1. [10] Let $a$ and $b$ be real numbers such that $\frac{a b}{a^{2}+b^{2}}=\frac{1}{4}$. Find all possible values of $\frac{\left|a^{2}-b^{2}\right|}{a^{2}+b^{2}}$.
Answer: $\frac{\sqrt{3}}{2}$ The hypothesis statement is equivalent to
$$
\begin{array}{c}
a^{2}+b^{2}=4 a b \\
1:(a+b)^{2}=6 a b \\
2:(a-b)^{2}=2 a b
\end{array}
$$

Multiplying equations 1 and 2 ,
$$
\begin{array}{l}
\left(a^{2}-b^{2}\right)^{2}=12(a b)^{2} \\
\left|a^{2}-b^{2}\right|= \pm \sqrt{12} a b
\end{array}
$$

Since the left hand side and $a b$ are both positive,
$$
\begin{array}{c}
\left|a^{2}-b^{2}\right|=\sqrt{12} a b \\
\frac{\left|a^{2}-b^{2}\right|}{a^{2}+b^{2}}=\frac{\sqrt{12} a b}{a^{2}+b^{2}}=\frac{\sqrt{12}}{4}=\frac{\sqrt{3}}{2}
\end{array}
$$
(It is clear that such $a$ and $b$ exist: for example, we can take $a=1$ and solve for $b$ by way of the quadratic formula.)