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 .
∴
=1 and
=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=()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,) lies on the parabola
(x−1)2=8(1−y)