The integral value of a which the point `(a^2+1,a)` lies in the angle between the lines x - 2y - 4 = 0 , 4x + 8y - 9 = 0 that contains origin is ……

Answer

To find the integral value of \( a \) such that the point \( (a^2 + 1, a) \) lies in the angle between the lines \( x - 2y - 4 = 0 \) and \( 4x + 8y - 9 = 0 \) that contains the origin, we will follow these steps:

### Step 1: Identify the lines and their equations
The equations of the lines are:
1. Line 1: \( L_1: x - 2y - 4 = 0 \)
2. Line 2: \( L_2: 4x + 8y - 9 = 0 \)

### Step 2: Rewrite the equations in slope-intercept form
To find the slopes of the lines, we can rewrite them in the form \( y = mx + b \).
1. For \( L_1 \):
\[ x - 2y - 4 = 0 \implies 2y = x - 4 \implies y = \frac{1}{2}x + 2 \]
The slope \( m_1 = \frac{1}{2} \).

2. For \( L_2 \):
\[ 4x + 8y - 9 = 0 \implies 8y = -4x + 9 \implies y = -\frac{1}{2}x + \frac{9}{8} \]
The slope \( m_2 = -\frac{1}{2} \).

### Step 3: Determine the angles between the lines
The lines intersect at an angle, and we need to check which angle contains the origin. The angle between the two lines can be determined by checking the signs of the slopes. Since \( m_1 > 0 \) and \( m_2 < 0 \), the angle between them is acute and contains the origin.

### Step 4: Substitute the point into the inequalities
The point \( (a^2 + 1, a) \) must satisfy the conditions of lying between the two lines. We will check this by substituting the point into the inequalities derived from the line equations.

1. For Line 1:
\[ a^2 + 1 - 2a - 4 < 0 \implies a^2 - 2a - 3 < 0 \]
Factoring gives:
\[ (a - 3)(a + 1) < 0 \]
The critical points are \( a = -1 \) and \( a = 3 \). The solution to this inequality is:
\[ -1 < a < 3 \]

2. For Line 2:
\[ 4(a^2 + 1) + 8a - 9 < 0 \implies 4a^2 + 8a + 4 - 9 < 0 \implies 4a^2 + 8a - 5 < 0 \]
Factoring gives:
\[ (2a - 1)(2a + 5) < 0 \]
The critical points are \( a = \frac{1}{2} \) and \( a = -\frac{5}{2} \). The solution to this inequality is:
\[ -\frac{5}{2} < a < \frac{1}{2} \]

### Step 5: Combine the inequalities
Now we have two inequalities:
1. From Line 1: \( -1 < a < 3 \)
2. From Line 2: \( -\frac{5}{2} < a < \frac{1}{2} \)

The intersection of these two intervals is:
\[ -1 < a < \frac{1}{2} \]

### Step 6: Find integral values of \( a \)
The integral values of \( a \) within the interval \( -1 < a < \frac{1}{2} \) are:
- \( a = 0 \)
- \( a = 1 \) (not included since \( 1 > \frac{1}{2} \))

Thus, the only integral value is:
\[ \text{Integral value of } a = 0 \]

### Final Answer
The integral value of \( a \) such that the point \( (a^2 + 1, a) \) lies in the angle between the lines containing the origin is \( 0 \).

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