# 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 $b^2$

Toolbox:
• Length of the perpendicular from a point $(x_1, y_1)$ is $d = \large\frac{|Ax_1+By_1+c|}{\sqrt{A^2+B^2}}$
The given equation is
$\large\frac{x}{a}$$\cos \theta + \large\frac{y}{b}$$ \sin \theta = 1$
This can be written as
$bx \cos \theta + ay \sin \theta = ab$
or $bx \cos \theta + ay \sin \theta - ab = 0$------(1)