Area enclosed between the curve `y^(2)=16x` and the chord BC, where `B-=(1, 4)` and `C-=(9,12)` is

Options

A. `(8)/(3)`
B. `(32)/(3)`
C. `(16)/(3)`
D. None of these

Answer

To find the area enclosed between the curve \( y^2 = 16x \) and the chord BC, where \( B = (1, 4) \) and \( C = (9, 12) \), we will follow these steps:

### Step 1: Understand the Curve
The equation \( y^2 = 16x \) represents a rightward-facing parabola. We can rewrite it in the form \( y = \pm 4\sqrt{x} \).

### Step 2: Verify Points B and C
We need to check if points B and C lie on the parabola.

- For point B \( (1, 4) \):
\[ y^2 = 4^2 = 16 \quad \text{and} \quad 16x = 16 \cdot 1 = 16 \quad \Rightarrow \quad \text{Point B lies on the parabola.} \]

- For point C \( (9, 12) \):
\[ y^2 = 12^2 = 144 \quad \text{and} \quad 16x = 16 \cdot 9 = 144 \quad \Rightarrow \quad \text{Point C lies on the parabola.} \]

### Step 3: Find the Equation of Chord BC
The slope \( m \) of the chord BC can be calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{12 - 4}{9 - 1} = \frac{8}{8} = 1. \]
Using point-slope form, the equation of the line through point B \( (1, 4) \) is:
\[ y - 4 = 1(x - 1) \quad \Rightarrow \quad y = x + 3. \]

### Step 4: Set Up the Area Calculation
We need to find the area between the parabola and the line from \( x = 1 \) to \( x = 9 \).

The area \( A \) can be calculated as:
\[ A = \int_{1}^{9} (y_{\text{parabola}} - y_{\text{line}}) \, dx. \]
Where:
- \( y_{\text{parabola}} = 4\sqrt{x} \)
- \( y_{\text{line}} = x + 3 \)

Thus, we have:
\[ A = \int_{1}^{9} (4\sqrt{x} - (x + 3)) \, dx. \]

### Step 5: Simplify the Integral
We simplify the integral:
\[ A = \int_{1}^{9} (4\sqrt{x} - x - 3) \, dx. \]

### Step 6: Compute the Integral
Now we compute the integral:
\[ A = \int_{1}^{9} 4x^{1/2} \, dx - \int_{1}^{9} x \, dx - \int_{1}^{9} 3 \, dx. \]

Calculating each part:
1. \( \int 4x^{1/2} \, dx = \frac{4 \cdot x^{3/2}}{3/2} = \frac{8}{3} x^{3/2} \)
2. \( \int x \, dx = \frac{x^2}{2} \)
3. \( \int 3 \, dx = 3x \)

Now evaluate from 1 to 9:
\[ A = \left[ \frac{8}{3} x^{3/2} \right]_{1}^{9} - \left[ \frac{x^2}{2} \right]_{1}^{9} - \left[ 3x \right]_{1}^{9}. \]

Calculating each:
1. For \( \frac{8}{3} x^{3/2} \):
\[ \frac{8}{3} (9^{3/2}) - \frac{8}{3} (1^{3/2}) = \frac{8}{3} (27 - 1) = \frac{8}{3} \cdot 26 = \frac{208}{3}. \]

2. For \( \frac{x^2}{2} \):
\[ \frac{9^2}{2} - \frac{1^2}{2} = \frac{81}{2} - \frac{1}{2} = \frac{80}{2} = 40. \]

3. For \( 3x \):
\[ 3(9) - 3(1) = 27 - 3 = 24. \]

### Step 7: Combine Results
Now combine the results:
\[ A = \frac{208}{3} - 40 - 24 = \frac{208}{3} - \frac{120}{3} - \frac{72}{3} = \frac{208 - 192}{3} = \frac{16}{3}. \]

### Final Answer
The area enclosed between the curve \( y^2 = 16x \) and the chord BC is:
\[ \boxed{\frac{16}{3}} \text{ square units.} \]