Question

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 $b^2$

Answer 1

Let $P_1$ and $p_2$ be the length of perpendiculars from $(\sqrt{(a^2-b^2},0)$ and $(-\sqrt{(a^2-b^2},0)$ to the line
$\frac{x}{a}cos\theta +\frac{y}{b}sin\theta =1$.

$\therefore p_1\left|\frac{\frac{\sqrt{a^2-b^2}cos\theta }{a}+\frac{0\times sin\theta }{b}-1}{\sqrt{{\left(\frac{cos\theta }{a}\right)}^2}+{\left(\frac{sin\theta }{b}\right)}^2}\right|$

$=\left|\frac{\frac{\sqrt{a^2-b^2}cos\theta }{a}-1}{\sqrt{\frac{co{s}^2\theta }{a^2}+\frac{si{n}^2\theta }{b^2}}}\right|$

$P_2=\left|\frac{-\frac{\sqrt{a^2-b^2}cos\theta }{a}+\frac{0\times sin\theta }{b}-1}{\sqrt{{\left(\frac{cos\theta }{a}\right)}^2}+{\left(\frac{sin\theta }{b}\right)}^2}\right|$

$=\left|\frac{\frac{-\sqrt{a^2-b^2}cos\theta }{a}-1}{\sqrt{\frac{co{s}^2\theta }{a^2}+\frac{si{n}^2\theta }{b^2}}}\right|$

Now $P_1{P}_2$

$=\left|\frac{\frac{\sqrt{a^2-b^2}cos\theta }{a}-1}{\sqrt{\frac{co{s}^2\theta }{a}+\frac{si{n}^2\theta }{b^2}}}\right|\left|\frac{\frac{-\sqrt{a^2-b^2}cos\theta }{a}-1}{\sqrt{\frac{co{s}^2\theta }{a^2}+\frac{si{n}^2\theta }{b^2}}}\right|$

$=\frac{\left|[\frac{\sqrt{a^2-b^2}cos\theta }{a}-1][\frac{\sqrt{a^2-b^2}cos\theta }{a}+1]\right|}{\frac{co{s}^2\theta }{a^2+}+\frac{si{n}^2\theta }{b^2+}}$

$=\frac{\left|[\frac{(a^2-b^2)co{s}^2\theta -1}{a^2}-1]\right|}{\frac{co{s}^2\theta }{a^2}+\frac{1-co{s}^2\theta }{b^2}}$

$=\frac{\left|[\frac{(a^2-b^2)co{s}^2\theta }{a^2}-1]\right|}{\frac{b^{2}co{s}^2\theta +a^2-a^{2}co{s}^2\theta }{a^2{b}^2}}$

$=\frac{|a^2-(a^2-b^2\left)co{s}^2\theta \right|}{\frac{a^2-(a^2-b^2)co{s}^2\theta }{b^2}}$

$=a^2-(a^2-b^2)co{s}^2\theta \times \frac{b^2}{a^2-(a^2-b^2)co{s}^2\theta }$

$=b^2$