Given: Difference of distance from points (3, 0) and (-3, 0) is 4 Consider A = (3, 0) and B = (-3, 0) Let a point be p(x, y) As we know distance is always positive So, |PA - PB| = 4 ⇒√(x−3)2+(y−0)2−√(x+3)2+(y−0)2=4 ⇒√(x−3)2+(y)2=4+√(x+3)2+(y)2 Squaring both sides, we get ⇒(√(x−3)2+(y)2)2=(4+√(x+3)2+(y)2)2 ⇒(x−3)2+y2=16+(x+3)2+y2+8√(x+3)2+(y)2 ⇒x2−6x+9=16+x2+6x+9+8√(x+3)2+(y)2 ⇒−12x−16=8√(x+3)2+(y)2⇒−3x−4=2√(x+3)2+(y)2 Squaring both sides, we get ⇒(−3x−4)2=(2√(x+3)2+(y)2)2 ⇒(3x+4)2=4((x+3)2+(y)2) ⇒9x2+16+24x=4(x2+6x+9+y2) ⇒5x2−4y2=20∴