If the points of intersection of two distinct conics $x ^ { 2 } + y ^ { 2 } = 4 b$ and $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1$ lie on the curve $y ^ { 2 } = 3 x ^ { 2 }$, then $3 \sqrt { 3 }$ times the area of the rectangle formed by the intersection points is $\_\_\_\_$ .