# Prove that the product of the lengths of the perpendiculars drawn from the points

Question:

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

Solution:

The equation of the given line is

$\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1$

Or, $b x \cos \theta+a y \sin \theta-a b=0$ $\ldots(1)$

Length of the perpendicular from point $\left(\sqrt{a^{2}-b^{2}}, 0\right)$ to line (1) is $p_{1}=\frac{\left|b \cos \theta\left(\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{\left|b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$ $\ldots(2)$

Length of the perpendicular from point $\left(-\sqrt{a^{2}-b^{2}}, 0\right)$ to line $(2)$ is

$p_{2}=\frac{\left|b \cos \theta\left(-\sqrt{a^{2}-b^{2}}\right)+a \sin \theta(0)-a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}=\frac{\left|b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right|}{\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}}$ $(3)$

On multiplying equations (2) and (3), we obtain

$p_{1} p_{2}=\frac{\left|b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right|\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right)\right|}{\left(\sqrt{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}\right)^{2}}$

$=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}-a b\right)\left(b \cos \theta \sqrt{a^{2}-b^{2}}+a b\right)\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}$

$=\frac{\left|\left(b \cos \theta \sqrt{a^{2}-b^{2}}\right)^{2}-(a b)^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}$

$=\frac{\left|b^{2} \cos ^{2} \theta\left(a^{2}-b^{2}\right)-a^{2} b^{2}\right|}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}$

$=\frac{\left|a^{2} b^{2} \cos ^{2} \theta-b^{4} \cos ^{2} \theta-a^{2} b^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$

$=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2}\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$

$=\frac{b^{2}\left|a^{2} \cos ^{2} \theta-b^{2} \cos ^{2} \theta-a^{2} \sin ^{2} \theta-a^{2} \cos ^{2} \theta\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta} \quad\left[\sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$=\frac{b^{2}\left|-\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)\right|}{b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta}$

$=\frac{b^{2}\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}{\left(b^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)}$

$=b^{2}$

Hence, proved.