Let $f\left ( x,y \right )=\dfrac{ax^{2}+by^{2}}{xy},$ where $a$ and $b$ are constants. If $\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial y}$ at $x = 1$ and $y = 2$, then

the relation between $a$ and $b$ is

- $a=\dfrac{b}{4}$
- $a=\dfrac{b}{2}$
- $a=2b$
- $a=4b$