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Question

If a variable straight line is drawn through the point of intersection of xa+yb=1 and xb+ya=1 meets the co-ordinate axes at A and B, then the locus of the mid-point of line segment AB is

A
(x+y)(a+b)=2xy(ab)
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B
(x+y)(ab)=2xy(ab)
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C
(x+y)ab=2xy(a+b)
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D
(x+y)ab=2xy(ab)
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Solution

The correct option is C (x+y)ab=2xy(a+b)
Given lines
L1:xa+yb1=0 and L2:xb+ya1=0
Line passing through the point of intersection of both the lines is
L1+λL2=0(xa+yb1)+λ(xb+ya1)=0(bx+ayab)+λ(ax+byab)=0(b+λa)x+(a+λb)yab(1+λ)=0

Finding A and B,
A=(ab(1+λ)b+aλ,0)B=(0,ab(1+λ)a+bλ)

We know that (h,k) is the mid-point of AB, we get
(h,k)=(ab(1+λ)2(b+aλ),ab(1+λ)2(a+bλ))12h+12k=a+b+aλ+bλab(1+λ)12h+12k=a(1+λ)+b(1+λ)ab(1+λ)12h+12k=a+babh+khk=2(a+b)ab

Hence, the required locus is
(x+y)ab=2xy(a+b)

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