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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 

a
x2+y2+ax−3a2=0
b
x2−y2−2ax−3a2=0
c
x2+y2−2ax−3a2=0
d
x2−y2+2ax−3a2=0

detailed solution

Correct option is C

Let PB make angle θ and PA make angle 2θ with x -axis.  Let P is (h,k) . slope tan⁡θ=kh+a-----(1)tan⁡2θ=kh−a----(2)From (2) 2tan⁡θ1−tan2⁡θ=kh−aLocus on solving is 2k(h+a)(h+a)2-k2=kh-a 2k(h2-a2)=k(h+a)2-k3h2+k2−2ah−3a2=0x2+y2−2ax−3a2=0

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