If the roots of  $\frac{a}{x-a}+\frac{b}{x-b}=k$ are equal in magnitude and opposite in sign, then

$\frac{a\left(x-b\right)+b\left(x-a\right)}{\left(x-a\right)\left(x-b\right)}=k$

$\frac{ax+bx-ab-ab}{x^2-bx-ax+ab}=k$

$ax+bx-2ab=k{x}^2-x\left(a+b\right)k+kab$

$k{x}^2-x\left(a+b\right)\left[\left(k+1\right)\right]+ab\left(k+2\right)=0$

$\left(\because \text{roots are equal in magnitude and opposite sign}\right)$

$\left(a+b\right)\left(k+1\right)=0$

$\left(a+b\right)=0\left(or\right)\left(k+1\right)=0$

$\left(a+b\right)=0\left(or\right)k=-1$