If ⁻¹ ( x^2 + y^2 x^2 - y^2 ) = 2a such that y dy dx = x f(a) then the value of f ( 2 3 ) is

Step-by-step solution

Given ⁻¹ ( x^2 + y^2 x^2 - y^2 ) = 2a x^2 + y^2 x^2 - y^2 = (2a) Applying componendo and dividendo: (x^2 + y^2) - (x^2 - y^2) (x^2 + y^2) + (x^2 - y^2) = (2a) - 1 (2a) + 1 2y^2 2x^2 = 1 (2a) - 1 1 (2a) + 1 y^2 x^2 = 1 - (2a) 1 + (2a) y^2 x^2 = 2 ^2(a) 2 ^2(a) = ^2(a) y^2 = x^2 ^2(a) Differentiating both sides with respect to x : 2y dy dx = 2x ^2(a) y dy dx = x ^2(a) Comparing with y dy dx = x f(a) , we get: f(a) = ^2(a) Substituting a = 2 3 : f ( 2 3 ) = ^2 ( 2 3 ) = (- 3 )^2 = 3 Answer: 3