Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12













Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Misc 8 Find the maximum area of an isosceles triangle inscribed in the ellipse ๐ฅ^2/๐^2 + ๐ฆ^2/๐^2 = 1 with its vertex at one end of the major axis. Given equation of ellipse is ๐ฅ^2/๐^2 +๐ฆ^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 , 0) An isosceles triangle inscribed in the ellipse its vertex at one end of major axis Let โ๐๐ด๐^โฒ be isosceles triangle We need to maximize area of triangle Let A be the area of isosceles โ๐๐ด๐โฒ A = 1/2 ร ๐ต๐๐ ๐ ร ๐ป๐๐๐โ๐ก A = 1/2 ร ใ๐๐ใ^โฒร ๐ด๐ Let (โ, ๐) ๐๐ ๐กโ๐ ๐๐๐๐๐๐๐๐ก๐ ๐๐ ๐๐๐๐๐ก ๐ Since, ellipse is symmetric So Coordinate of Pโ is (โ , โ๐) Since point M lie on the ๐ฅ โ axis So, coordinate of point M = (โ , 0) Since point (โ, ๐) ๐๐๐ ๐๐ ๐กโ๐ ๐๐๐๐๐๐ ๐ โด (โ , ๐) ๐ค๐๐๐ satisfy the equation of ellipse Putting ๐ฅ = โ & y = k in equation ๐ฅ^2/๐^2 +๐ฆ^2/๐^2 =1 (โ)^2/๐^2 +(๐)^2/๐^2 =1 (๐^2 โ^2 + ๐^2 ๐^2)/(๐^2 ๐^2 )=1 ๐2โ2+๐2๐2 = ๐2๐2 ๐2๐2 = ๐2๐2 โ ๐2โ2 ๐2 = (๐2๐2" โ " ๐2โ2" " )/๐2 ๐ = โ((๐2๐2" โ " ๐2โ2" " )/๐2) ๐ = 1/๐ โ((๐๐)^2โ(๐โ)^2 ) Now we have P = (โ , ๐) , ๐^โฒ=(โ , โ๐) A = (๐ , 0) & M = (โ,0) Finding PPโ PPโ = โ((โโโ)^2+(โ๐โ๐)^2 ) PPโ = โ((โ2๐)^2 ) PPโ = โ(ใ4๐ใ^2 ) PPโ = 2๐ Finding AM AM = โ((๐โโ)^2+(0โ0)^2 ) AM = โ((๐โโ)^2 ) AM = a โ h Thus, Area of Triangle A = 1/2 ร๐ด๐ ร ๐๐^โฒ A = 1/2 ร (๐โโ) ร2๐ A = (๐โโ) ร1/๐ โ((๐๐)^2โ๐^2 โ^2 ) Let Z = A2 Z = ((๐โโ) ร 1/๐ โ((๐๐)^2โ(๐ โ)^2 ))^2 Z = (๐โโ)^2 ร 1/๐^2 ((๐๐)^2โ(๐^2 โ^2 )) Z = (๐โโ)^2 ร 1/๐^2 ร๐^2 (๐^2โโ^2 ) Z = (๐โโ)^2 ร (๐/๐)^2 (๐^2โโ^2 ) A is maximum if Z is maximum So, maximizing Z Z = (๐โโ)^2 ร(๐/๐)^2 (๐^2โโ^2 ) Z = (๐/๐)^2 (๐โโ)^2 (๐^2โโ^2 ) Differentiating w.r.t h ๐๐/๐โ= ๐((๐/๐)^2 (๐ โ โ)^2 (๐^2 โ โ^2 ))/๐โ ๐๐/๐โ= ๐((๐/๐)^2 (๐ โ โ)^2 (๐^2 โ โ^2 ))/๐โ ๐๐/๐โ=(๐/๐)^2. [(๐(๐โโ)^2)/๐โ.(๐^2โโ^2 )+๐(๐^2โ โ^2 )/๐โ . (๐โโ)^2 ] ๐๐/๐โ=(๐/๐)^2 [2(๐โโ). ๐(โโ)/๐โ (๐^2โโ^2 )+(0โ2โ) (๐โโ)^2 ] ๐๐/๐โ= (๐/๐)^2 [โ2(๐โโ)(๐^2โโ^2 )โ2โ(๐โโ)^2 ] ๐๐/๐โ= (๐/๐)^2 (โ2)[(๐โโ)(๐^2โโ^2 )+โ(๐โโ)^2 ] ๐๐/๐โ= ใโ2(๐/๐)ใ^2 [(๐โโ)(๐^2โโ^2+โ(๐โโ))] ๐๐/๐โ= ใโ2(๐/๐)ใ^2 (๐โโ)[๐^2โโ^2+โ๐โโ^2 ] ๐๐/๐โ= ใโ2(๐/๐)ใ^2 (๐โโ)[๐^2+โ๐โ2โ^2 ] Putting ๐๐/๐โ= 0 โ2(๐/๐)^2 (๐โโ)(๐^2+๐โโ2โ^2 )=0 (๐โโ)(๐^2+๐โโ2โ^2 )=0 Thus, h = a & h = (โ๐)/2 If h = a then k = 1/๐ โ((๐๐)^2โ(๐โ)^2 )= 1/๐ โ((๐๐)^2โ(๐๐)^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+2โ)โโ(๐+2โ)=0 (๐โโ)(๐+2โ)=0 So, h = a and h = (โ๐)/2 Hence, h = (โ๐)/2 only Finding (๐ ^๐ ๐)/(๐ ๐^๐ ) ๐๐/๐โ=โ2(๐/๐)^2 ((๐โโ)(๐^2+๐โโ2โ^2 )) Differentiating w.r.t โ (๐^2 ๐)/(๐โ^2 ) = โ2(๐/๐)^2. ๐((๐โโ)(๐^2+๐โโ2โ^2 ))/๐โ =โ2(๐/๐)^2 [๐/๐โ (๐โโ)(๐^2+๐โโ2โ^2 )+๐(๐^2+๐โโ2โ^2 )/๐โ.(๐โโ)] Using product rule as (๐ข๐ฃ)=๐ข^โฒ ๐ฃ+๐ฃ^โฒ ๐ข = โ2(๐/๐)^2 [(0โ1)(๐^2+๐โโ2โ^2 )+(0+๐ โ4โ)(๐โโ)] = โ2(๐/๐)^2 [(โ1)(๐^2+๐โโ2โ^2 )+(๐ โ4โ)(๐โโ)] = โ2(๐/๐)^2 [โ๐^2โ๐โ+2โ^2+๐^2โ๐โโ4๐โ+4โ^2 ] = โ2(๐/๐)^2 [โ2๐โโ4๐โ+6โ^2 ] Putting h = (โ๐)/2 = โ2(๐/๐)^2 [โ2๐(((โ๐)/2)โ4๐((โ๐)/2)+6((โ๐)/2)^2 )] = โ2(๐/๐)^2 [๐^2+2๐^2+6/4 ๐^2 ] = โ2 (๐/๐)^2 [3๐^2+3/2 ๐^2 ] = โ2((๐^2 )/๐^2 )((9๐^2)/2) = โ9b^2 = < 0 โด (๐^2 ๐ง)/(๐โ^2 )<0 at h = (โ๐)/2 โ Z is maximum when h = (โ๐)/2 โ Area is maximum when h = (โ๐)/2 From (1) k = 1/๐ โ((๐๐)^2โ(๐โ)^2 ) k = 1/๐ โ((๐๐)^2โ(๐((โ๐)/2))^2 ) ("Putting h = " (โ๐)/2) k = 1/๐ โ((๐๐)^2โ((๐๐)^2/4) ) k = 1/๐ โ((4(๐๐)^2โ(๐๐)^2)/4) k = 1/๐ โ((3(๐๐)^2)/4) k = 1/๐ ร (โ3 ร๐๐)/2 k = โ3๐/2 Hence โ = (โ๐)/2 & k = โ3๐/2 Maximum value of A A = 1/2 ร (๐โโ)(2๐) = 1/2 ร(๐โ((โ๐)/2))๐โ3 = 1/2 (๐+๐/2)๐โ3 = 1/2 ((2๐ + ๐)/2)๐โ3 = 1/2 (3๐/2)๐โ3 = (3โ3 ๐๐)/4 Hence, maximum area is (๐โ๐ ๐๐)/๐
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