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Question

$$\mathrm{A}$$ variable line drawn through the point of intersection of the lines $$\displaystyle \frac{x}{a}+\frac{y}{b}=1,\frac{x}{b}+\frac{y}{a}=1$$ meets the coordinate axes in $${A}$$ and $${B}$$. Then the locus of the mid point of $$AB$$ is 


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

The correct option is A $$2xy({a}+{b})={a}{b}({x}+{y})$$
Let $$C$$ be the point of intersection of the lines. 
Then, $$\displaystyle C=\left( \frac { ab }{ a+b } +\frac { ab }{ a+b }  \right) $$
Let $$M(h,k)$$ be the mid- point of $$AB$$
Then, the equation of $$A$$ may be written as $$\displaystyle \frac { x }{ 2h } +\frac { y }{ 2k } =1$$
Since $$AB$$ passes through $$C,$$ we have
$$\displaystyle \frac { ab }{ 2h\left( a+b \right)  } +\frac { ab }{ 2k\left( a+b \right)  } =1\Rightarrow \frac { 1 }{ h } +\frac { 1 }{ k } =\frac { 2\left( a+b \right)  }{ ab } $$
Therefore locus of $$M(h,k)$$ is $$\displaystyle \frac { 1 }{ x } +\frac { 1 }{ y } =\frac { 2\left( a+b \right)  }{ ab } $$
$$\Rightarrow 2xy\left( a+b \right) =ab\left( x+y \right) $$

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