Let Q(h,k) be the point of intersection of the axis AS with the directrix. The point A(1,1) is the midpoint of QS . ∴
h+1
2
=1 and
k−1
2
=1 h=1 and k=3 Thus, the coordinates of the point Q(1,3). So, the directrix passes through point (1,3) and has the gradient of 0. The equation of directrix is. y−3=0 It is assumed the point p(x,y) a point on the parabola and M be the foot of the perpendicular drawn from P on the directrix. PS=PM PS2=PM2 Substitute the value we get. (x−1)2+(y+1)2=(
y−√3
√1
)2 (x−1)2+(y+1)2=(y−3)2 (x−1)2=8(1−y) By going through option a it satisfy the above equation. Thus, the point (3,