Let $a$ and $b$ be real numbers such that $ \frac {ab}{a^2 + b^2} = \frac {1}{4} $. Find all possible values of $ \frac {|a^2-b^2|}{a^2+b^2} $.
1. Given the equation:
   \[
   \frac{ab}{a^2 + b^2} = \frac{1}{4}
   \]
   we start by multiplying both sides by \(4(a^2 + b^2)\):
   \[
   4ab = a^2 + b^2
   \]

2. We need to find the value of:
   \[
   \frac{|a^2 - b^2|}{a^2 + b^2}
   \]

3. From the given equation \(4ab = a^2 + b^2\), we can rewrite \(a^2 + b^2\) in terms of \(ab\):
   \[
   a^2 + b^2 = 4ab
   \]

4. Now, consider the expression \(|a^2 - b^2|\). We can use the identity:
   \[
   a^2 - b^2 = (a + b)(a - b)
   \]

5. Substitute \(a^2 + b^2 = 4ab\) into the expression we need to find:
   \[
   \frac{|a^2 - b^2|}{a^2 + b^2} = \frac{|(a + b)(a - b)|}{4ab}
   \]

6. To proceed, we need to express \(a + b\) and \(a - b\) in terms of \(a\) and \(b\). Let's solve for \(a\) and \(b\) using the quadratic equation:
   \[
   t^2 - (a + b)t + ab = 0
   \]
   The roots of this equation are \(a\) and \(b\). Given \(a^2 + b^2 = 4ab\), we can use the identity:
   \[
   (a + b)^2 = a^2 + 2ab + b^2 = 6ab
   \]
   Therefore:
   \[
   a + b = \sqrt{6ab}
   \]

7. Similarly, we can find \(a - b\) using:
   \[
   (a - b)^2 = a^2 - 2ab + b^2 = 2ab
   \]
   Therefore:
   \[
   a - b = \sqrt{2ab}
   \]

8. Substitute these into the expression:
   \[
   \frac{|(a + b)(a - b)|}{4ab} = \frac{|\sqrt{6ab} \cdot \sqrt{2ab}|}{4ab} = \frac{\sqrt{12(ab)^2}}{4ab} = \frac{\sqrt{12}ab}{4ab} = \frac{\sqrt{12}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}
   \]

The final answer is \(\boxed{\frac{\sqrt{3}}{2}}\).