Question

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

Answer 1

Let 

$$d_1\text{ and }{d}_2$$ be the perpendicular distances of line $$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$$  from points $$(\sqrt{a^2-b^2},0)\text{ and }(-\sqrt{a^2-b^2},0)$$ ,respectively.

$$\therefore d_1=\left|\frac{\frac{\sqrt{a^2-b^2}}{a}cos\theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|=b\left|\frac{\sqrt{a^2-b^2}cos\theta -a}{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}\right|$$

Similarly,

$$d_1=\left|\frac{-\frac{\sqrt{a^2-b^2}}{a}cos\theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\right|=b\left|\frac{-\sqrt{a^2-b^2}cos\theta -a}{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}\right|=b\left|\frac{\sqrt{a^2-b^2}cos\theta +a}{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}\right|$$

Now,

$$d_1{d}_2=b\left|\frac{\sqrt{a^2-b^2}cos\theta -a}{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}\right|\times b\left|\frac{\sqrt{a^2-b^2}cos\theta +a}{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}\right|$$

$$\Rightarrow d_1{d}_2=b^2\left|\frac{(a^2-b^2){\cos }^2\theta -a^2}{a^2{\sin }^2\theta +b^2{\cos }^2\theta }\right|$$

$$\Rightarrow d_1{d}_2=b^2\left|\frac{a^2({\cos }^2\theta -1){-b}^2{\cos }^2\theta }{a^2{\sin }^2\theta +b^2{\cos }^2\theta }\right|$$

$$\Rightarrow d_1{d}_2=b^2\left|\frac{-a^2{\sin }^2\theta {-b}^2{\cos }^2\theta }{a^2{\sin }^2\theta +b^2{\cos }^2\theta }\right|=b^2\left|\frac{a^2{\sin }^2\theta {+b}^2{\cos }^2\theta }{a^2{\sin }^2\theta +b^2{\cos }^2\theta }\right|=b^2$$.