TS EAMCET 5 May 2018 Shift 1 Solved Paper

The given function is expressed as,
|PA−PB|=3
√(h−1)2+(k−2)2 −√(h−2)2+(k−1)2=3
√(h−1)2+(k−2)2 =3+√(h−2)2+(k−1)2
Square both sides of the above expression and simplify it,
2h−2k−9 =6√(h−2)2+(k−1)2
Again square the above expression, then
(2h−2k−9)2=36[(h−2)2 +(k−1)2]
32h2+32k2+8hk−108h −108k+99=0
Thus, the locus of p(h,k) is,
32x2+32y2+8xy−108x −108y+99=0