Line joining A(bcosα,bsinα) and  B(acosβ,asinβ) is produced to the point M(x,y) such that AM:MB = b:a then xcosα+β2+ysinα+β2=

M divides AB in the ratio b : a externally∴(x,y)=b(acosβ)−a(bcosα)b−a,b(asinβ)−a(bsinα)b−axy=cosβ−cosαsinβ−sinα=−2sinβ+α2sinβ−α22cosβ+α2sinβ−α2⇒xcosα+β2+ysinα+β2=0