If the point of intersection of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the circle $x^2+y^2=4b$ ,$b>4$ lies on the curve  $y^2=3x^2$, then b =

$\begin{array}{l}\frac{x^2}{16}+\frac{y^2}{b^2}=1\to \left(1\right)\\ x^2+y^2=4b\to \left(2\right)\\ y^2=3x^2\to \left(3\right)\\ \end{array}$

From equation (2)&(3) $x^2=band{y}^2=3b$

From equation (1) we get 

$$\begin{gathered} \frac{b}{16}+\frac{3b}{b^2}=1\Rightarrow b^2+48=16b\Rightarrow b=4 or 12 \\ \therefore b=12(\because b>4) \end{gathered}$$