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 `x/a``costheta``+``y/b``sintheta=1`is `b^2`.

Explanation

To 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 \(b^2\), we will follow these steps:

### Step 1: Identify the points and the line equation
The points from which we need to draw the perpendiculars are:
- \(P_1 = (\sqrt{a^2 - b^2}, 0)\)
- \(P_2 = (-\sqrt{a^2 - b^2}, 0)\)

The line equation can be rewritten in the standard form:
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