bx/a+ay/b=a2+b2 x/y=2ab

bxa+ayb=a2+b2 1xy=2ab 2From 2 x= 2abySubstituting the value of x in 1 , we getb2 abya+ayb=a2+b22b2y+ayb=a2+b22b3+ayb=a2+b2y=ba2+b22b3+aSo, x=2abba2+b22b3+a=2ab2a2+b22b3+aSo, value of x=2ab2a2+b22b3+a and y=ba2+b22b3+a

bx/a+ay/b=a²+b²
x/y=2ab

\frac{bx}{a} + \frac{ay}{b} = a^{2} + b^{2} . . . (1) \frac{x}{y} = 2 ab . . . (2) From (2) x = 2 aby Substituting the value of x in (1), we get \frac{b (2 aby)}{a} + \frac{ay}{b} = a^{2} + b^{2} 2 b^{2} y + \frac{ay}{b} = a^{2} + b^{2} \frac{(2 b^{3} + a) y}{b} = a^{2} + b^{2} y = \frac{b (a^{2} + b^{2})}{(2 b^{3} + a)} So, x = 2 ab (\frac{b (a^{2} + b^{2})}{(2 b^{3} + a)}) = \frac{2 {ab}^{2} (a^{2} + b^{2})}{(2 b^{3} + a)} So, value of x = \frac{2 {ab}^{2} (a^{2} + b^{2})}{(2 b^{3} + a)} and y = \frac{b (a^{2} + b^{2})}{(2 b^{3} + a)} .

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