1. Given real numbers $a, b$ satisfy $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{4}{a^{2}+b^{2}}$. Then the value of $\left(\frac{b}{a}\right)^{2008}-\left(\frac{a}{b}\right)^{2009}$ is ( ).
(A) -2
(B) 0
(C) 2
(D) 0 or 2
-、1.D.
Given the equation, after eliminating the denominator and rearranging, we get $\left(a^{2}-b^{2}\right)^{2}=0$. Therefore, $a=-b$ or $a=b$.
When $a=-b$, the original expression $=1+1=2$;
When $a=b$, the original expression $=1-1=0$.