The line joining A (bcosa, bsina) and B (a cosb, a sinb ) is produced to the point M (x, y) so that AM:MB=b:a,, then xcosα+β2+ ysinα+β2=
-1
0
1
a2+b2
As M divides AB externally in the ratio b : ax=b(acosβ)−a(bcosα)b−a and y=b(asinβ)−a(bsinα)b−a⇒ xy=cosβ−cosαsinβ−sinα=2sinα+β2sinα−β22cosα+β2sinβ−α2⇒ xcosα+β2+ysinα+β2=0