If x^k+y^k=a^k(a, k>0) and d y d x + ( y x )^ 1 b -0 , then k has the value

Step-by-step solution

x^ k +y^ k = a ^ k Differentiating w.r.t. x , we get aligned & k x^ k-1 + k y^ k -1 d y ~d x =0 & d y ~d x =- k x^ k -1 k y^ k -1 & d y ~d x =- ( x y )^ k -1 & d y ~d x + ( y x )^ 1- k =0 & But d y ~d x + ( y x )^ 1 3 =0 ...[Given] aligned Comparing above equations, we get aligned 1-k & = 1 3 & 1- 1 3 =k k & = 2 3 aligned