Question

The product of perpendiculars drawn 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:

• A. $a^2-b^2$
• B. $a^2+b^2$
• C. $a^2$
• D. $b^2$

Answer 1

The correct option is D $b^2$

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

$⟹bx\cos \theta +ay\sin \theta =ab$

$⟹bx\cos \theta +ay\sin \theta -ab=0$ (equation $1$)

Length of perpendicular from the point $(\sqrt{a^2-b^2},0)$

$P_1=\frac{|b\cos \theta (\sqrt{a^2-b^2}+a\sin \theta (0)-ab)|}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}=\frac{|b\cos \theta \sqrt{a^2-b^2}-ab|}{\sqrt{b^2{\cos }^2\theta +a^2{\sin }^2\theta }}$ (equation $2$)

Length of perpendicular from the point $(-\sqrt{a^2-b^2},0)$ is,

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

Product $(P_1,P_2)$

$=\frac{|b^2{\cos }^2\theta (a^2-b^2)-a^2{b}^2|}{b^2{\cos }^2\theta }$

$=\frac{a^2{b}^2{\cos }^2\theta -b^4{\cos }^2\theta -a^2{b}^2}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$

$=\frac{b^2|(a^2{\cos }^2\theta -b^2{\cos }^2\theta -a^2)|}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$

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

$=\frac{b^2|-a^2{\sin }^2\theta -b^2{\cos }^2\theta |}{b^2{\cos }^2\theta +a^2{\sin }^2\theta }$

$=b^2\frac{(a^2{\sin }^2\theta +b^2{\cos }^2\theta )}{(b^2{\cos }^2\theta +a^2{\sin }^2\theta )}$

$=b^2$