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 $ \large\frac{x}{a}$$ \cos \theta + \large\frac{y}{b}$$ \sin \theta = 1$ is $b^2$

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