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

The points common to the circle $$x^{2}+y^{2}=4b\quad -(1)$$ and the ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\quad -(2)$$ are stated to lie on the pair of straight lines $$y^{2}=3x^{2}\; \Longrightarrow\; y=\pm\sqrt{3}\,x\quad -(3)$$.

Substitute $$y^{2}=3x^{2}$$ from $$(3)$$ in the circle $$(1)$$:

$$x^{2}+3x^{2}=4b\;\Longrightarrow\;4x^{2}=4b\;\Longrightarrow\;x^{2}=b\;\Longrightarrow\;|x|=\sqrt{b}$$.

Using $$(3)$$ again, $$y^{2}=3x^{2}=3b\;\Longrightarrow\;|y|=\sqrt{3b}$$.

Hence the four intersection points are $$\bigl(\pm\sqrt{b},\,\pm\sqrt{3b}\bigr)$$, one in each quadrant.

These points must also satisfy the ellipse $$(2)$$. Substitute $$x^{2}=b,\;y^{2}=3b$$ in $$(2)$$:

$$\frac{b}{16}+\frac{3b}{b^{2}}=1 \;\Longrightarrow\;\frac{b}{16}+\frac{3}{b}=1 \;\Longrightarrow\;b^{2}-16b+48=0 \;\Longrightarrow\;(b-12)(b-4)=0$$.

This gives $$b=12$$ or $$b=4$$. When $$b=4$$ the ellipse becomes $$\dfrac{x^{2}}{16}+\dfrac{y^{2}}{16}=1 \;\Longrightarrow\;x^{2}+y^{2}=16$$, identical to the circle. The problem states the conics are distinct, so we reject $$b=4$$ and take $$b=12$$.

Width of the rectangle formed by the four points: $$2|x|=2\sqrt{b}=2\sqrt{12}=4\sqrt{3}$$.
Height of the rectangle: $$2|y|=2\sqrt{3b}=2\sqrt{36}=12$$.

Area of the rectangle $$=4\sqrt{3}\times12=48\sqrt{3}$$.

Required expression: $$3\sqrt{3}\times(\text{area})=3\sqrt{3}\times48\sqrt{3}=3\sqrt{3}\times48\sqrt{3}=3\times48\times3=432$$.

Hence, $$3\sqrt{3}$$ times the required area equals $$432$$.