Question

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

• A. $(2 x-1)^{2}+4(y+1)^{2}=5$
• B. $(2 x-1)^{2}+(y+1)^{2}=5$
• C. $(2 x+1)^{2}+4(y-1)^{2}=5$
• D. $(2 x-1)^{2}+4(y-1)^{2}=5$

Answer 1

Answer: (D)

Given family of lines is $(4 a+3) x-(a+1) y-(2 a+1)=0$
or $(3 x-y-1)+a(4 x-y-2)=0$
Family of lines passes through the fixed point $P$ which is the intersection of
$3 x-y=1 $ and $ 4 x-y=2$
Solving we get $P(1,2)$.

Now let $(h, k)$ be the foot of perpendicular on each of the family.
$\therefore \frac{k}{h} \cdot \frac{k-2}{h-1}=-1$
$\therefore $ Locus is $x(x-1)+y(y-2)=0$
or $(2 x-1)^{2}+4(y-1)^{2}=5$