Question

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

Answer 1

Equation of the given line is
$x\left(b\cos \theta \right)+y\left(a\sin \theta \right)-ab=0$
$\therefore p_1=\frac{\sqrt{(a^2-b^2)}\left(b\cos \theta \right)-ab}{\surd (b^2{\cos }^2\theta +a^2{\sin }^2\theta )}$
$\therefore p_1=\frac{-\sqrt{(a^2-b^2)}\left(b\cos \theta \right)-ab}{\surd (b^2{\cos }^2\theta +a^2{\sin }^2\theta )}$
$\therefore p_1{p}_2=-\frac{(a^2-b^2)b^2{\cos }^2\theta -a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$
$\because (L+M)(L-M)=L^2-M^2$
$=b^2\frac{[a^2-a^2{\cos }^2\theta +b^2{\cos }^2\theta ]}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$
$=b^2\frac{[a^2{\sin }^2\theta +b^2{\cos }^2\theta ]}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }=b^2$