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Questions  

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=

a
-1
b
0
c
1
d
a2+b2

detailed solution

Correct option is B

As M divides AB externally in the ratio b : ax=b(acos⁡β)−a(bcos⁡α)b−a and           y=b(asin⁡β)−a(bsin⁡α)b−a⇒ xy=cos⁡β−cos⁡αsin⁡β−sin⁡α=2sin⁡α+β2sin⁡α−β22cos⁡α+β2sin⁡β−α2⇒ xcos⁡α+β2+ysin⁡α+β2=0

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