Question

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

Answer 1

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



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

Similarly,

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

Now,

$$\begin{gathered} d_1{d}_2=b\left|\frac{{\displaystyle \sqrt{a^2-b^2}\mathrm{cos\theta }-a}}{\sqrt{{\displaystyle a^2{\sin }^2\mathrm{\theta }}+{\displaystyle b^2{\cos }^2\mathrm{\theta }}}}\right|\times b\left|\frac{{\displaystyle \sqrt{a^2-b^2}\mathrm{cos\theta }+a}}{\sqrt{{\displaystyle a^2{\sin }^2\mathrm{\theta }}+{\displaystyle b^2{\cos }^2\mathrm{\theta }}}}\right| \\ \Rightarrow d_1{d}_2=b^2\left|\frac{{{\displaystyle \left(a^2-b^2\right){\cos }^2\theta -a}}^2}{a^2{\sin }^2\mathrm{\theta }+b^2{\cos }^2\mathrm{\theta }}\right| \\ \Rightarrow d_1{d}_2=b^2\left|\frac{a^2\left({\cos }^2\mathrm{\theta }-1\right){{\displaystyle -b}}^2{\cos }^2\mathrm{\theta }}{a^2{\sin }^2\mathrm{\theta }+b^2{\cos }^2\mathrm{\theta }}\right| \\ \Rightarrow d_1{d}_2=b^2\left|\frac{-a^2{\sin }^2\mathrm{\theta }{{\displaystyle -b}}^2{\cos }^2\mathrm{\theta }}{a^2{\sin }^2\mathrm{\theta }+b^2{\cos }^2\mathrm{\theta }}\right|=b^2\left|\frac{a^2{\sin }^2\mathrm{\theta }{{\displaystyle +b}}^2{\cos }^2\mathrm{\theta }}{a^2{\sin }^2\mathrm{\theta }+b^2{\cos }^2\mathrm{\theta }}\right|=b^2 \end{gathered}$$