Question

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

• A. $xy(a+b)=ab(x+y)$
• B. $2xy(a-b)=ab(x+y)$
• C. $2xy(a-b)=ab(x-y)$
• D. $2xy(a+b)=ab(x+y)$

Answer 1

The correct option is D $2xy(a+b)=ab(x+y)$

we have,

$\frac{x}{a}+\frac{y}{b}=1and\frac{x}{b}+\frac{y}{a}=1$
$\Rightarrow bx+ay-ab=0⟶\left(1\right)$
$and,ax+by-ab=0⟶\left(2\right)$

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

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

where $k$ is constant

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

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

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

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

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

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

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

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

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

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

From (3) and (4) we get

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

To get equation of locus we take $h\to xandm\to y$

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

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