Let `T` and `C` respectively be the transverse and conjugate axes of the hyperbola `16x^(2)-y^(2)+64x+4y+44=0`.Then the area of the region above the parabola `x^(2)=y+4`,below the transverse axis `T` and on the right of the conjugate axis `C` is:

Options

A. `4sqrt(6)+(44)/(3)`
B. `4sqrt(6)-(28)/(3)`
C. `4sqrt(6)+(28)/(3)`
D. `4sqrt(6)-(44)/(3)`

Answer

To solve the problem, we need to follow these steps:

### Step 1: Rewrite the equation of the hyperbola
The given equation of the hyperbola is:
\[ 16x^2 - y^2 + 64x + 4y + 44 = 0 \]

First, we rearrange it:
\[ 16x^2 + 64x - y^2 + 4y + 44 = 0 \]

### Step 2: Complete the square for \(x\) and \(y\)
For \(x\):
\[ 16(x^2 + 4x) = 16((x + 2)^2 - 4) = 16(x + 2)^2 - 64 \]

For \(y\):
\[ - (y^2 - 4y) = -((y - 2)^2 - 4) = -(y - 2)^2 + 4 \]

Substituting these into the equation gives:
\[ 16(x + 2)^2 - 64 - (y - 2)^2 + 4 + 44 = 0 \]
\[ 16(x + 2)^2 - (y - 2)^2 - 16 = 0 \]
\[ 16(x + 2)^2 - (y - 2)^2 = 16 \]

Dividing through by 16:
\[ \frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1 \]

This is the standard form of the hyperbola:
\[ \frac{(x + 2)^2}{1} - \frac{(y - 2)^2}{16} = 1 \]

### Step 3: Identify the transverse and conjugate axes
From the standard form, we can identify:
- Transverse axis \(T\): \(y = 2\) (the line where \(y\) takes the value of the center)
- Conjugate axis \(C\): \(x = -2\) (the line where \(x\) takes the value of the center)

### Step 4: Determine the area of the region
We need to find the area above the parabola \(x^2 = y + 4\) (which can be rewritten as \(y = x^2 - 4\)), below the transverse axis \(y = 2\), and to the right of the conjugate axis \(x = -2\).

### Step 5: Find the points of intersection
Set \(y = 2\) equal to the parabola:
\[ 2 = x^2 - 4 \]
\[ x^2 = 6 \]
\[ x = \sqrt{6} \quad \text{and} \quad x = -\sqrt{6} \]

### Step 6: Set up the integral for the area
The area \(A\) can be calculated as:
\[ A = \int_{-2}^{\sqrt{6}} \left(2 - (x^2 - 4)\right) \, dx \]
This simplifies to:
\[ A = \int_{-2}^{\sqrt{6}} \left(6 - x^2\right) \, dx \]

### Step 7: Calculate the integral
Calculating the integral:
\[ A = \left[ 6x - \frac{x^3}{3} \right]_{-2}^{\sqrt{6}} \]

Calculating at the limits:
1. At \(x = \sqrt{6}\):
\[ 6\sqrt{6} - \frac{(\sqrt{6})^3}{3} = 6\sqrt{6} - \frac{6\sqrt{6}}{3} = 6\sqrt{6} - 2\sqrt{6} = 4\sqrt{6} \]

2. At \(x = -2\):
\[ 6(-2) - \frac{(-2)^3}{3} = -12 + \frac{8}{3} = -12 + \frac{8}{3} = -\frac{36}{3} + \frac{8}{3} = -\frac{28}{3} \]

### Step 8: Combine results
Thus,
\[ A = 4\sqrt{6} - \left(-\frac{28}{3}\right) = 4\sqrt{6} + \frac{28}{3} \]

### Final Answer
The area of the region is:
\[ 4\sqrt{6} + \frac{28}{3} \]