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
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of \( P_1 \) and \( P_2 \) are orthogonal to each other. The polarizer \( P_3 \) covers both the slits with its transmission axis at \( 45^\circ \) to those of \( P_1 \) and \( P_2 \). An unpolarized light of wavelength \( \lambda \) and intensity \( I_0 \) is incident on \( P_1 \) and \( P_2 \). The intensity at a point after \( P_3 \), where the path difference between the light waves from \( S_1 \) and \( S_2 \) is \( \frac{\lambda}{3} \), is:
