Correct Answer - C
The two parabolas intersect at (0, 0) and `(4a^("1/3")b^("2/3"),4a^("2/3")b^("1/3"))`. The tangents at (0, 0) to `y^(2)=4ax` and `x^(2)=4ab` are y-axis respectively. So, the angle of intersection of the two parabolas at (0, 0) is `pi/1". So, pi/3` is the angle of intersection at `P(4a^("1/3")b^("2/3"),4a^("2/3")b^("1/3"))`.
Now, `:." "((dy)/(dx))_(P)=(2a)/(4a^("2/3")b^("1/3"))=a^("1/3")/(2b^("1/3"))`
`:." "m_(1)=("Slope of tngent to "y^(2)=4ax" at "P)=a^("1/3")/(2b^("1/3"))" and, "x^(2)=4by`
`rArr" "(dy)/(dx)=x/(2b)rArr((dy)/(dx))_(P)=(4a^("1/3")b^("1/3"))/(2b)=(2a^("1/3"))/(b^("1/3"))`
`:." "m_(2)=("Slope of tangent to "x^(2)=4by" at point P")=(2a^"1/3")/(b^("1/3"))`
The angle of intersection at P is `pi/3`.
`:." tan"pi/3=|(m_(1)-m_(2))/(1+m_(1)m_(2))|`
`rArr" "sqrt3=|(a^("1/3")/(2b^("1/3"))-(2a^("1/3"))/b^("1/3"))/(1+(2a^("1/3"))/(b^("1/3"))xxa^("1/3")/(2b^("1/3")))|`
`rArr" "sqrt3=3/2{(a^("1/3")b^("1/3"))/(a^("2/3")+b^("2/3"))}rArr(a/b)^("1/3")+(b/a)^("1/3")=sqrt3/2`