Prove that the product of the lengths of the perpendiculars drawn from the points $(\sqrt{a^{2}-b^{2}}, 0)$ and $(-\sqrt{a^{2}-b^{2}}, 0)$ to the line $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$ is $b^{2}$.

The equation of the given line is $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$.
Multiplying by $ab$,we get $bx \cos \theta+ay \sin \theta-ab=0$.....$(1)$
Let $p_{1}$ be the length of the perpendicular from $(\sqrt{a^{2}-b^{2}}, 0)$ to line $(1)$:
$p_{1}=\frac{|b \cos \theta(\sqrt{a^{2}-b^{2}})+a \sin \theta(0)-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$.....$(2)$
Let $p_{2}$ be the length of the perpendicular from $(-\sqrt{a^{2}-b^{2}}, 0)$ to line $(1)$:
$p_{2}=\frac{|b \cos \theta(-\sqrt{a^{2}-b^{2}})+a \sin \theta(0)-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|-b \cos \theta \sqrt{a^{2}-b^{2}}-ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}+ab|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$.....$(3)$
Multiplying $p_{1}$ and $p_{2}$:
$p_{1} p_{2}=\frac{|b \cos \theta \sqrt{a^{2}-b^{2}}-ab| \cdot |b \cos \theta \sqrt{a^{2}-b^{2}}+ab|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$
$p_{1} p_{2}=\frac{|(b \cos \theta \sqrt{a^{2}-b^{2}})^{2}-(ab)^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} = \frac{|b^{2} \cos ^{2} \theta(a^{2}-b^{2})-a^{2}b^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$
$p_{1} p_{2}=\frac{|a^{2}b^{2} \cos ^{2} \theta-b^{4} \cos ^{2} \theta-a^{2}b^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} = \frac{b^{2}|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$
Using $a^{2} = a^{2}(\sin^{2}\theta + \cos^{2}\theta)$,the numerator becomes $b^{2}|a^{2} \cos^{2}\theta - b^{2} \cos^{2}\theta - a^{2} \sin^{2}\theta - a^{2} \cos^{2}\theta| = b^{2}|-(b^{2} \cos^{2}\theta + a^{2} \sin^{2}\theta)|$.
Thus,$p_{1} p_{2} = \frac{b^{2}(b^{2} \cos^{2}\theta + a^{2} \sin^{2}\theta)}{b^{2} \cos^{2}\theta + a^{2} \sin^{2}\theta} = b^{2}$.