The correct option is A tan−1(35)
The slope of the tangent to y2=4x at (16,8) is given by
m1=(dydx)(16,8)=(42y)(16,8)=14
The slope of the tangent to x2=32y at (16,8) is given by
m2=(dydx)(16,8)=(2x32)(16,8)=1
∴tan θ=(1−(1/4)1+(1/4)∣∣
∣∣=35⇒θ=tan−1(35) (acute angle)⇒or θ=π − tan−1(35) (obtuse angle)