Question

If $p,q$ are the lengths of the perpendiculars from the origin to the straight lines

$x\sec \alpha +y\text{cosec}\alpha =a$ and $x\cos \alpha -y\sin \alpha =a\cos 2\alpha$ , prove that $4p^2+q^2=a^2$

Answer 1

Length of perpendicular from origin,

$=\frac{-c}{\sqrt{a^2+b^2}}$

$x\sec \alpha +y\csc \alpha =a$

$\frac{x}{\cos \alpha }+\frac{y}{\sin \alpha }=a$

$\Rightarrow x\sin \alpha +y\cos \alpha -a\cos \alpha \sin \alpha =0$

$p=-\frac{-a\cos \alpha \sin \alpha }{\sqrt{{\sin }^2\alpha +{\cos }^2\alpha }}=a\cos \alpha \sin \alpha$

$x\cos \alpha +y\sin \alpha =a\cos 2\alpha \Rightarrow q=a\cos 2\alpha$

$\therefore p=a\cos \alpha \sin \alpha ,q=a\cos 2\alpha$

$\therefore 4p^2+q^2$

$=4(a\cos \alpha \sin \alpha {)}^2+(a\cos 2\alpha {)}^2$

$=4a^2{\cos }^2\alpha {\sin }^2\alpha +a^2{\cos }^{2}2\alpha$

$=4a^2{\cos }^2\alpha {\sin }^2\alpha +a^2-4a^2{\sin }^2\alpha {\cos }^2\alpha$

$=a^2$

$\therefore 4p^2+q^2=a^2$