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Misc 12 - A point on hypotenuse is at distance a and b from - Miscellaneous

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Misc 12 A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum length of the hypotenuse is (𝑎﷮ 2﷮3﷯﷯ + 𝑏﷮ 2﷮3﷯﷯) ﷮ 3﷮2﷯﷯ Let P be the point of the hypotenuse AC of ∆ABC Given point P is at distance a & b from sides of triangle Construct PL ⊥ AB & PM ⊥ BC ∴ PL = a & PM = b Let ∠ ACB = θ Thus, ∠ APL = θ We need to find maximum length of the hypotenuse Let l be the length of the hypotenuse , ⇒ l = AP + PC Now l = AP + PC l = a sec θ + b cosec θ Differentiating w.r.t θ 𝑑𝑙﷮𝑑𝜃﷯= 𝑑 𝑎 sec﷮𝜃 + 𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃﷯﷯﷮𝑑𝜃﷯ 𝑑𝑙﷮𝑑𝜃﷯= a sec θ tan θ – b cosec θ cot θ Putting 𝑑𝑙﷮𝑑𝜃﷯=0 a sec θ tan θ – b cosec θ cot θ = 0 a 1﷮ cos﷮𝜃﷯﷯ . sin﷮𝜃﷯﷮ cos﷮𝜃﷯﷯−𝑏 . 1﷮ sin﷮𝜃 ﷯﷯ × cos﷮𝜃﷯﷮ sin﷮𝜃﷯﷯=0 𝑎 sin﷮𝜃﷯﷮ cos﷮2﷯﷮𝜃﷯﷯− 𝑏 cos﷮𝜃﷯﷮ sin﷮2﷯﷮𝜃﷯﷯=0 𝑎 sin﷮𝜃﷯﷮ cos﷮2﷯﷮𝜃﷯﷯ = 𝑏 cos﷮𝜃﷯﷮ sin﷮2﷯﷮𝜃﷯﷯ a sin θ × sin2 θ = b cos θ cos2 θ a sin3 θ = b cos3 θ sin﷮3﷯﷮𝜃﷯﷮ cos﷮3﷯﷮𝜃﷯﷯= 𝑏﷮𝑎﷯ tan3 θ = 𝑏﷮𝑎﷯ tan θ = 𝑏﷮𝑎﷯﷯﷮ 1﷮3﷯﷯ Finding 𝑑﷮2﷯𝑙﷮ 𝑑﷮2﷯𝜃﷯ 𝑑𝑙﷮𝑑𝜃﷯=𝑎 sec﷮𝑡𝑎𝑛𝜃−𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃 cot﷮𝜃﷯﷯ Differentiating w.rt θ 𝑑﷮2﷯𝑙﷮ 𝑑﷮2﷯𝜃﷯= 𝑑 𝑎 sec﷮𝜃 tan﷮𝜃 − 𝑏 𝑐𝑜𝑠𝑒𝑐𝜃 cot﷮𝜃﷯﷯﷯﷯﷮𝑑𝜃﷯ = a 𝑑 sec﷮𝜃 tan﷮𝜃﷯﷯﷯﷮𝑑𝜃﷯−𝑏 𝑑 𝑐𝑜𝑠𝑒𝑐 𝜃 cot﷮𝜃﷯﷯﷮𝑑𝜃﷯ = a sec﷮𝜃﷯﷯﷮′﷯ tan﷮𝜃﷯+ tan﷮𝜃﷯﷯﷮′﷯ sec﷮𝜃﷯﷯−𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃﷯﷮′﷯ cot﷮𝜃﷯+𝑐𝑜𝑠𝑒𝑐 𝜃﷯ = a sec﷮𝜃﷯. tan﷮𝜃. tan﷮𝜃﷯+ sec﷮2﷯﷮𝜃. sec﷮𝜃﷯﷯﷯﷯−𝑏 −𝑐𝑜𝑠𝑒𝑐 𝜃. cot﷮𝜃﷯+ −𝑐𝑜𝑠𝑒 𝑐﷮2﷯𝜃﷯.𝑐𝑜𝑠𝑒𝑐 𝜃﷯ = a sec﷮𝜃 tan﷮2﷯﷮𝜃+ sec﷮2﷯﷮𝜃﷯﷯﷯﷯−𝑏 −𝑐𝑜𝑠𝑒𝑐 𝜃 cot﷮2﷯﷮𝜃−𝑐𝑜𝑠𝑒 𝑐﷮3﷯𝜃﷯﷯ = a sec θ tan﷮2﷯﷮𝜃+ sec﷮2﷯﷮𝜃﷯﷯﷯+𝑏 𝑐𝑜𝑠𝑒𝑐 𝜃 cot﷮2﷯𝜃﷮+𝑐𝑜𝑠𝑒 𝑐﷮2﷯𝜃﷯﷯ = a sec θ + b cosec θ Since θ is acute , i.e. 0 < θ < 𝜋﷮2﷯ ⇒ θ lie in 1st quadrant So, sec θ & cosec θ will be positive ⇒ 𝑑﷮2﷯𝑙﷮ 𝑑﷮2﷯𝜃﷯ > 0 ⇒ 𝑑﷮2﷯𝑙﷮ 𝑑﷮2﷯𝜃﷯ > 0 at tan θ = 𝑏﷮𝑎﷯﷯﷮ 1﷮3﷯﷯ ⇒ l is least when tan θ = 𝑏﷮𝑎﷯﷯﷮ 1﷮3﷯﷯ Now, tan θ = 𝑏﷮ 1﷮3﷯﷯﷮ 𝑎﷮ 1﷮3﷯﷯﷯ tan θ = ℎ𝑒𝑖𝑔ℎ𝑡﷮𝑏𝑎𝑠𝑒﷯ Height = 𝑏﷮ 1﷮3﷯﷯ & base 𝑎﷮ 1﷮3﷯﷯ Using Pythagoras theorem 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠 𝑒﷮2﷯=𝐻𝑒𝑖𝑔ℎ 𝑡﷮2﷯+𝐵𝑎𝑠 𝑒﷮2﷯ 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠 𝑒﷮2﷯= 𝑏﷮ 1﷮3﷯﷯﷯﷮2﷯+ 𝑎﷮ 1﷮3﷯﷯﷯﷮2﷯ 𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 = ﷮ 𝑎﷮ 2﷮3﷯﷯ + 𝑏﷮ 2﷮3﷯﷯﷯ Least value of l = a sec θ + b cosec θ = a × ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ﷮𝑏𝑎𝑠𝑒 ﷯+𝑏 × ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 ﷮ℎ𝑒𝑖𝑔ℎ𝑡 ﷯ = a × ﷮ 𝑎﷮ 2﷮3﷯﷯ + 𝑏﷮ 2﷮3﷯﷯﷯﷮ 𝑎﷮ 1﷮3﷯﷯﷯+b × ﷮ 𝑎﷮ 2﷮3﷯﷯ + 𝑏﷮ 2﷮3﷯﷯﷯﷮ 𝑏﷮ 1﷮3﷯﷯﷯ l = ﷮ 𝑎﷮ 2﷮3﷯﷯+ 𝑏﷮ 2﷮3﷯﷯﷯ 𝑎﷮1 − 1﷮3﷯﷯+ 𝑏﷮1 − 1﷮3﷯﷯﷯ l = ﷮ 𝑎﷮ 2﷮3﷯﷯+ 𝑏﷮ 2﷮3﷯﷯﷯ 𝑎﷮ 2﷮3﷯﷯+ 𝑏﷮ 2﷮3﷯﷯﷯ l = 𝑎﷮ 2﷮3﷯﷯+ 𝑏﷮ 2﷮3﷯﷯﷯﷮ 1﷮2﷯ + 1﷯ l = 𝑎﷮ 2﷮3﷯﷯+ 𝑏﷮ 2﷮3﷯﷯﷯﷮ 3﷮2﷯﷯ Hence l = 𝒂﷮ 𝟐﷮𝟑﷯﷯+ 𝒃﷮ 𝟐﷮𝟑﷯﷯﷯﷮ 𝟑﷮𝟐﷯﷯ Hence proved..

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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.
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