VITEEE 2010 Paper

Let coordinates of P be (h,k) then
h=

2(10‌cos‌θ)+3(5)

2+3

=4‌cos‌θ+3
and k=

2(10‌sin‌θ)+3(0)

2+3

=4‌sin‌θ
[Using the internal section formula]
⇒

h−3

4

=cos‌θ‌and‌

k

4

=sin‌θ
Squaring and adding both of these equations,

(h−3)2

16

+

k2

16

=cos2θ+sin2θ
⇒ (h−3)2+k2=16
Therefore, locus of point P is
(x−3)2+y2=16 which is a circle.