Two straight lines rotate about two fixed points (-a, O) and (a,0) in anti clockwise direction. If they start from their position of coincidence such that one rotates at a rate double of the other, then locus of curve is 

$\text{ Let }PB\text{ make angle }\theta \text{ and }PA\text{ make angle }2\theta \text{ with }x\text{ -axis. Let }P\text{ is }(h,k)\text{ . }$

$slope \tan \theta =\frac{k}{h+a}-----\left(1\right)$

$\tan 2\theta =\frac{k}{h-a}----\left(2\right)$

From (2) 

$\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{k}{h-a}$

Locus on solving is 

$\frac{2k(h+a)}{{(h+a)}^2-k^2}=\frac{k}{h-a}$

$2k(h^2-a^2)=k{(h+a)}^2-k^3$

$\begin{array}{r}{h}^2+k^2-2ah-3a^2=0\\ x^2+y^2-2ax-3a^2=0\end{array}$