If the points of intersection of two distinct conics x ^ 2 + y ^ 2...

Given conics are ${x}^{2} + {y}^{2} = 4b\text{and}\dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{{b}^{2}} = 1$.

The intersection point of these conics lie on ${y}^{2} = 3{x}^{2}$.

So, putting ${y}^{2} = 3{x}^{2}$ in both the conics.

$⇒ {x}^{2} + 3{x}^{2} = 4b, \dfrac{{x}^{2}}{16} + \dfrac{3{x}^{2}}{{b}^{2}} = 1$

$⇒ {x}^{2} = b, \dfrac{b}{16} + \dfrac{3b}{{b}^{2}} = 1$

$⇒ \dfrac{b}{16} + \dfrac{3}{b} = 1$

$⇒ {b}^{2} + 48 = 16b$

$⇒ {b}^{2} - 16b + 48 = 0$

$⇒ {b}^{2} - 12b - 4b + 48 = 0$

$⇒ \left(b - 4\right)\left(b - 12\right)$

$⇒ b = 4, 12$ ( $b = 4$is rejected because curves coincide)

$⇒ b = 12$

$⇒ x = \pm \sqrt{12}$

$⇒ 12 + {y}^{2} = 48$

$⇒ y = \pm 6$

Hence points of intersection are $( \pm \sqrt{12}, \pm 6 )$.

So, length and breadth of rectangle are $2\sqrt{12}\text{and}12$. 

So, the area of rectangle is given by, $A = 2\sqrt{12} \times 12 = 48\sqrt{3}$

$⇒ 3\sqrt{3}A = 48\sqrt{3} \times 3\sqrt{3}$

$⇒ 3\sqrt{3}A = 432$