$(a)$ Let $x, y$ be integers, not both zero. Find the minimum possible value of $|5x^2 + 11xy - 5y^2|$.
$(b)$ Find all positive real numbers $t$ such that $\frac{9t}{10}=\frac{[t]}{t - [t]}$.
1. Let \( f(x, y) = 5x^2 + 11xy - 5y^2 \). We need to find the minimum possible value of \( |f(x, y)| \) for integers \( x \) and \( y \) not both zero.
2. Note that \( -f(x, y) = f(y, -x) \). Therefore, it suffices to prove that \( f(x, y) = k \) for \( k \in \{1, 2, 3, 4\} \) has no integral solution.
3. First, consider \( f(x, y) \) modulo 2. If \( f(x, y) \) is even, then both \( x \) and \( y \) must be even. This implies \( 4 \mid f(x, y) \), ruling out \( k = 2 \). Additionally, if \( f(x, y) = 4 \) has a solution, then \( f(x, y) = 1 \) must also have a solution.
4. Next, consider \( f(x, y) \) modulo 3. If \( f(x, y) \equiv 0 \pmod{3} \), then \( x^2 + xy - y^2 \equiv 0 \pmod{3} \). If neither \( x \) nor \( y \) is divisible by 3, then \( x^2 - y^2 \equiv 0 \pmod{3} \) implies \( xy \equiv 0 \pmod{3} \), a contradiction. Therefore, if 3 divides one of \( x \) or \( y \), it must divide both, implying \( 9 \mid f(x, y) \), ruling out \( k = 3 \).
5. Now, consider \( f(x, y) = 1 \). This equation is equivalent to:
   \[
   (10x + 11y)^2 - 221y^2 = 20
   \]
   Let \( A = 10x + 11y \). Then:
   \[
   A^2 \equiv 20 \pmod{13}
   \]
   However, \( 20 \) is a quadratic nonresidue modulo 13, as shown by:
   \[
   \left( \frac{20}{13} \right) = \left( \frac{5}{13} \right) = \left( \frac{13}{5} \right) = \left( \frac{3}{5} \right) = -1
   \]
   Therefore, \( 20 \) is not a quadratic residue modulo 13, and no solution exists for \( f(x, y) = 1 \).
6. Consequently, \( |f(x, y)| \geq 5 \). The minimum value is achieved when \( x = 1 \) and \( y = 0 \), giving \( f(1, 0) = 5 \).

The final answer is \( \boxed{5} \).