Given that $$x * y = \frac{x^2 + 4y^2}{xy}$$ and $$x \triangle y = \frac{x^2}{y}$$ for any real numbers x and y. If $$a * b = 4 \triangle 4$$ then b =

A

2a

B

$$\frac{a}{2}$$

C

a

D

$$\frac{a}{3}$$