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Misc 8 - Find maximum area of an isosceles triangle inscribed

Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 3
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 4
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 5
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 6
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 7
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 8
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 9
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 10
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 11
Misc 8 - Chapter 6 Class 12 Application of Derivatives - Part 12

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Misc 8 Find the maximum area of an isosceles triangle inscribed in the ellipse 𝑥^2/𝑎^2 + 𝑦^2/𝑏^2 = 1 with its vertex at one end of the major axis.Given equation of ellipse is 𝑥^2/𝑎^2 +𝑦^2/𝑏^2 =1 where Major axis of ellipse is along x-axis Here, Coordinate of A = (a, 0) Coordinate of A’ = (−a, 0) An isosceles triangle inscribed in the ellipse its vertex at one end of major axis Let ∆𝑷𝑨𝑷^′ be isosceles triangle Let coordinates of point P be (𝒉, 𝒌) Since, ellipse is symmetric So, coordinates of point P’ is (𝒉 , −𝒌) Since point M lies on the 𝒙-axis So, coordinates of point M = (𝒉 , 𝟎) We need to Maximize Area of triangle Let A be the area of isosceles ∆𝑃𝐴𝑃′ A = 1/2 × 𝐵𝑎𝑠𝑒 × 𝐻𝑒𝑖𝑔ℎ𝑡 A = 𝟏/𝟐 × PP’ × AM By Distance Formula, AM = a − h PP’ = 2k Since h and k are two variables, we write h in terms of k Now, Point (𝒉, 𝒌) lies on the ellipse ∴ (ℎ , 𝑘) will satisfy the equation of ellipse Putting 𝑥 = ℎ & y = k in equation 𝑥^2/𝑎^2 +𝑦^2/𝑏^2 =1 (ℎ)^2/𝑎^2 +(𝑘)^2/𝑏^2 =1 (𝑏^2 ℎ^2 + 𝑎^2 𝑘^2)/(𝑎^2 𝑏^2 )=1 𝑏2ℎ2+𝑎2𝑘2 = 𝑎2𝑏2 𝒂𝟐𝒌𝟐 = 𝑎2𝑏2 – 𝑏2ℎ2 𝑘2 = (𝑎2𝑏2" – " 𝑏2ℎ2" " )/𝑎2 𝑘 = √((𝑎2𝑏2" – " 𝑏2ℎ2" " )/𝑎2) 𝒌 = 𝟏/𝒂 √((𝒂𝒃)^𝟐−(𝒃𝒉)^𝟐 ) Thus, Area of Triangle A = 1/2 × AM × PP’ A = 1/2 × (𝑎−ℎ) ×2𝑘 A = (𝒂−𝒉) ×𝟏/𝒂 √((𝒂𝒃)^𝟐−𝒃^𝟐 𝒉^𝟐 ) We need to maximise A, but A has a square root Which will be difficult to differentiate Let Z = A2 Z = ((𝑎−ℎ) × 1/𝑎 √((𝑎𝑏)^2−(𝑏 ℎ)^2 ))^2 Z = (𝑎−ℎ)^2 × 1/𝑎^2 ((𝑎𝑏)^2−(𝑏^2 ℎ^2 )) Z = (𝑎−ℎ)^2 × 1/𝑎^2 × 𝑏^2 (𝑎^2−ℎ^2 ) Z = (𝒂−𝒉)^𝟐 × (𝒃/𝒂)^𝟐 (𝒂^𝟐−𝒉^𝟐 ) Z = (𝑏/𝑎)^2 (𝑎−ℎ)^2 (𝑎^2−ℎ^2 ) Since A is positive, A is maximum if A2 is maximum So, we maximise Z = A2 Differentiating w.r.t h 𝑑𝑍/𝑑ℎ= 𝑑((𝑏/𝑎)^2 (𝑎 − ℎ)^2 (𝑎^2 − ℎ^2 ))/𝑑ℎ 𝑑𝑍/𝑑ℎ=(𝑏/𝑎)^2. [(𝑑(𝑎−ℎ)^2)/𝑑ℎ.(𝑎^2−ℎ^2 )+𝑑(𝑎^2− ℎ^2 )/𝑑ℎ . (𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ=(𝑏/𝑎)^2 [2(𝑎−ℎ). 𝑑(−ℎ)/𝑑ℎ (𝑎^2−ℎ^2 )+(0−2ℎ) (𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ= (𝑏/𝑎)^2 [−2(𝑎−ℎ)(𝑎^2−ℎ^2 )−2ℎ(𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ= (𝑏/𝑎)^2 (−2)[(𝑎−ℎ)(𝑎^2−ℎ^2 )+ℎ(𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ= 〖−2(𝑏/𝑎)〗^2 [(𝑎−ℎ)(𝑎^2−ℎ^2+ℎ(𝑎−ℎ))] 𝑑𝑍/𝑑ℎ= 〖−2(𝑏/𝑎)〗^2 (𝑎−ℎ)[𝑎^2−ℎ^2+ℎ𝑎−ℎ^2 ] 𝑑𝑍/𝑑ℎ= 〖−2(𝑏/𝑎)〗^2 (𝑎−ℎ)[𝑎^2+ℎ𝑎−2ℎ^2 ] 𝑑𝑍/𝑑ℎ=(𝑏/𝑎)^2. [(𝑑(𝑎−ℎ)^2)/𝑑ℎ.(𝑎^2−ℎ^2 )+𝑑(𝑎^2− ℎ^2 )/𝑑ℎ . (𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ=(𝑏/𝑎)^2 [2(𝑎−ℎ). 𝑑(−ℎ)/𝑑ℎ (𝑎^2−ℎ^2 )+(0−2ℎ) (𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ= (𝑏/𝑎)^2 [−2(𝑎−ℎ)(𝑎^2−ℎ^2 )−2ℎ(𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ= (𝑏/𝑎)^2 (−2)[(𝑎−ℎ)(𝑎^2−ℎ^2 )+ℎ(𝑎−ℎ)^2 ] 𝑑𝑍/𝑑ℎ= 〖−2(𝑏/𝑎)〗^2 [(𝑎−ℎ)(𝑎^2−ℎ^2+ℎ(𝑎−ℎ))] 𝑑𝑍/𝑑ℎ= 〖−2(𝑏/𝑎)〗^2 (𝑎−ℎ)[𝑎^2−ℎ^2+ℎ𝑎−ℎ^2 ] 𝑑𝑍/𝑑ℎ= 〖−2(𝑏/𝑎)〗^2 (𝑎−ℎ)[𝑎^2+ℎ𝑎−2ℎ^2 ] a – h = 0 h = a a2 + ah – 2h2 = 0 a2 + 2ah – ah – 2h2 = 0 a(𝑎+2ℎ)−ℎ(𝑎+2ℎ)=0 (𝑎−ℎ)(𝑎+2ℎ)=0 So, h = a &h = (−𝒂)/𝟐 If h = a then k = 1/𝑎 √((𝑎𝑏)^2−(𝑏ℎ)^2 )= 1/𝑎 √((𝑎𝑏)^2−(𝑏𝑎)^2 )= 0 Which is not possible Hence, h = (−𝒂)/𝟐 only Finding (𝒅^𝟐 𝒁)/(𝒅𝒉^𝟐 ) 𝑑𝑍/𝑑ℎ=−2(𝑏/𝑎)^2 ((𝑎−ℎ)(𝑎^2+𝑎ℎ−2ℎ^2 )) Differentiating w.r.t ℎ (𝑑^2 𝑍)/(𝑑ℎ^2 ) = −2(𝑏/𝑎)^2. 𝑑((𝑎−ℎ)(𝑎^2+𝑎ℎ−2ℎ^2 ))/𝑑ℎ = −2(𝑏/𝑎)^2 [𝑎^2+2𝑎^2+6/4 𝑎^2 ] = –2 (𝑏/𝑎)^2 [3𝑎^2+3/2 𝑎^2 ] = –2((𝑏^2 )/𝑎^2 )((9𝑎^2)/2) = −9b^2 < 0 Since 𝐙^′′ < 0 for h = (−𝑎)/2 ∴ Z is maximum when h = (−𝑎)/2 Thus, A is maximum at h = (−𝒂)/𝟐 Maximum value of Area A = (𝒂−𝒉) ×𝟏/𝒂 √((𝒂𝒃)^𝟐−𝒃^𝟐 𝒉^𝟐 ) = (𝑎−((−𝑎)/2)) ×1/𝑎 √((𝑎𝑏)^2−𝑏^2 ((−𝑎)/2)^2 ) = 3a/2 ×1/𝑎 √(𝑎^2 𝑏^2−(𝑎^2 𝑏^2)/4) = 3a/2 ×1/𝑎 √((3𝑎^2 𝑏^2)/4) = 3a/2 ×1/𝑎 × (√3 𝑎𝑏)/2 = (3√3 𝑎𝑏)/4 Hence, maximum area is (𝟑√𝟑 𝒂𝒃)/𝟒 square units

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.