A variable straight line drawn through point of intersection of the straight lines  $\frac{x}{a}+\frac{y}{b}=1and\frac{x}{b}+\frac{y}{a}=1$ meets the axes in A and B. The locus of midpoint of AB is

Given lines

$\begin{array}{l}bx+ay-ab=0--\left(1\right)\\ ax+by-ab=0--\left(2\right)\\ Solving\left(1\right)\&\left(2\right)\\ x y 1\\ \begin{array}{cccc}a& -ab& b& a\\ b& -ab& a& b\end{array}\end{array}$

$\begin{array}{l}\Rightarrow \frac{x}{-a^{2}b+a{b}^2}=\frac{y}{-a^{2}b+a{b}^2}=\frac{1}{b^2-a^2}\\ \Rightarrow \frac{x}{ab\left(b-a\right)}=\frac{y}{ab\left(b-a\right)}=\frac{1}{\left(b+a\right)\left(b-a\right)}\\ \Rightarrow x=\frac{ab}{a+b},y=\frac{ab}{a+b}\\ \therefore P.I=\left(\frac{ab}{a+b},\frac{ab}{a+b}\right)\end{array}$

Let the Equation of line $\frac{x}{\alpha }+\frac{y}{\beta }=1$

 $A\left(\alpha ,0\right)B\left(0,\beta \right)$

$Let\left(x_1,y_1\right)$ be the point on lotus

$\begin{array}{l}\therefore \left(x_1,y_1\right)=\left(\frac{\alpha +0}{2},\frac{0+\beta }{2}\right)\\ \Rightarrow \alpha =2x_1,\beta =2y_1\\ \therefore Theline\frac{x}{2x_1}+\frac{y}{2y_1}=1\\ Ifpassesthrough\left(\frac{ab}{a+b},\frac{ab}{a+b}\right)\\ \Rightarrow \frac{ab}{2x_1\left(a+b\right)}+\frac{ab}{2y_1\left(a+b\right)}=1\\ \Rightarrow ab\left(x_1+y_1\right)=2x_1{y}_1\left(a+b\right)\\ \therefore Thelocusisab\left(x+y\right)=2\left(a+b\right)xy\end{array}$