If x - y = x^2 + y^2 and ^2 a^2 + ^2 b^2 = 1x^2 + y^2 , then the correct relation is?

Answer: Option B

$$\begin{array}{rl}& x\cos \theta -y\sin \theta =\sqrt{x^2+y^2}.....(i)\\ & \frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}=\frac{1}{x^2+y^2}.....(ii)\\ & \frac{x}{\sqrt{x^2+y^2}}\cos \theta +\frac{-y}{\sqrt{x^2+y^2}}\sin \theta =1\\ & \text{from equation (i)}\\ & \Rightarrow \sin \theta =\frac{-y}{\sqrt{x^2+y^2}}\\ & \Rightarrow \cos \theta =\frac{x}{\sqrt{x^2+y^2}}\\ & \text{Put value in equation (ii)}\\ & \therefore \frac{\text{co}{\text{s}}^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}=\frac{1}{x^2+y^2}\\ & \Rightarrow \frac{x^2}{\left(x^2+y^2\right)a^2}+\frac{y^2}{\left(x^2+y^2\right)b^2}=\frac{1}{x^2+y^2}\\ & \Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\end{array}$$