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` (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`

Options

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

C

Explanation

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 =2`
Solving we get `P(1,2)`.

Now let (h,k) be the foot of perpendicular on each of the family.
`:. (k)/(h).(k-2)/(h-1) =-1`
`:.` Locus is `x (x-1) +y(y-2) =0`
or `(2x -1)^(2) +4(y-1)^(2) =5`