The smaller area bounded by `x^2/16+y^2/9=1` and the line `3x+4y=12` is

Options

A. (a) `3pi` sq. units
B. (b) `(3pi - 6)` sq. units
C. (c) `(3pi - 2)` sq. units
D. (d) `(3pi - 4)` sq. units

Answer

To find the smaller area bounded by the ellipse \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \) and the line \( 3x + 4y = 12 \), we can follow these steps:

### Step 1: Identify the ellipse and the line equations
The equation of the ellipse is given by:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
This can be rewritten as:
\[ x^2 = 16(1 - \frac{y^2}{9}) \quad \text{or} \quad y^2 = 9(1 - \frac{x^2}{16}) \]

The equation of the line is:
\[ 3x + 4y = 12 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ y = 3 - \frac{3}{4}x \]

### Step 2: Find the points of intersection
To find the points of intersection between the ellipse and the line, we substitute \( y = 3 - \frac{3}{4}x \) into the ellipse equation:
\[ \frac{x^2}{16} + \frac{(3 - \frac{3}{4}x)^2}{9} = 1 \]
Expanding the second term:
\[ (3 - \frac{3}{4}x)^2 = 9 - 2 \cdot 3 \cdot \frac{3}{4}x + \left(\frac{3}{4}x\right)^2 = 9 - \frac{9}{2}x + \frac{9}{16}x^2 \]
Substituting this back into the ellipse equation:
\[ \frac{x^2}{16} + \frac{9 - \frac{9}{2}x + \frac{9}{16}x^2}{9} = 1 \]
Multiply through by 144 (the least common multiple of the denominators):
\[ 9x^2 + 16(9 - \frac{9}{2}x + \frac{9}{16}x^2) = 144 \]
This simplifies to:
\[ 9x^2 + 144 - 72x + 9x^2 = 144 \]
Combining like terms:
\[ 18x^2 - 72x = 0 \]
Factoring out \( 18x \):
\[ 18x(x - 4) = 0 \]
Thus, \( x = 0 \) or \( x = 4 \).

### Step 3: Find corresponding \( y \) values
For \( x = 0 \):
\[ y = 3 - \frac{3}{4}(0) = 3 \quad \Rightarrow \quad (0, 3) \]
For \( x = 4 \):
\[ y = 3 - \frac{3}{4}(4) = 0 \quad \Rightarrow \quad (4, 0) \]

### Step 4: Set up the integral for the area
The area \( A \) between the curves from \( x = 0 \) to \( x = 4 \) can be calculated using:
\[ A = \int_{0}^{4} \left( \text{Upper curve} - \text{Lower curve} \right) \, dx \]
Here, the upper curve is the ellipse and the lower curve is the line:
\[ A = \int_{0}^{4} \left( \frac{3}{4}\sqrt{16 - x^2} - \left(3 - \frac{3}{4}x\right) \right) \, dx \]

### Step 5: Simplify the integral
This simplifies to:
\[ A = \int_{0}^{4} \left( \frac{3}{4}\sqrt{16 - x^2} - 3 + \frac{3}{4}x \right) \, dx \]
We can split this into two integrals:
\[ A = \frac{3}{4} \int_{0}^{4} \sqrt{16 - x^2} \, dx - \int_{0}^{4} 3 \, dx + \frac{3}{4} \int_{0}^{4} x \, dx \]

### Step 6: Evaluate the integrals
1. The integral \( \int \sqrt{16 - x^2} \, dx \) can be evaluated using the formula:
\[ \int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2} \sin^{-1}\left(\frac{x}{a}\right) + C \]
Here, \( a = 4 \):
\[ \int_{0}^{4} \sqrt{16 - x^2} \, dx = \left[ \frac{x}{2}\sqrt{16 - x^2} + 8 \sin^{-1}\left(\frac{x}{4}\right) \right]_{0}^{4} \]
Evaluating this gives:
\[ = 0 + 8 \cdot \frac{\pi}{2} - 0 = 4\pi \]

2. The integral \( \int_{0}^{4} 3 \, dx = 3 \cdot 4 = 12 \).

3. The integral \( \int_{0}^{4} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{4} = 8 \).

### Step 7: Combine the results
Putting it all together:
\[ A = \frac{3}{4}(4\pi) - 12 + \frac{3}{4}(8) \]
\[ A = 3\pi - 12 + 6 = 3\pi - 6 \]

### Final Answer
Thus, the smaller area bounded by the ellipse and the line is:
\[ \boxed{3\pi - 6} \text{ square units} \]