If x^2+y^2=t+ 1 t and x^4+y^4=t^2+ 1 t^2 then d y d x =

Step-by-step solution

Given x^2 + y^2 = t + 1 t and x^4 + y^4 = t^2 + 1 t^2 . Squaring the first equation: (x^2 + y^2)^2 = (t + 1 t )^2 x^4 + y^4 + 2x^2y^2 = t^2 + 1 t^2 + 2 Substituting x^4 + y^4 = t^2 + 1 t^2 into the expanded equation: (t^2 + 1 t^2 ) + 2x^2y^2 = t^2 + 1 t^2 + 2 2x^2y^2 = 2 x^2y^2 = 1 Differentiating x^2y^2 = 1 with respect to x : d dx (x^2y^2) = d dx (1) x^2(2y dy dx ) + y^2(2x) = 0 2x^2y dy dx = -2xy^2 dy dx = - 2xy^2 2x^2y = - y x Answer: - y x