The correct option is D 4a2
Here O is the vertex,
Equation of tangent at (at2,2at) is yt=x+at2 ⋯(1)
Distance
OP=|0−0−at2|√1+t2⇒OP=|at2|√1+t2
Equation of perpendicular to line (1)from O(0,0) is
y+tx=0 ⋯(2)
Now, intersection of y2=4ax and line (2) is
Q(4at2,−4at)
Distance
OQ=√(4at2)2+(−4at)2⇒OQ=|4a|√1t4+1t2⇒OQ=|4a|√1+t2t2
Now,
OP⋅OQ=|at2|√1+t2×|4a|√1+t2t2∴OP⋅OQ=4a2