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

As $M$ divides $AB$ externally in the ratio $b : a$
$x=\frac{b(a\cos \beta )-a(b\cos \alpha )}{b-a}$ and 
$\begin{array}{l} y=\frac{b(a\sin \beta )-a(b\sin \alpha )}{b-a}\\ \Rightarrow \frac{x}{y}=\frac{\cos \beta -\cos \alpha }{\sin \beta -\sin \alpha }=\frac{2\sin \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}}{2\cos \frac{\alpha +\beta }{2}\sin \frac{\beta -\alpha }{2}}\\ \Rightarrow x\cos \frac{\alpha +\beta }{2}+y\sin \frac{\alpha +\beta }{2}=0\end{array}$