**Misc 8**

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 8 Find the maximum area of an isosceles triangle inscribed in the ellipse 𝑥2 𝑎2 + 𝑦2 𝑏2 = 1 with its vertex at one end of the major axis. Given equation of ellipse is 𝑥2 𝑎2+ 𝑦2 𝑏2=1 Where Major axis of ellipse is AA’ (along x-axis) Length of major axis = 2a ⇒ AA’ = 2a And of OA’ = OA = a Hence coordinate of A = 𝑎 , 0 An isosceles triangle inscribed in the ellipse its vertex at one end of major axis Let ∆𝑃𝐴 𝑃′ be isosceles triangle We need to maximize area of triangle Let A be the area of isosceles ∆𝑃𝐴𝑃′ A = 12 ×𝑏𝑎𝑠𝑒 ×ℎ𝑒𝑖𝑔ℎ𝑡 A = 12 × 𝑃𝑃′×𝐴𝑀 Let ℎ, 𝑘 𝑏𝑒 𝑡ℎ𝑒 𝑐𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑖𝑛𝑡 𝑃 Since, ellipse is symmetric So Coordinate of P’ is ℎ , −𝑘 Since point M lie on the 𝑥 – axis So, coordinate of point M = ℎ , 0 Since point ℎ, 𝑘 𝑙𝑖𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑒𝑙𝑙𝑖𝑝𝑠𝑒 ⇒ ℎ , 𝑘 𝑤𝑖𝑙𝑙 satisfy the equation of ellipse Putting 𝑥 = ℎ & y = k in equation 𝑥2 𝑎2+ 𝑦2 𝑏2=1 ℎ2 𝑎2+ 𝑘2 𝑏2=1 𝑏2 ℎ2 + 𝑎2 𝑘2 𝑎2 𝑏2=1 𝑏2ℎ2+𝑎2𝑘2 = 𝑎2𝑏2 𝑎2𝑘2 = 𝑎2𝑏2 – 𝑏2ℎ2 𝑘2 = 𝑎2𝑏2 – 𝑏2ℎ2 𝑎2 𝑘 = 𝑎2𝑏2 – 𝑏2ℎ2 𝑎2 𝑘 = 1𝑎 𝑎𝑏2− 𝑏ℎ2 Now we have P = ℎ , 𝑘 , 𝑃′= ℎ , −𝑘 A = 𝑎 , 0 & M = ℎ,0 Finding PP’ PP’ = ℎ−ℎ2+ −𝑘−𝑘2 PP’ = −2𝑘2 PP’ = 4𝑘2 PP’ = 2𝑘 Finding AM AM = 𝑎−ℎ2+ 0−02 AM = 𝑎−ℎ2 AM = a – h Thus, Area of Triangle A = 12 ×𝐴𝑀 × 𝑃 𝑃′ A = 12 × 𝑎−ℎ ×2𝑘 A = 𝑎−ℎ × 1𝑎 𝑎𝑏2− 𝑏2 ℎ2 Take Z = A2 Z = 𝑎−ℎ × 1𝑎 𝑎𝑏2− 𝑏 ℎ22 Z = 𝑎−ℎ2 × 1 𝑎2 𝑎𝑏2− 𝑏2 ℎ2 = 𝑎−ℎ2 × 1 𝑎2 × 𝑏2 𝑎2− ℎ2 = 𝑎−ℎ2 × 𝑏𝑎2 𝑎2− ℎ2 A is maximum if Z is maximum So, maximizing Z Z = 𝑎−ℎ2 × 𝑏𝑎2 𝑎2− ℎ2 Z = 𝑏𝑎2 𝑎−ℎ2 𝑎2− ℎ2 Differentiate w.r.t h 𝑑𝑍𝑑ℎ= 𝑑 𝑏𝑎2 𝑎 − ℎ2 𝑎2 − ℎ2𝑑ℎ 𝑑𝑍𝑑ℎ= 𝑑 𝑏𝑎2 𝑎 − ℎ2 𝑎2 − ℎ2𝑑ℎ 𝑑𝑍𝑑ℎ= 𝑏𝑎2 . 𝑑 𝑎−ℎ2𝑑ℎ. 𝑎2− ℎ2+ 𝑑 𝑎2− ℎ2𝑑ℎ . 𝑎−ℎ2 𝑑𝑍𝑑ℎ= 𝑏𝑎2 2 𝑎−ℎ. 𝑑 −ℎ𝑑ℎ 𝑎2− ℎ2+ 0−2ℎ 𝑎−ℎ2 𝑑𝑍𝑑ℎ= 𝑏𝑎2 −2 𝑎−ℎ 𝑎2− ℎ2−2ℎ 𝑎−ℎ2 𝑑𝑍𝑑ℎ= 𝑏𝑎2 −2 𝑎−ℎ 𝑎2− ℎ2+ℎ 𝑎−ℎ2 𝑑𝑍𝑑ℎ= −2 𝑏𝑎2 𝑎−ℎ 𝑎2− ℎ2+ℎ 𝑎−ℎ 𝑑𝑍𝑑ℎ= −2 𝑏𝑎2 𝑎−ℎ 𝑎2− ℎ2+ℎ𝑎− ℎ2 𝑑𝑍𝑑ℎ= −2 𝑏𝑎2 𝑎−ℎ 𝑎2+ℎ𝑎−2 ℎ2 Putting 𝑑𝑍𝑑ℎ= 0 –2 𝑏𝑎2 𝑎−ℎ 𝑎2+𝑎ℎ−2 ℎ2=0 𝑎−ℎ 𝑎2+𝑎ℎ−2 ℎ2=0 Thus, h = a & h = −𝑎2 If h = a then k = 1𝑎 𝑎𝑏2− 𝑏ℎ2= 1𝑎 𝑎𝑏2− 𝑏𝑎2= 0 Which is not possible Hence h = −𝑎2 only Finding 𝑑2𝑍𝑑 ℎ2 𝑑𝑍𝑑ℎ=−2 𝑏𝑎2 𝑎−ℎ 𝑎2+𝑎ℎ−2 ℎ2 Differentiating w.r.t ℎ 𝑑2𝑍𝑑 ℎ2 = −2 𝑏𝑎2. 𝑑 𝑎−ℎ 𝑎2+𝑎ℎ−2 ℎ2𝑑ℎ =−2 𝑏𝑎2 𝑑𝑑ℎ 𝑎−ℎ 𝑎2+𝑎ℎ−2 ℎ2+ 𝑑 𝑎2+𝑎ℎ−2 ℎ2𝑑ℎ. 𝑎−ℎ = −2 𝑏𝑎2 0−1 𝑎2+𝑎ℎ−2 ℎ2+ 0+𝑎 −4ℎ 𝑎−ℎ = −2 𝑏𝑎2 −1 𝑎2+𝑎ℎ−2 ℎ2+ 𝑎 −4ℎ 𝑎−ℎ = −2 𝑏𝑎2 − 𝑎2−𝑎ℎ+2 ℎ2+ 𝑎2−𝑎ℎ−4𝑎ℎ+4 ℎ2 = −2 𝑏𝑎2 −2𝑎ℎ−4𝑎ℎ+6 ℎ2 Putting h = −𝑎2 = −2 𝑏𝑎2 −2𝑎 −𝑎2−4𝑎 −𝑎2+6 −𝑎22 = −2 𝑏𝑎2 𝑎2+2 𝑎2+ 64 𝑎2 = –2 𝑏𝑎2 3 𝑎2+ 32 𝑎2 = –2 𝑏2 𝑎2 9 𝑎22 = −9 b2 = < 0 ∴ 𝑑2𝑧𝑑 ℎ2<0 at h = −𝑎2 ⇒ Z is maximum when h = −𝑎2 ⇒ Area is maximum when h = −𝑎2 From (1) k = 1𝑎 𝑎𝑏2− 𝑏ℎ2 k = 1𝑎 𝑎𝑏2− 𝑏 −𝑎22 k = 1𝑎 𝑎𝑏2− 𝑎𝑏24 k = 1𝑎 4 𝑎𝑏2− 𝑎𝑏24 k = 1𝑎 3 𝑎𝑏24 k = 1𝑎 × 3 ×𝑎𝑏2 k = 3𝑏2 Hence ℎ = −𝑎2 & k = 3𝑏2 Maximum value of A A = 12 × 𝑎−ℎ 2𝑘 = 12 × 𝑎− −𝑎2𝑏 3 = 12 𝑎+ 𝑎2𝑏 3 = 12 2𝑎 + 𝑎2𝑏 3 = 12 3𝑎2𝑏 3 = 3 3 𝑎𝑏4 Hence, maximum area is 𝟑 𝟑 𝒂𝒃𝟒

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.