The locus of the foot of the perpendicular from the origin on each member of the family $(4a+3)x-(a+1)y-$$(2a+1)=0\text{ is }$

detailed solution

Correct option is D

Given family of lines is (4a + 3)x - (a + 1)y - (2a + 1) = 0     or     (3x-y - 1) + a (4x-y -2) = 0 Family of lines passes through the fixed point P which is the intersection of 3x−y=1 and 4x−y=2Solving we get P(1,2).Now let (h, k) be the foot of perpendicular drawn from 0,0 on each of the family. The line passing through origin is perpendicular to the line passing through (1,2)kh⋅k−2h−1=−1k(k-2)=-h(h-1)∴  Locus of h,k is x(x−1)+y(y−2)=0 or (2x−1)2+4(y−1)2=5