Line joining $A(b\cos \alpha ,b\sin \alpha )$ and  $B(a\cos \beta ,a\sin \beta )$ is produced to the point $M(x,y)$ such that AM:MB = b:a then $x\cos \frac{\alpha +\beta }{2}+y\sin \frac{\alpha +\beta }{2}=$

M divides AB in the ratio b : a externally

$\therefore (x,y)=\left(\frac{{\displaystyle b\left(a\cos \beta \right)-a\left(b\cos \alpha \right)}}{{\displaystyle b-a}},\frac{{\displaystyle b\left(a\sin \beta \right)-a\left(b\sin \alpha \right)}}{{\displaystyle b-a}}\right)$
$\frac{x}{y}=\frac{\cos \beta -\cos \alpha }{\sin \beta -\sin \alpha }$$=\frac{-2\sin \left(\frac{\beta +\alpha }{2}\right)\sin \left(\frac{\beta -\alpha }{2}\right)}{2\cos \left(\frac{\beta +\alpha }{2}\right)\sin \left(\frac{\beta -\alpha }{2}\right)}$

$\Rightarrow x\cos \left(\frac{{\displaystyle \alpha +\beta }}{{\displaystyle 2}}\right)+y\sin \left(\frac{{\displaystyle \alpha +\beta }}{{\displaystyle 2}}\right)=0$