The smaller area between the ellipse `(x^(2))/(9)+(y^(2))/(16)=1` at the line `(x)/(3)+(y)/(4)=1` is

Options

A. `pi-2`
B. `3(pi-2)`
C. `3(pi+2)`
D. `3(3pi+2)`

Answer

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

### Step 1: Identify the equations
We have the equations of the ellipse and the line:
- Ellipse: \(\frac{x^2}{9} + \frac{y^2}{16} = 1\)
- Line: \(\frac{x}{3} + \frac{y}{4} = 1\)

### Step 2: Find the intersection points
To find the intersection points of the ellipse and the line, we can express \(x\) in terms of \(y\) from the line equation:
\[ x = 3(1 - \frac{y}{4}) = 3 - \frac{3y}{4} \]
Now substitute this expression for \(x\) into the ellipse equation:
\[ \frac{(3 - \frac{3y}{4})^2}{9} + \frac{y^2}{16} = 1 \]
Expanding this:
\[ \frac{(3 - \frac{3y}{4})^2}{9} = \frac{9 - 2 \cdot 3 \cdot \frac{3y}{4} + \frac{9y^2}{16}}{9} = 1 - \frac{3y}{4} + \frac{y^2}{16} \]
Now, simplifying:
\[ 1 - \frac{3y}{4} + \frac{y^2}{16} + \frac{y^2}{16} = 1 \]
This simplifies to:
\[ \frac{9}{16}y^2 - \frac{3}{4}y = 0 \]
Factoring out \(y\):
\[ y(\frac{9}{16}y - \frac{3}{4}) = 0 \]
Thus, \(y = 0\) or \(\frac{9}{16}y - \frac{3}{4} = 0\) gives \(y = 2\).

Now substituting back to find \(x\):
- For \(y = 0\): \(x = 3\)
- For \(y = 2\): \(x = 3(1 - \frac{2}{4}) = 3(1 - 0.5) = 1.5\)

So the intersection points are \((3, 0)\) and \((1.5, 2)\).

### Step 3: Calculate the area of the ellipse
The area \(A\) of the ellipse is given by the formula:
\[ A = \pi \cdot a \cdot b \]
where \(a = 3\) and \(b = 4\):
\[ A = \pi \cdot 3 \cdot 4 = 12\pi \]
The area of the quarter ellipse (since we are only considering one quadrant) is:
\[ \text{Area of quarter ellipse} = \frac{12\pi}{4} = 3\pi \]

### Step 4: Calculate the area of the triangle
The triangle formed by the points \((0, 0)\), \((3, 0)\), and \((1.5, 2)\) has a base of 3 and a height of 2. The area \(A_t\) of the triangle is given by:
\[ A_t = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 3 \cdot 2 = 3 \]

### Step 5: Calculate the smaller area
The smaller area between the ellipse and the line is:
\[ \text{Smaller Area} = \text{Area of quarter ellipse} - \text{Area of triangle} = 3\pi - 3 \]

### Final Answer
The smaller area between the ellipse and the line is:
\[ \boxed{3\pi - 3} \]