Question

Show that the product of perpendiculars on the line $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ from the points $(\pm \sqrt{a^2-b^2},0)$ is $b^2$.

Answer 1

Let $p_1$ and $p_1$ be the lengths of perpendicular draw from point $P(\sqrt{a^2-b^2},0)$

$Q(-\sqrt{a^2-b^2},0)$ on the line $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$.

Then,

$P_1=\left|\frac{\frac{\sqrt{a^2-b^2}}{a}\cos \theta +\frac{0}{b}\sin \theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|=\left|\frac{\frac{\sqrt{a^2-b^2}}{a}\cos \theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|$

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

$\therefore P_1{P}_2=\left|\frac{\frac{\sqrt{a^2-b^2}}{a}\cos \theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|\times \left|\frac{\frac{\sqrt{a^2-b^2}}{a}\cos \theta +1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|$

$\Rightarrow P_1{P}_2=\left|\frac{\left(\frac{\sqrt{a^2-b^2}}{a}\cos \theta -1\right)\left(\frac{\sqrt{a^2-b^2}}{a}\cos \theta +1\right)}{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}\right|$

$\Rightarrow P_1{P}_2=\left|\frac{\frac{a^2-b^2}{a}{\cos }^2\theta -1}{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}\right|=\left|\frac{(a^2-b^2){\cos }^2\theta -a^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|b^2$

$\Rightarrow P_1{P}_2=\left|\frac{-b^2{\cos }^2\theta -a^2{\sin }^2\theta }{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|b^2=b^2$