1. Given positive real numbers $a, b (a>b)$, the product of the distances from the points $\left(\sqrt{a^{2}-b^{2}}, 0\right)$ and $\left(-\sqrt{a^{2}-b^{2}}, 0\right)$ to the line
$$
\frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1
$$
is $\qquad$
$-1 . b^{2}$.
The equation of the straight line can be written as
$$
\begin{array}{l}
b \cos \theta \cdot x + a \sin \theta \cdot y - a b = 0 \text{.} \\
\text{Then } d_{1} d_{2} \\
=\frac{\left|b \sqrt{a^{2}-b^{2}} \cos \theta - a b\right|}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta} \text{.} \\
\left|-b \sqrt{a^{2}-b^{2}} \cos \theta - a b\right| \\
=\frac{\left|a^{2} b^{2} - b^{2}\left(a^{2}-b^{2}\right) \cos ^{2} \theta\right|}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta} \\
=\frac{\left|a^{2} b^{2}\left(1-\cos ^{2} \theta\right) + b^{4} \cos ^{2} \theta\right|}{b^{2} \cos ^{2} \theta + a^{2} \sin ^{2} \theta} = b^{2} \text{.} \\
\end{array}
$$