To solve the problem, we need to find the intersection points of the given conics and verify that they lie on the curve \(y^2 = 3x^2\). Analyze each step as follows:
The final product, \(3\sqrt{3}\) times the area, is 432.
From \(y^2 = 3x^2\), substitute into \(x^2 + y^2 = 4b\) and \(\frac{x^2}{16} + \frac{y^2}{b^2} = 1\) to find values of \(b\).
By substituting, we get:
\[ x^2 = b \quad \text{and} \quad \frac{b}{16} + \frac{3}{b} = 1 \]
Solving this equation gives \(b = 4\) or \(b = 12\). Since \(b = 4\) makes the curves coincide, we reject it, so \(b = 12\).
With \(b = 12\), the points of intersection are \(\left(\pm \sqrt{12}, \pm 6\right)\).
The area of the rectangle formed by these points is:
\[ \text{Area} = 2 \cdot \sqrt{12} \times 2 \cdot 6 = 4 \cdot \sqrt{12} \cdot 6 = 432 \]
So, the correct answer is: 432
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
