The correct option is A θ=tan−138
Given parabolas,
y2=4ax and x2=4by
⇒y416a2=4by⇒y=4a2/3b1/3
Intersection point of the given parabolas is (0,0) and (4a1/3b2/3,4a2/3b1/3)
Slope of the tangent on the parabola y2=4ax at (4a1/3b2/3,4a2/3b1/3) is
m1=2ay=2a4a2/3b1/3=a1/32b1/3
Slope of the tangent on the parabola x2=4by at (4a1/3b2/3,4a2/3b1/3) is
m2=x2b=4a1/3b2/32b=2a1/3b1/3
So, the angle between the parabola is
tanθ=∣∣∣m1−m21+m1m2∣∣∣=∣∣
∣
∣
∣
∣
∣∣a132b13−2a13b131+a23b23∣∣
∣
∣
∣
∣
∣∣=32×1(ab)13+(ba)13⇒tanθ=32×14⇒θ=tan−138