Question

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

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

Answer 1

The correct option is D $(2x-1{)}^2+4(y-1{)}^2=5$

$(3x-y-1)+a(4x-y-2)=0$
This family of lines passes through a fixed point P which is the intersection of 3x-y=1 and 4x-y=2
Solving we get P=(1, 2)
Let (h, k) be the foot of perpendicular on each of the family
$\frac{k}{h}.\frac{k-2}{h-1}=-1$
Locus is
$x.(x-1)+y(y-2)=0$
$(2x-1{)}^2+4(y-1{)}^2=5$