Misc 8 - Chapter 6 Class 12 Application of Derivatives

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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 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 = 1﷮2﷯ ×𝑏𝑎𝑠𝑒 ×ℎ𝑒𝑖𝑔ℎ𝑡 A = 1﷮2﷯ × 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 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﷯﷯ Now we have P = ℎ , 𝑘﷯ , 𝑃﷮′﷯= ℎ , −𝑘﷯ A = 𝑎 , 0﷯ & M = ℎ,0﷯ Finding PP’ PP’ = ﷮ ℎ−ℎ﷯﷮2﷯+ −𝑘−𝑘﷯﷮2﷯﷯ PP’ = ﷮ −2𝑘﷯﷮2﷯﷯ PP’ = ﷮ 4𝑘﷮2﷯﷯ PP’ = 2𝑘 Finding AM AM = ﷮ 𝑎−ℎ﷯﷮2﷯+ 0−0﷯﷮2﷯﷯ AM = ﷮ 𝑎−ℎ﷯﷮2﷯﷯ AM = a – h Thus, Area of Triangle A = 1﷮2﷯ ×𝐴𝑀 × 𝑃 𝑃﷮′﷯ A = 1﷮2﷯ × 𝑎−ℎ﷯ ×2𝑘 A = 𝑎−ℎ﷯ × 1﷮𝑎﷯ ﷮ 𝑎𝑏﷯﷮2﷯− 𝑏﷮2﷯ ℎ﷮2﷯﷯ Take 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﷯﷯﷯ = 𝑎−ℎ﷯﷮2﷯ × 1﷮ 𝑎﷮2﷯﷯ × 𝑏﷮2﷯ 𝑎﷮2﷯− ℎ﷮2﷯﷯ = 𝑎−ℎ﷯﷮2﷯ × 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑎﷮2﷯− ℎ﷮2﷯﷯ A is maximum if Z is maximum So, maximizing Z Z = 𝑎−ℎ﷯﷮2﷯ × 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑎﷮2﷯− ℎ﷮2﷯﷯ Z = 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑎−ℎ﷯﷮2﷯ 𝑎﷮2﷯− ℎ﷮2﷯﷯ Differentiate w.r.t h 𝑑𝑍﷮𝑑ℎ﷯= 𝑑 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑎 − ℎ﷯﷮2﷯ 𝑎﷮2﷯ − ℎ﷮2﷯﷯﷯﷮𝑑ℎ﷯ 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−2 ℎ﷮2﷯﷯ Putting 𝑑𝑍﷮𝑑ℎ﷯= 0 –2 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑎−ℎ﷯ 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯=0 𝑎−ℎ﷯ 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯=0 Thus, 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﷮𝑑 ℎ﷮2﷯﷯ 𝑑𝑍﷮𝑑ℎ﷯=−2 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑎−ℎ﷯ 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯﷯ Differentiating w.r.t ℎ 𝑑﷮2﷯𝑍﷮𝑑 ℎ﷮2﷯﷯ = −2 𝑏﷮𝑎﷯﷯﷮2﷯. 𝑑 𝑎−ℎ﷯ 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯﷯﷮𝑑ℎ﷯ =−2 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑑﷮𝑑ℎ﷯ 𝑎−ℎ﷯ 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯+ 𝑑 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯﷮𝑑ℎ﷯. 𝑎−ℎ﷯﷯ = −2 𝑏﷮𝑎﷯﷯﷮2﷯ 0−1﷯ 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯+ 0+𝑎 −4ℎ﷯ 𝑎−ℎ﷯﷯ = −2 𝑏﷮𝑎﷯﷯﷮2﷯ −1﷯ 𝑎﷮2﷯+𝑎ℎ−2 ℎ﷮2﷯﷯+ 𝑎 −4ℎ﷯ 𝑎−ℎ﷯﷯ = −2 𝑏﷮𝑎﷯﷯﷮2﷯ − 𝑎﷮2﷯−𝑎ℎ+2 ℎ﷮2﷯+ 𝑎﷮2﷯−𝑎ℎ−4𝑎ℎ+4 ℎ﷮2﷯﷯ = −2 𝑏﷮𝑎﷯﷯﷮2﷯ −2𝑎ℎ−4𝑎ℎ+6 ℎ﷮2﷯﷯ Putting h = −𝑎﷮2﷯ = −2 𝑏﷮𝑎﷯﷯﷮2﷯ −2𝑎 −𝑎﷮2﷯﷯−4𝑎 −𝑎﷮2﷯﷯+6 −𝑎﷮2﷯﷯﷮2﷯﷯﷯ = −2 𝑏﷮𝑎﷯﷯﷮2﷯ 𝑎﷮2﷯+2 𝑎﷮2﷯+ 6﷮4﷯ 𝑎﷮2﷯﷯ = –2 𝑏﷮𝑎﷯﷯﷮2﷯ 3 𝑎﷮2﷯+ 3﷮2﷯ 𝑎﷮2﷯﷯ = –2 𝑏﷮2﷯ ﷮ 𝑎﷮2﷯﷯﷯ 9 𝑎﷮2﷯﷮2﷯﷯ = −9 b﷮2﷯ = < 0 ∴ 𝑑﷮2﷯𝑧﷮𝑑 ℎ﷮2﷯﷯<0 at h = −𝑎﷮2﷯ ⇒ Z is maximum when h = −𝑎﷮2﷯ ⇒ Area is maximum when h = −𝑎﷮2﷯ From (1) k = 1﷮𝑎﷯ ﷮ 𝑎𝑏﷯﷮2﷯− 𝑏ℎ﷯﷮2﷯﷯ k = 1﷮𝑎﷯ ﷮ 𝑎𝑏﷯﷮2﷯− 𝑏 −𝑎﷮2﷯﷯﷯﷮2﷯﷯ k = 1﷮𝑎﷯ ﷮ 𝑎𝑏﷯﷮2﷯− 𝑎𝑏﷯﷮2﷯﷮4﷯﷯﷯ k = 1﷮𝑎﷯ ﷮ 4 𝑎𝑏﷯﷮2﷯− 𝑎𝑏﷯﷮2﷯﷮4﷯﷯ k = 1﷮𝑎﷯ ﷮ 3 𝑎𝑏﷯﷮2﷯﷮4﷯﷯ k = 1﷮𝑎﷯ × ﷮3﷯ ×𝑎𝑏﷮2﷯ k = ﷮3𝑏﷯﷮2﷯ Hence ℎ = −𝑎﷮2﷯ & k = ﷮3𝑏﷯﷮2﷯ Maximum value of A A = 1﷮2﷯ × 𝑎−ℎ﷯ 2𝑘﷯ = 1﷮2﷯ × 𝑎− −𝑎﷮2﷯﷯﷯𝑏 ﷮3﷯ = 1﷮2﷯ 𝑎+ 𝑎﷮2﷯﷯𝑏 ﷮3﷯ = 1﷮2﷯ 2𝑎 + 𝑎﷮2﷯﷯𝑏 ﷮3﷯ = 1﷮2﷯ 3𝑎﷮2﷯﷯𝑏 ﷮3﷯ = 3 ﷮3﷯ 𝑎𝑏﷮4﷯ Hence, maximum area is 𝟑 ﷮𝟑﷯ 𝒂𝒃﷮𝟒﷯