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Question

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


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

The correct option is D $$2xy (a + b) = ab(x + y)$$
we have,
$$\dfrac { x }{ a } +\dfrac { y }{ b } =1\quad and\quad \dfrac { x }{ b } +\dfrac { y }{ a } =1$$
$$ \Rightarrow bx+ay-ab=0\longrightarrow \left( 1 \right) $$
$$and, \, ax+by-ab=0\longrightarrow \left( 2 \right) \\ $$

Equation of line through their point of intersection will be given by:

$$\left( bx+ay-ab \right) +k\left( ax+by-ab \right) =0$$

where $$k$$ is constant

$$\Rightarrow x\left( b+ak \right) +y\left( a+bk \right) -ab\left( k-1 \right) =0$$

Line meet $$x$$-axis at $$A$$, hence for $$A$$, $$y=0$$

$$\Rightarrow x\left( b+ak \right) -ab\left( k-1 \right) =0$$
$$\Rightarrow x=\dfrac { ab\left( k-1 \right)  }{ \left( b+ak \right)  } $$

so, coordinate of $$A$$ is $$\left( \dfrac { ab\left( k-1 \right)  }{ \left( b+ak \right)  } ,0 \right) $$

Line meets $$y$$-axis at $$B$$, hence for $$B$$, $$x=0$$

$$\Rightarrow y\left( a+bk \right) -ab\left( k-1 \right) =0$$

$$\Rightarrow y=\dfrac { ab\left( k-1 \right)  }{ \left( a+bk \right)  } $$

so, coordinate of $$B$$ is $$\left( 0,\dfrac { ab\left( k-1 \right)  }{ \left( a+bk \right)  }  \right) $$

Let $$(h,m)$$ be midpoint of $$AB$$, hence

$$h=\dfrac { \dfrac { ab\left( k-1 \right)  }{ \left( b+ak \right)  } +0 }{ 2 } $$
$$\Rightarrow k=\dfrac { b }{ a } \left( \dfrac { 2h+a }{ b-2h }  \right) \longrightarrow (3)\\ and,\quad m=\dfrac { 0+\dfrac { ab\left( k-1 \right)  }{ a+bk }  }{ 2 } $$
$$\Rightarrow k=\dfrac { a }{ b } \left( \dfrac { 2m+b }{ a-2m }  \right) \longrightarrow \left( 4 \right)  $$

From (3) and (4) we get

$$\Rightarrow\dfrac { b }{ a } \left( \dfrac { 2h+a }{ b-2h }  \right) =\dfrac { a }{ b } \left( \dfrac { 2m+b }{ a-2m }  \right) $$

To get equation of locus we take $$h\rightarrow x\quad and\quad m\rightarrow y$$

$$\Rightarrow \dfrac { b }{ a } \left( \dfrac { 2x+a }{ b-2x }  \right) =\dfrac { a }{ b } \left( \dfrac { 2y+b }{ a-2y }  \right) $$

$$\Rightarrow 2xy\left( a+b \right) =ab\left( x+y \right) $$

Maths

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