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Misc 15 If (𝑥 – 𝑎)^2+ (𝑦 – 𝑏)^2= 𝑐2, for some 𝑐 > 0, prove that 〖[1 + (𝑑𝑦/𝑑𝑥)^2 ]/((𝑑^2 𝑦)/〖𝑑𝑥〗^2 )〗^(3/2)is a constant independent of a and b.First we will calculate 𝑑𝑦/𝑑𝑥 (𝑥 – 𝑎)^2+ (𝑦 – 𝑏)^2= 𝑐2 Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑((𝑥 – 𝑎)^2+ (𝑦 – 𝑏)^2 )/𝑑𝑥 = 𝑑(𝑐^2 )/𝑑𝑥 𝑑((𝑥 – 𝑎)^2 )/𝑑𝑥 +" " 𝑑((𝑦 – 𝑏)^2 )/𝑑𝑥 = 0 2(𝑥 – 𝑎). 𝑑(𝑥 − 𝑎)/𝑑𝑥 + 2 (𝑦 – 𝑏). 𝑑(𝑦 − 𝑏)/𝑑𝑥 = 0 2 (𝑥 – 𝑎) (1 −0) + 2(𝑦 – 𝑏) . (𝑑𝑦/𝑑𝑥 −0) = 0 2 (𝑥 – 𝑎) + 2(𝑦 – 𝑏) . (𝑑𝑦/𝑑𝑥) = 0 2(𝑦 – 𝑏) . 𝑑𝑦/𝑑𝑥 = −2 (𝑥 – 𝑎) 𝑑𝑦/𝑑𝑥 = (−2 (𝑥 – 𝑎))/2(𝑦 – 𝑏) 𝒅𝒚/𝒅𝒙 = (−(𝒙 − 𝒂))/(𝒚 − 𝒃) Again Differentiating 𝑤.𝑟.𝑡.𝑥. 𝑑/𝑑𝑥 (𝑑𝑦/𝑑𝑥) = 𝑑/𝑑𝑥 ((−(𝑥 − 𝑎))/(𝑦 − 𝑏)) (𝑑^2 𝑦)/(𝑑𝑥^2 ) = − 𝑑/𝑑𝑥 ((𝑥 − 𝑎)/(𝑦 − 𝑏)) (𝑑^2 𝑦)/(𝑑𝑥^2 )= − ((𝑑(𝑥 – 𝑎)/𝑑𝑥 (𝑦 – 𝑏) − 𝑑(𝑦 – 𝑏)/𝑑𝑥 . (𝑥 – 𝑎))/(𝑦 − 𝑏)^2 ) (𝑑^2 𝑦)/(𝑑𝑥^2 ) = − (((1 − 0) (𝑦 – 𝑏) − (𝑑𝑦/𝑑𝑥 − 0)(𝑥 – 𝑎))/(𝑦 − 𝑏)^2 ) Using Quotient rule As (𝑢/𝑣)′ = (𝑢^′ 𝑣 − 𝑣^′ 𝑢)/𝑣^2 where u = x − 𝑎 & v = y − b (𝑑^2 𝑦)/(𝑑𝑥^2 ) = − (((𝑦 – 𝑏) − (𝑑𝑦/𝑑𝑥)(𝑥 – 𝑎))/(𝑦 − 𝑏)^2 ) (𝑑^2 𝑦)/(𝑑𝑥^2 ) = − (((𝑦 – 𝑏) − (− (𝑥 – 𝑎))/((𝑦 – 𝑏) ) (𝑥 – 𝑎))/(𝑦 − 𝑏)^2 ) (𝑑^2 𝑦)/(𝑑𝑥^2 )= − (((𝑦 – 𝑏)^2 + (𝑥 – 𝑎)^2)/((𝑦 − 𝑏)^2 (𝑦 − 𝑏) )) (𝒅^𝟐 𝒚)/(𝒅𝒙^𝟐 )= (−𝒄^𝟐)/(𝒚 − 𝒃)^𝟑 Now, finding value of 〖[𝟏+ (𝒅𝒚/𝒅𝒙)^𝟐 ]/((𝒅^𝟐 𝒚)/〖𝒅𝒙〗^𝟐 )〗^(𝟑/𝟐) (Given (𝑥 – 𝑎)^2+ (𝑦 – 𝑏)^2= 𝑐2) 〖[𝟏+ (𝒅𝒚/𝒅𝒙)^𝟐 ]/((𝒅^𝟐 𝒚)/〖𝒅𝒙〗^𝟐 )〗^(𝟑/𝟐) Putting values = 〖[1+ ((−(𝑥 – 𝑎))/(𝑦 – 𝑏))^2 ]/((−𝑐^2)/(𝑦 − 𝑏)^3 )〗^(3/2) = − 〖[((𝑦 − 𝑏)^2 + (𝑥 – 𝑎)^2)/(𝑦 – 𝑏)^2 ]/(𝑐^2/(𝑦 − 𝑏)^3 )〗^(3/2) = − 〖[𝑐^2/(𝑦 – 𝑏)^2 ]/(𝑐^2/(𝑦 − 𝑏)^3 )〗^(3/2) = − [𝑐^2/(𝑦 – 𝑏)^2 ]^(3/2) × (𝑦 − 𝑏)^3/𝑐^2 = − (𝑐/(𝑦 – 𝑏))^(2 × 3/2) × (𝑦 − 𝑏)^3/𝑐^2 "= −" (𝑐/(𝑦 – 𝑏))^3 " × " (𝑦 − 𝑏)^3/𝑐^2 "= −" 𝑐^3/𝑐^2 × (𝑦 − 𝑏)^3/(𝑦 − 𝑏)^3 = −𝒄 = Constant Which is constant independent of a & b Hence proved

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