(2) Consider a tangent line to the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$, which intersects the two symmetry axes of the ellipse at points $A$ and $B$, respectively. Then the minimum length of the line segment $AB$ is
(2) 8 Hint: Let the point of tangency be $P(5 \cos \theta, 3 \sin \theta)$, then the equation of the tangent line to the ellipse at point $P$ is $\frac{\cos \theta}{5} x+\frac{\sin \theta}{3} y=1$, which intersects the $x$-axis at $A\left(\frac{5}{\cos \theta}, 0\right)$ and the $y$-axis at $B\left(0, \frac{3}{\sin \theta}\right)$, so
$$
A B^{2}=\frac{5^{2}}{\cos ^{2} \theta}+\frac{3^{2}}{\sin ^{2} \theta} ;
$$

By the Cauchy-Schwarz inequality,
$$
\begin{aligned}
\frac{5^{2}}{\cos ^{2} \theta}+\frac{3^{2}}{\sin ^{2} \theta} & =\left(\frac{5^{2}}{\cos ^{2} \theta}+\frac{3^{2}}{\sin ^{2} \theta}\right)\left(\cos ^{2} \theta+\sin ^{2} \theta\right) \\
& \geqslant(5+3)^{2}=8^{2},
\end{aligned}
$$

Therefore, $|A B| \geqslant 8$ (the equality holds when $\tan ^{2} \theta=\frac{3}{5}$).