The correct option is A 6√3 units
y2=4x
a=1
Let normal be drawn at point P(t2,2t)and other end of normal be Q(t0)
Vertex O=(0,0)
Slope of OP=2t−0t2−0=2t
Slope of OQ=2t0−0t20−0=2t0
∠POQ=90∘⇒2t⋅2t0=−1⇒tt0=−4 ⋯(1)
Also,
t0=−t−2t⇒−4t=−t−2t⇒2t=t⇒t2=2⇒t=±√2⇒t0=∓2√2
The coordinates of P are (2,2√2) or (2,−2√2)
And coordinates of Q are (8,−4√2) or (8,4√2)
∴ Length of PQ
=√62+(6√2)2=6√3 units