Question

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

Answer 1

Given line

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

Let $p_1$ is the perpendicular distance of given line from point $(\sqrt{a^2-b^2},0)$

Hence by formula $p=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$p_1=\left|\frac{\frac{\cos \theta \sqrt{a^2-b^2}}{a}-1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|$

$p_1=\left|\frac{\cos \theta \sqrt{a^2-b^2}-a}{\frac{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}{b}}\right|$

Let $p_2$ is the perpendicular distance of given line from point $(-\sqrt{a^2-b^2},0)$

Hence by formula $p=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$p_2=\left|\frac{\frac{-\cos \theta \sqrt{a^2-b^2}}{a}-1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|$

$p_2=\frac{\cos \theta \sqrt{a^2-b^2}+a}{\frac{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}{b}}$

Product of length $p_1{p}_2$

$p_1{p}_2=\left(\frac{\cos \theta \sqrt{a^2-b^2}-a}{\frac{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}{b}}\right)\left(\frac{\cos \theta \sqrt{a^2-b^2}+a}{\frac{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}{b}}\right)$

$p_1{p}_2=\left|\left(\frac{b^2\left(\cos {\theta }^2\right(a^2-b^2)-a^2)}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right)\right|$

Since $si{n}^2\theta =1-{\cos }^2\theta$

$p_1{p}_2=\left|\left(\frac{b^2\left(\cos {\theta }^2\right(a^2-b^2)-a^2)}{b^2{\cos }^2\theta +a^2-a^2{\cos }^2\theta }\right)\right|$

$p_1{p}_2=\left(\frac{b^2\left(\cos {\theta }^2\right(a^2-b^2)-a^2)}{{\cos }^2\theta (a^2-b^2)-a^2}\right)$

$p_1{p}_2=b^2$