Find the angle of intersection of the curves $y^{2}=4ax$ and $x^{2}=4by$.

(A) Given the curves $y^{2}=4ax \dots (i)$ and $x^{2}=4by \dots (ii)$.
Solving $(i)$ and $(ii)$,we substitute $y = \frac{x^{2}}{4b}$ into $(i)$:
$\left(\frac{x^{2}}{4b}\right)^{2} = 4ax \Rightarrow \frac{x^{4}}{16b^{2}} = 4ax \Rightarrow x^{4} = 64ab^{2}x$.
This gives $x(x^{3} - 64ab^{2}) = 0$,so $x = 0$ or $x = 4a^{1/3}b^{2/3}$.
The points of intersection are $(0,0)$ and $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$.
For $(i)$,$\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$. For $(ii)$,$\frac{dy}{dx} = \frac{2x}{4b} = \frac{x}{2b}$.
At $(0,0)$,the tangent to $(i)$ is vertical $(x=0)$ and to $(ii)$ is horizontal $(y=0)$,so the angle is $\frac{\pi}{2}$.
At $(4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})$,slopes are $m_{1} = \frac{2a}{4a^{2/3}b^{1/3}} = \frac{1}{2}(\frac{a}{b})^{1/3}$ and $m_{2} = \frac{4a^{1/3}b^{2/3}}{2b} = 2(\frac{a}{b})^{1/3}$.
The angle $\theta$ is given by $\tan \theta = |\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}| = |\frac{2(a/b)^{1/3} - 0.5(a/b)^{1/3}}{1 + 2(a/b)^{1/3} \cdot 0.5(a/b)^{1/3}}| = \frac{1.5(a/b)^{1/3}}{1 + (a/b)^{2/3}} = \frac{3a^{1/3}b^{1/3}}{2(a^{2/3}+b^{2/3})}$.
Thus,$\theta = \tan^{-1}\left(\frac{3a^{1/3}b^{1/3}}{2(a^{2/3}+b^{2/3})}\right)$.