To solve the problem, we need to determine the integral value of \( a \) such that the point \( (a^2 + 1, a) \) lies in the angle between the lines defined by the equations \( x - 2y - 4 = 0 \) and \( 4x + 8y - 9 = 0 \) that contains the origin.
### Step-by-Step Solution:
1. **Identify the Lines**:
- The first line is given by \( x - 2y - 4 = 0 \).
- The second line is given by \( 4x + 8y - 9 = 0 \).
2. **Check if the Origin Lies on the Same Side**:
- For the first line, substituting \( (0, 0) \):
\[
0 - 2(0) - 4 = -4 < 0
\]
This means the origin lies on one side of the line.
- For the second line, substituting \( (0, 0) \):
\[
4(0) + 8(0) - 9 = -9 < 0
\]
This also indicates that the origin lies on the same side of the second line.
3. **Check the Point \( (a^2 + 1, a) \)**:
- For the first line:
\[
(a^2 + 1) - 2a - 4 < 0 \implies a^2 - 2a - 3 < 0
\]
Factoring gives:
\[
(a - 3)(a + 1) < 0
\]
The critical points are \( a = -1 \) and \( a = 3 \). The intervals to check are \( (-\infty, -1) \), \( (-1, 3) \), and \( (3, \infty) \). Testing these intervals, we find:
- For \( a \in (-1, 3) \), the inequality holds.
4. **Check the Second Line**:
- For the second line:
\[
4(a^2 + 1) + 8a - 9 < 0 \implies 4a^2 + 8a - 5 < 0
\]
Factoring gives:
\[
(2a - 1)(2a + 5) < 0
\]
The critical points are \( a = -\frac{5}{2} \) and \( a = \frac{1}{2} \). The intervals to check are \( (-\infty, -\frac{5}{2}) \), \( (-\frac{5}{2}, \frac{1}{2}) \), and \( (\frac{1}{2}, \infty) \). Testing these intervals, we find:
- For \( a \in (-\frac{5}{2}, \frac{1}{2}) \), the inequality holds.
5. **Find the Common Interval**:
- The first line gives \( a \in (-1, 3) \).
- The second line gives \( a \in (-\frac{5}{2}, \frac{1}{2}) \).
- The common interval is \( (-1, \frac{1}{2}) \).
6. **Determine Integral Values**:
- The only integer in the interval \( (-1, \frac{1}{2}) \) is \( 0 \).
### Final Answer:
The integral value of \( a \) is \( 0 \).
