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MHT CET202121 Sep 2021Morning ShiftMathematicsApplication of DerivativesActual

The curves x^2 a^2 + y^2 4 =4 and y^3=16 x intersect each other orthogonally, then a ^2=

Options

  1. A2
  2. B3 4
  3. C1 2
  4. D4 3

Correct answer

D. 4 3

Step-by-step solution

aligned & x^2 a^2 + y^2 4 =1 & 1 a^2 2 x+ 1 4 2 y d y d x =0 d y d x = ( -x a^2 ) ( 4 y ) aligned Also y ^3=16 x 3 y^2 d y d x =16 d y d x = 16 3 y^2 Since curves intersect orthogonally, from (1) and (2), we write ( -x a^2 ) ( 4 y ) ( 16 3 y^2 )=-1 64 x 3 a^2 y^3 =1 and we have y^3=16 x 64 x 3 a ^2(16 x ) =1 a ^2= 4 3

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