The ratio in which the area bounded by the curves \(\mathrm{y}^{2}=12 \mathrm{x}\) and \(\mathrm{x}^{2}=12 \mathrm{y}\) is divided by the line \(x=3\) is (1) \(15: 49\) (2) \(13: 480\) (3) \(13: 37\) (4) \(1: 1\)

The intersection of curves is a fundamental concept in coordinate geometry. To find where two curves intersect, we need to solve their equations simultaneously. For this problem, we have the curves \(\text{y}^{2}=12\text{x}\) and \(\text{x}^{2}=12\text{y}\). Solving these equations together implies finding common solutions for \(x\) and \(y\). By equating the two equations, we get \(\text{y}^{2}=12\text{x}=12\text{y}^{1/2}\) and simplifying this, we find the points of intersection to be \(0,0\) and \((6,6)\). These points are crucial because they mark the boundaries of the areas we need to calculate later.

Integrals help us calculate the area under a curve. To find the area bounded by \(\text{y}^{2}=12\text{x}\), we need to set up an appropriate integral from \(x=0\) to \(x=6\). This involves integrating the expression for \(\text{y}\) across the specified limit. The same process is applied to \(\text{x}^{2}=12\text{y}\), integrating with respect to \(\text{y}\). Let's examine the integrals for both curves:
1. For \(\text{y}^{2}=12\text{x}\), the integral is: \(\frac{1}{3} x^{3}\) evaluated from \(0\) to \(3\), yielding \(27\).
2. For \(\text{x}^{2}=12\text{y}\), we integrate \(3 \text{x} \text{y^{2}}\) with respect to \(y\), also resulting in an area of \(27\). Integrals convert complex area calculations into manageable computations, providing a clear and comprehensive approach to find bounded areas.

To calculate the area between curves and a line, follow these steps:

• First, solve for the points of intersection between the curves (here, at \(0,0\) and \((6,6)\)).
• Next, integrate to find the area under each curve between these points.
• Finally, use the line \(x=3\) to divide the total area between the specified ranges.

The given line splits the bounded area into two equal parts:
From \(x=0\) to \(x=3\) and from \(x=3 \)to \(x=6\). Integrating each part shows they are equal, resulting in the ratio \1:1\. This straightforward division demonstrates the effectiveness of integrating curves and the importance of understanding how lines can segment areas for comparison.
To sum up, grasping curve intersection, using integrals, and dividing areas allows for precise area calculations essential for understanding complex geometrical problems, as found in IIT JEE mathematics.