AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Consider the given equation.
√(x−2)2+y2+√(x+2)2+y2=4
Sum of the distance of point P(x,y) with respect to points
A(2,0) and B(−2,0) is equal to the distance between the
‌ points ‌A(2,0)‌ and ‌B(−2,0)
AP+BP=AB
Therefore, the locus of point P(x,y) is a line segment.