Let the point P be (at2,2at)
Equation of normal at P is
y=−tx+2at+at3
at3+(2a−x)t−y=0.......(i)
It intersect the axis at G(2a+at2,0)
Now G divides PQ in 2:1
Let the point q be (h,k)
(2h+at23,2k+2at3)≡(2a+at2,0)⇒h=3a+at2,k=−at
So the coordinates of Q are (3a+at2,−at)
Now (i) is cubic in t
⇒t1t2t3=ya
It passes through Q so y=−at and t1=t
tt2t3=−atat2t3=−1
So the other two normals are perpendicular