The correct option is B (21,−30)
Parabola is y2=4x
let the points be P(at21,2at1) and Q(at22,2at2).
a=1 then points will be P(t21,2t1) and Q(t22,2t2)
Normals are drawn at P and Q
Given that ordinates of P and Q are 4 and 6 so,
2t1=4⇒t1=2 and 2t2=6⇒t2=3
Now, point of intersection of normals at P(at21,2at1) and Q(at22,2at2) is R≡(2a+a(t21+t22+t1t2),−at1t2(t1+t2))
substitute t1=2 and t2=3, a=1
R≡(21,−30)