Question

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

Answer 1

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1bx\cos \theta +ay\sin \theta =abbx\cos \theta +ay\sin \theta -ab=0$

Let $p_1$ be the perpendicular from $(\sqrt{a^2-b^2},0)$

$p_1=\frac{|b\cos \theta \sqrt{a^2-b^2}-0(ay\sin \theta )-ab|}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}{p}_1=\frac{|b\cos \theta \sqrt{a^2-b^2}-ab|}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}$

$p_2$ be the perpendicular from $(-\sqrt{a^2-b^2},0)$

$p_2=\frac{|-b\cos \theta \sqrt{a^2-b^2}-0(ay\sin \theta )-ab|}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}{p}_2=\frac{|b\cos \theta \sqrt{a^2-b^2}+ab|}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}$

$p_1{p}_2=\left|\left(\frac{b\cos \theta \sqrt{a^2-b^2}-ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}\right)\left(\frac{b\cos \theta \sqrt{a^2-b^2}+ab}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}\right)\right|p_1{p}_2=\left|\frac{{\left(b\cos \theta \sqrt{a^2-b^2}\right)}^2-{\left(ab\right)}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|p_1{p}_2=\left|\frac{b^2{\cos }^2\theta (a^2-b^2)-a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|p_1{p}_2=\left|\frac{b^2\{{\cos }^2\theta (a^2-b^2)-a^2\}}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|p_1{p}_2=b^2\left|\frac{\{a^2{\cos }^2\theta -b^2{\cos }^2\theta -a^2\}}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|p_1{p}_2=b^2\left|\frac{\left\{a^2\right({\cos }^2\theta -1)-b^2{\cos }^2\theta \}}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|p_1{p}_2=b^2\left|\frac{{-b}^2{\cos }^2\theta -a^2{\sin }^2\theta }{b^2{\cos }^2\theta +a^2{\sin }^2\theta }\right|p_1{p}_2=b^2$