The area of the ellipse
`(x^(2))/(9)+ (y^(2))/(4)=1` in first quadrant is `6pi` sq. units.
The ellipse is rotated about its centre in anti-clockwise direction till its major axis coincides with y-axis. Now the area of the ellipse in first quadrant is........ `pi` sq. units.

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to find the area of the ellipse in the first quadrant after it has been rotated. Let's go through the solution step by step.

### Step 1: Identify the Area of the Ellipse
The area \( A \) of an ellipse is given by the formula:
\[ A = \pi \times a \times b \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 2: Determine the Semi-Major and Semi-Minor Axes
From the given equation of the ellipse:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
we can identify:
- \( a^2 = 9 \) which gives \( a = 3 \) (semi-major axis)
- \( b^2 = 4 \) which gives \( b = 2 \) (semi-minor axis)

### Step 3: Calculate the Total Area of the Ellipse
Using the area formula:
\[ A = \pi \times 3 \times 2 = 6\pi \]
This is the total area of the ellipse.

### Step 4: Find the Area in the First Quadrant
Since the ellipse is symmetric about both axes, the area in the first quadrant is:
\[ \text{Area in 1st Quadrant} = \frac{1}{4} \times \text{Total Area} = \frac{1}{4} \times 6\pi = \frac{3\pi}{2} \]

### Step 5: Rotation of the Ellipse
The problem states that the ellipse is rotated about its center until its major axis coincides with the y-axis. However, the area of the ellipse does not change with rotation.

### Step 6: Area in the First Quadrant After Rotation
Since the area remains unchanged, the area of the ellipse in the first quadrant after rotation will still be:
\[ \text{Area in 1st Quadrant} = \frac{3\pi}{2} \]

### Final Answer
Thus, the area of the ellipse in the first quadrant after rotation remains:
\[ \text{Area in 1st Quadrant} = 6\pi \]