The correct option is C π2
We are given y2=4ax
Let any point on parabola beP(at2,2at)
Tangent atP:y=xt+at
⇒ ty=x+at2⋯(1)
And tangent at vertex is x=0⋯(2)
Solving (1) and (2), we get
ty=at2
⇒y=at
So, coordinates of Q≡(0,at) and S=(a,0)
Now, slope of QP=2at−atat2−0=1t=m (say)
Slope of QS=0−ata−0=−t=m′ say
Now, mm′=1t×−t=−1
Hence ∠SQP=π2