The correct option is A {(x,1);x≥3}
y2−2y−4x+5=0⇒(y−1)2=4(x−1)
Axis of parabola is
y−1=0⇒y=1
Equation of normal is
(y−1)=m(x−1)−2m−m3
Let (h,1) be any point on its axis, then
0=m(h−1)−2m−m3⇒m=0, m2=h−3
For three normals to exist,
⇒h≥3
So, x≥3 and y=1