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The locus of the foot of the perpendicul...

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`

Text Solution

Verified by Experts

The correct Answer is:
C
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`
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