž+2โ„Ž)โˆ’โ„Ž(๐‘Ž+2โ„Ž)=0 (๐‘Žโˆ’โ„Ž)(๐‘Ž+2โ„Ž)=0 So, h = a and h = (โˆ’๐‘Ž)/2 Hence, h = (โˆ’๐‘Ž)/2 only Finding (๐’…^๐Ÿ ๐’)/(๐’…๐’‰^๐Ÿ ) ๐‘‘๐‘/๐‘‘โ„Ž=โˆ’2(๐‘/๐‘Ž)^2 ((๐‘Žโˆ’โ„Ž)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )) Differentiating w.r.t โ„Ž (๐‘‘^2 ๐‘)/(๐‘‘โ„Ž^2 ) = โˆ’2(๐‘/๐‘Ž)^2. ๐‘‘((๐‘Žโˆ’โ„Ž)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 ))/๐‘‘โ„Ž =โˆ’2(๐‘/๐‘Ž)^2 [๐‘‘/๐‘‘โ„Ž (๐‘Žโˆ’โ„Ž)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )+๐‘‘(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )/๐‘‘โ„Ž.(๐‘Žโˆ’โ„Ž)] Using product rule as (๐‘ข๐‘ฃ)=๐‘ข^โ€ฒ ๐‘ฃ+๐‘ฃ^โ€ฒ ๐‘ข = โˆ’2(๐‘/๐‘Ž)^2 [(0โˆ’1)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )+(0+๐‘Ž โˆ’4โ„Ž)(๐‘Žโˆ’โ„Ž)] = โˆ’2(๐‘/๐‘Ž)^2 [(โˆ’1)(๐‘Ž^2+๐‘Žโ„Žโˆ’2โ„Ž^2 )+(๐‘Ž โˆ’4โ„Ž)(๐‘Žโˆ’โ„Ž)] = โˆ’2(๐‘/๐‘Ž)^2 [โˆ’๐‘Ž^2โˆ’๐‘Žโ„Ž+2โ„Ž^2+๐‘Ž^2โˆ’๐‘Žโ„Žโˆ’4๐‘Žโ„Ž+4โ„Ž^2 ] = โˆ’2(๐‘/๐‘Ž)^2 [โˆ’2๐‘Žโ„Žโˆ’4๐‘Žโ„Ž+6โ„Ž^2 ] Putting h = (โˆ’๐‘Ž)/2 = โˆ’2(๐‘/๐‘Ž)^2 [โˆ’2๐‘Ž(((โˆ’๐‘Ž)/2)โˆ’4๐‘Ž((โˆ’๐‘Ž)/2)+6((โˆ’๐‘Ž)/2)^2 )] = โˆ’2(๐‘/๐‘Ž)^2 [๐‘Ž^2+2๐‘Ž^2+6/4 ๐‘Ž^2 ] = โ€“2 (๐‘/๐‘Ž)^2 [3๐‘Ž^2+3/2 ๐‘Ž^2 ] = โ€“2((๐‘^2 )/๐‘Ž^2 )((9๐‘Ž^2)/2) = โˆ’9b^2 = < 0 โˆด (๐‘‘^2 ๐‘ง)/(๐‘‘โ„Ž^2 )<0 at h = (โˆ’๐‘Ž)/2 โ‡’ Z is maximum when h = (โˆ’๐‘Ž)/2 โ‡’ Area is maximum when h = (โˆ’๐‘Ž)/2 From (1) k = 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘โ„Ž)^2 ) k = 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’(๐‘((โˆ’๐‘Ž)/2))^2 ) ("Putting h = " (โˆ’๐‘Ž)/2) k = 1/๐‘Ž โˆš((๐‘Ž๐‘)^2โˆ’((๐‘Ž๐‘)^2/4) ) k = 1/๐‘Ž โˆš((4(๐‘Ž๐‘)^2โˆ’(๐‘Ž๐‘)^2)/4) k = 1/๐‘Ž โˆš((3(๐‘Ž๐‘)^2)/4) k = 1/๐‘Ž ร— (โˆš3 ร—๐‘Ž๐‘)/2 k = โˆš3๐‘/2 Hence โ„Ž = (โˆ’๐‘Ž)/2 & k = โˆš3๐‘/2 Maximum value of A A = 1/2 ร— (๐‘Žโˆ’โ„Ž)(2๐‘˜) = 1/2 ร—(๐‘Žโˆ’((โˆ’๐‘Ž)/2))๐‘โˆš3 = 1/2 (๐‘Ž+๐‘Ž/2)๐‘โˆš3 = 1/2 ((2๐‘Ž + ๐‘Ž)/2)๐‘โˆš3 = 1/2 (3๐‘Ž/2)๐‘โˆš3 = (3โˆš3 ๐‘Ž๐‘)/4 Hence, maximum area is (๐Ÿ‘โˆš๐Ÿ‘ ๐’‚๐’ƒ)/๐Ÿ’

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
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.